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OCA Oracle Database 11g:
SQL Fundamentals I
Exam Guide (Exam 1Z0-051)

ABOUT THE AUTHORS

John Watson (Oxford, UK) works for BPLC Management Consultants, teaching
and consulting throughout Europe and Africa. He was with Oracle University for
several years in South Africa, and before that worked for a number of companies,
government departments, and NGOs in England and Europe. He is OCP qualified in
both database and Application Server administration. John is the author of several
books and numerous articles on technology and has 25 years of experience in IT.
Roopesh Ramklass (South Africa), OCP, is an independent Oracle specialist with
over 10 years of experience in a wide variety of IT environments. These include
software design and development, systems analysis, courseware development, and
lecturing. He has worked for Oracle Support and taught at Oracle University in
South Africa for several years. Roopesh is experienced in managing and executing
IT development projects, including infrastructure systems provisioning, software
development, and systems integration.

About the Technical Editor
Bruce Swart (South Africa) works for 2Cana Solutions and has over 14 years of
experience in IT. Whilst maintaining a keen interest for teaching others, he has
performed several roles including developer, analyst, team leader, administrator,
project manager, consultant, and lecturer. He is OCP qualified in both database and
developer roles. He has taught at Oracle University in South Africa for several years
and has also spoken at numerous local Oracle User Group conferences. His passion
is helping others achieve greatness.

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

®

OCA Oracle Database 11g:
SQL Fundamentals I
Exam Guide (Exam 1Z0-051)
John Watson
Roopesh Ramklass

This publication and CD may be used in assisting students to prepare for the OCP
Oracle Database 11g: SQL Fundamentals I exam. Neither Oracle Corporation nor
The McGraw-Hill Companies warrant that use of this publication and CD will ensure
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DOI: 10.1036/0071597867

With thanks to Silvia for making life worth living.
—John
Ameetha, you have been wonderfully supportive and I want to
thank you for every moment that you share with me.
—Roopesh

This page intentionally left blank

CONTENTS AT A GLANCE

1

Oracle Server Technologies and the Relational Paradigm

..........

1

2

Data Retrieval Using the SQL SELECT Statement

...............

51

3

Restricting and Sorting Data

.................................

103

4

Single-Row Functions

......................................

169

5

Using Conversion Functions and Conditional Expressions

6

Reporting Aggregated Data Using the Group Functions

7

.........

227

...........

273

Displaying Data from Multiple Tables

..........................

309

8

Using Subqueries to Solve Problems

...........................

357

9

Using the Set Operators

.....................................

381

10

Manipulating Data

.........................................

403

11

Using DDL Statements to Create and Manage Tables

12

Creating Other Schema Objects

.............

449

..............................

487

................................................

533

.................................................

537

....................................................

555

Appendix
Glossary
Index

vii

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For more information about this title, click here

CONTENTS

Introduction

1

........................................

Oracle Server Technologies and
the Relational Paradigm . . . . . . . . . . . . . . . . . . . . . . . .
Position the Server Technologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Oracle Server Architecture . . . . . . . . . . . . . . . . . . . . . . .
The Oracle Application Server . . . . . . . . . . . . . . . . . . . . . . . .
Oracle Enterprise Manager . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Grid Computing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 1-1: Investigate Your Database and
Application Environment . . . . . . . . . . . . . . . . . . . . . . . . . .
Development Tools and Languages . . . . . . . . . . . . . . . . . . . . .
Understand Relational Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Rows and Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Data Normalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 1-2: Perform an Extended
Relational Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Summarize the SQL Language . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
SQL Standards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
SQL Commands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A Set-oriented Language . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Use the Client Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
SQL*Plus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
SQL Developer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Create the Demonstration Schemas . . . . . . . . . . . . . . . . . . . . . . . . . . .
Users and Schemas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The HR and OE Schemas . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Demonstration Schema Creation . . . . . . . . . . . . . . . . . . . . . .
✓ Two-Minute Drill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q&A Self Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xix

1
3
3
5
7
8
9
9
10
11
14
20
23
23
23
25
26
26
33
37
38
38
42
45
46

ix

x

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

Lab Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self Test Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

Data Retrieval Using the SQL SELECT Statement

...

List the Capabilities of SQL SELECT Statements . . . . . . . . . . . . . . . .
Introducing the SQL SELECT Statement . . . . . . . . . . . . . . . .
The DESCRIBE Table Command . . . . . . . . . . . . . . . . . . . . . .
Exercise 2-1: Describing the Human
Resources Schema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Capabilities of the SELECT Statement . . . . . . . . . . . . . . . . . .
Execute a Basic SELECT Statement . . . . . . . . . . . . . . . . . . . . . . . . . . .
Syntax of the Primitive SELECT Statement . . . . . . . . . . . . . .
Rules Are Meant to be Followed . . . . . . . . . . . . . . . . . . . . . . .
Exercise 2-2: Answering Our First
Questions with SQL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
SQL Expressions and Operators . . . . . . . . . . . . . . . . . . . . . . . .
NULL Is Nothing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 2-3: Experimenting with
Expressions and the DUAL Table . . . . . . . . . . . . . . . . . . . .
✓ Two-Minute Drill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q&A Self Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self Test Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

Restricting and Sorting Data

48
49
50

51
52
52
53
55
57
59
59
64
67
70
81
86
90
92
94
96
98

. . . . . . . . . . . . . . . . . . . . . 103

Limit the Rows Retrieved by a Query . . . . . . . . . . . . . . . . . . . . . . . . . .
The WHERE clause . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Comparison Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 3-1: Using the LIKE Operator . . . . . . . . . . . . . . .
Boolean Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Precedence Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Sort the Rows Retrieved by a Query . . . . . . . . . . . . . . . . . . . . . . . . . . .
The ORDER BY Clause . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 3-2: Sorting Data Using
the ORDER BY Clause . . . . . . . . . . . . . . . . . . . . . . . . . . . .

104
104
113
124
126
132
136
136
141

Contents

Ampersand Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Substitution Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Define and Verify . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 3-3: Using Ampersand Substitution . . . . . . . . . . .
✓ Two-Minute Drill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q&A Self Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self Test Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4

Single-Row Functions

142
143
149
155
158
160
162
164
166

. . . . . . . . . . . . . . . . . . . . . . . . . . . 169

Describe Various Types of Functions Available in SQL . . . . . . . . . . . .
Defining a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Types of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Use Character, Number, and Date Functions
in SELECT Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Using Character Case Conversion Functions . . . . . . . . . . . . .
Exercise 4-1: Using the Case Conversion Functions . . . . .
Using Character Manipulations Functions . . . . . . . . . . . . . . .
Exercise 4-2: Using the Case Manipulation Functions . . .
Using Numeric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Working with Dates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Using Date Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 4-3: Using the Date Functions . . . . . . . . . . . . . . .
✓ Two-Minute Drill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q&A Self Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self Test Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

xi

170
170
174
177
177
181
183
194
196
202
206
210
219
221
223
224
225

Using Conversion Functions and
Conditional Expressions . . . . . . . . . . . . . . . . . . . . . . . 227
Describe Various Types of Conversion
Functions Available in SQL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Conversion Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Use the TO_CHAR, TO_NUMBER,
and TO_DATE Conversion Functions . . . . . . . . . . . . . . . . . . . . . . .
Using the Conversion Functions . . . . . . . . . . . . . . . . . . . . . . .

228
228
231
232

xii

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

Exercise 5-1: Converting Dates into Characters
Using the TO_CHAR Function . . . . . . . . . . . . . . . . . . . . .
Apply Conditional Expressions in a SELECT Statement . . . . . . . . . . .
Nesting Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 5-2: Using NULLIF and NVL2
for Simple Conditional Logic . . . . . . . . . . . . . . . . . . . . . . .
Conditional Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 5-3: Using the DECODE Function . . . . . . . . . . .
✓ Two-Minute Drill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q&A Self Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self Test Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

239
245
245
251
254
260
263
265
267
268
269

Reporting Aggregated Data Using
the Group Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 273
Describe the Group Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Definition of Group Functions . . . . . . . . . . . . . . . . . . . . . . . . .
Types and Syntax of Group Functions . . . . . . . . . . . . . . . . . . .
Identify the Available Group Functions . . . . . . . . . . . . . . . . . . . . . . . .
Using the Group Functions . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 6-1: Using the Group Functions . . . . . . . . . . . . .
Nested Group Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Group Data Using the GROUP BY Clause . . . . . . . . . . . . . . . . . . . . . .
Creating Groups of Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The GROUP BY Clause . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Grouping by Multiple Columns . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 6-2: Grouping Data Based
on Multiple Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Include or Exclude Grouped Rows Using the HAVING Clause . . . . . .
Restricting Group Results . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The HAVING Clause . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 6-3: Using the HAVING Clause . . . . . . . . . . . . .
✓ Two-Minute Drill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q&A Self Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

274
274
275
279
279
284
285
287
287
288
291
292
294
294
296
298
301
303

Contents

Lab Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self Test Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

Displaying Data from Multiple Tables

xiii
305
306
308

. . . . . . . . . . . . . . 309

Write SELECT Statements to Access Data from
More Than One Table Using Equijoins and Nonequijoins . . . . . . . .
Types of Joins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Joining Tables Using SQL:1999 Syntax . . . . . . . . . . . . . . . . . .
Qualifying Ambiguous Column Names . . . . . . . . . . . . . . . . . .
The NATURAL JOIN Clause . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 7-1: Using the NATURAL JOIN . . . . . . . . . . . .
The Natural JOIN USING Clause . . . . . . . . . . . . . . . . . . . . .
The Natural JOIN ON Clause . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 7-2: Using the NATURAL
JOIN…ON Clause . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
N-Way Joins and Additional Join Conditions . . . . . . . . . . . . .
Nonequijoins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Join a Table to Itself Using a Self-Join . . . . . . . . . . . . . . . . . . . . . . . . . .
Joining a Table to Itself Using the JOIN…ON Clause . . . . . .
Exercise 7-3: Performing a Self-Join . . . . . . . . . . . . . . . . . .
View Data that Does Not Meet a Join Condition
by Using Outer Joins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Inner versus Outer Joins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Left Outer Joins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Right Outer Joins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Full Outer Joins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 7-4: Performing an Outer-Join . . . . . . . . . . . . . . .
Generate a Cartesian Product of Two or More Tables . . . . . . . . . . . . . .
Creating Cartesian Products Using Cross Joins . . . . . . . . . . . .
Exercise 7-5: Performing a Cross-Join . . . . . . . . . . . . . . . .
✓ Two-Minute Drill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q&A Self Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self Test Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

310
311
316
317
319
320
321
322
324
327
329
331
331
332
334
335
336
337
338
340
342
342
344
348
350
352
354
355

xiv

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

8

Using Subqueries to Solve Problems

. . . . . . . . . . . . . . 357

Define Subqueries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 8-1: Types of Subquery . . . . . . . . . . . . . . . . . . . . .
Describe the Types of Problems That the Subqueries Can Solve . . . . .
Use of a Subquery Result Set for Comparison Purposes . . . . .
Star Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Generate a Table from Which to SELECT . . . . . . . . . . . . . . .
Generate Values for Projection . . . . . . . . . . . . . . . . . . . . . . . .
Generate Rows to be Passed to a DML Statement . . . . . . . . .
Exercise 8-2: More Complex Subqueries . . . . . . . . . . . . . .
List the Types of Subqueries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Single- and Multiple-Row Subqueries . . . . . . . . . . . . . . . . . . .
Correlated Subqueries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 8-3: Investigate the Different
Types of Subquery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Write Single-Row and Multiple-Row Subqueries . . . . . . . . . . . . . . . . .
Exercise 8-4: Write a Query That Is Reliable
and User Friendly . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
✓ Two-Minute Drill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q&A Self Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self Test Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9

Using the Set Operators

358
359
360
360
361
362
362
363
363
364
365
366
367
369
370
374
375
378
379
380

. . . . . . . . . . . . . . . . . . . . . . . . 381

Describe the Set Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Sets and Venn Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Set Operator General Principles . . . . . . . . . . . . . . . . . . . . . . .
Exercise 9-1: Describe the Set Operators . . . . . . . . . . . . . .
Use a Set Operator to Combine Multiple Queries
into a Single Query . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The UNION ALL Operator . . . . . . . . . . . . . . . . . . . . . . . . . . .
The UNION Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The INTERSECT Operator . . . . . . . . . . . . . . . . . . . . . . . . . . .
The MINUS Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
More Complex Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 9-2: Using the Set Operators . . . . . . . . . . . . . . . .

382
383
384
385
386
387
387
388
389
390
392

Contents

xv

Control the Order of Rows Returned . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 9-3: Control the Order of Rows Returned
.....
✓ Two-Minute Drill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q&A Self Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self Test Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

393
394
396
397
398
400
401

10 Manipulating Data

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403

Describe Each Data Manipulation Language (DML) Statement . . . . .
INSERT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
UPDATE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
DELETE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
MERGE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
TRUNCATE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
DML Statement Failures . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Insert Rows into a Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 10-1: Use the INSERT Command . . . . . . . . . . . .
Update Rows in a Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 10-2: Use the UPDATE Command . . . . . . . . . . .
Delete Rows from a Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Removing Rows with DELETE . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 10-3: Use the DELETE Command . . . . . . . . . . .
Removing Rows with TRUNCATE . . . . . . . . . . . . . . . . . . . . .
MERGE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Control Transactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Database Transactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Transaction Control Statements . . . . . . . . . . . . . . . . . . .
Exercise 10-4: Use the COMMIT and
ROLLBACK Commands . . . . . . . . . . . . . . . . . . . . . . . . . . .
✓ Two-Minute Drill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q&A Self Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self Test Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

404
405
406
407
408
409
409
413
417
419
421
423
423
424
425
426
427
428
431
433
439
441
444
446
448

xvi

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

11 Using DDL Statements to
Create and Manage Tables

. . . . . . . . . . . . . . . . . . . . . 449

Categorize the Main Database Objects . . . . . . . . . . . . . . . . . . . . . . . . .
Object Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Users and Schemas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Naming Schema Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Object Namespaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 11-1: Determine What Objects
Are Accessible to Your Session . . . . . . . . . . . . . . . . . . . . . .
Review the Table Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 11-2: Investigate Table Structures . . . . . . . . . . . .
List the Data Types That Are Available for Columns . . . . . . . . . . . . . .
Exercise 11-3: Investigate the Data Types
in the HR schema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Create a Simple Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Creating Tables with Column Specifications . . . . . . . . . . . . .
Creating Tables from Subqueries . . . . . . . . . . . . . . . . . . . . . . .
Altering Table Definitions after Creation . . . . . . . . . . . . . . . .
Dropping and Truncating Tables . . . . . . . . . . . . . . . . . . . . . . .
Exercise 11-4: Create Tables . . . . . . . . . . . . . . . . . . . . . . .
Explain How Constraints Are Created
at the Time of Table Creation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Types of Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Defining Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 11-5: Work with Constraints . . . . . . . . . . . . . . . .
✓ Two-Minute Drill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q&A Self Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self Test Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12 Creating Other Schema Objects

450
450
452
453
455
455
456
457
457
461
462
462
464
465
466
467
469
470
473
476
478
479
482
483
484

. . . . . . . . . . . . . . . . . . 487

Create Simple and Complex Views . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Why Use Views at All? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Simple and Complex Views . . . . . . . . . . . . . . . . . . . . . . . . . . .
CREATE VIEW, ALTER VIEW, and DROP VIEW . . . . . . .
Exercise 12-1: Create Views . . . . . . . . . . . . . . . . . . . . . . . .

488
489
492
493
496

Contents

Retrieve Data from Views . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 12-2: Use Views . . . . . . . . . . . . . . . . . . . . . . . . . .
Create Private and Public Synonyms . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 12-3: Create and Use Synonyms . . . . . . . . . . . . .
Create, Maintain, and Use Sequences . . . . . . . . . . . . . . . . . . . . . . . . . .
Creating Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Using Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercise 12-4: Create and Use Sequences . . . . . . . . . . . . .
Create and Maintain Indexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
What Indexes Are For . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Types of Index
.....................................
Creating and Using Indexes . . . . . . . . . . . . . . . . . . . . . . . . . . .
Modifying and Dropping Indexes . . . . . . . . . . . . . . . . . . . . . . .
Exercise 12-5: Creating Indexes . . . . . . . . . . . . . . . . . . . . .
✓ Two-Minute Drill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Q&A Self Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self Test Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lab Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xvii
497
497
498
500
501
502
504
507
509
510
511
515
517
518
521
523
527
529
531

Appendix

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533

Glossary

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537

Index

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555

This page intentionally left blank

INTRODUCTION

T

here is an ever-increasing demand for staff with IT industry certification. The benefits
to employers are significant—they can be certain that staff have a certain level of
competence—and the benefits to the individuals, in terms of demand for their services,
are equally great. Many employers are now requiring technical staff to have certifications, and many
IT purchasers will not buy from firms that do not have certified staff. The Oracle certifications
are among the most sought after. But apart from rewards in a business sense, knowing that you
are among a relatively small pool of elite Oracle professionals and that you have proved your
competence is a personal reward well worth attaining.
There are several Oracle certification tracks—this book is concerned with the
Oracle Database Administration certification track, specifically for release 11g of the
database. There are three levels of DBA certification: Certified Associate (OCA),
Certified Professional (OCP), and Certified Master (OCM). The OCA qualification
is based on two examinations, the first of which is covered in this book. The OCP
qualification requires passing a third examination. These examinations can be taken
at any Prometric Center and consist of 60 to 70 questions to be completed in 90
minutes. The OCM qualification requires completing a further two-day evaluation
at an Oracle testing center, involving simulations of complex environments and use
of advanced techniques.
The exam content is also likely to be the starting point if you intend to study for
the Oracle PL/SQL and Oracle Forms Developer track, though (as of the time of
writing) this is at release 10g with no announcement of an 11g release.
To prepare for the first OCA examination, you can attend an Oracle University
instructor-led training course, you can study Oracle University online learning
material, or you can read this book. In all cases, you should also refer to the Oracle
Documentation Library for details on syntax. This book will be a valuable addition
to other study methods, but it is also sufficient by itself. It has been designed
with the examination objectives in mind, though it also includes a great deal
of information that will be useful to you in the course of your work. For readers
working in development, the subject matter of this book is also the starting point for
studying Oracle Corporation’s development tools: SQL, PL/SQL, and the Internet
application development kits shipped with the Oracle Application Server.

xix
Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

xx

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

However, it is not enough to buy the book, place it under your pillow, and assume
that knowledge will permeate the brain by a process of osmosis: you must read it
thoroughly, work through the exercises and sample questions, and experiment
further with various commands. As you become more familiar with the Oracle
environment, you will realize that there is one golden rule:
When in doubt, try it out.
In a multitude of cases, you will find that a simple test that takes a couple of
minutes can save hours of speculation and poring through manuals. If anything
is ever unclear, construct an example and see what happens. This book was
developed using Windows and Linux, but to carry out the exercises and your further
investigations you can use any platform that is supported for Oracle.
Your initiation into the magical world of Oracle database administration is about
to begin. This is a subject you can study and enjoy for the rest of your life.

In This Book
This book is organized to serve as an in-depth review for the OCA Oracle Database
11g: SQL Fundamentals I Exam for Oracle professionals. Each chapter covers a major
aspect of the exam; all the OCA official certification objectives are carefully covered
in the book.

On the CD
The CD-ROM contains the entire contents of the book in electronic form, as well
as a practice test that simulates the real Oracle Database 11g OCA certification test.
For more information on the CD-ROM, please see the appendix.

Exam Readiness Checklist
At the end of this introduction, you will find an Exam Readiness Checklist. This
table is the official exam objectives. The checklist also allows you to gauge your level
of expertise on each objective at the outset of your studies. This should allow you to
check your progress and make sure you spend the time you need on more difficult
or unfamiliar sections. Each objective is exactly as Oracle Corporation presents it,
including the chapter and page reference.

Introduction

xxi

In Every Chapter
This book includes a set of chapter components that call your attention to important
items, reinforce important points, and provide helpful exam-taking hints. Take a look
at what you’ll find in every chapter:
■ Exam Watch notes call attention to information about the exam, as well

as potential pitfalls.

The redo log stream
includes all changes: those applied to data

segments and to undo segments, for both
committed and uncommitted transactions.

■ Exercises are interspersed throughout the chapters, and they allow you to

get the hands-on experience you need in order to pass the exams. They help
you master skills that are likely to be an area of focus on the exam. Don’t
just read through the exercises; they are hands-on practice that you should
be comfortable completing. Learning by doing is an effective way to increase
your competency with a product.
■ On the Job notes describe the issues that come up most often in real-world

settings. They provide a valuable perspective on certification- and
product-related topics. They point out common mistakes and address
questions that have arisen from on-the-job discussions and experience.
■ Inside the Exam sections toward the end of each chapter are designed to

anticipate what the exam will emphasize. These are pointers regarding key
topics to focus on based on the authors’ experiences of sitting through many
production and beta examinations and having been on the Oracle internal
group that validates examination questions.
■ The Certification Summary is a succinct review of the chapter and

a restatement of salient points regarding the exam.

✓■

The Two-Minute Drill at the end of every chapter is a checklist of the main
points of the chapter. You can use it for a quick, last-minute review before
the test.

xxii

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

Q&A ■

The Self Test offers questions similar to those found on the certification exam.
The answers to these questions, as well as explanations of the answers, can
be found at the end of each chapter. By taking the Self Test after completing
each chapter, you’ll reinforce what you’ve learned from that chapter, while
becoming familiar with the structure of the exam questions.

■ The Lab Questions at the end of the Self Test sections offer a unique and

challenging question format that, in order to answer correctly, require the
reader to understand multiple chapter concepts. These questions are more
complex and more comprehensive than the other questions, as they test your
ability to take all the knowledge you have gained from reading the chapter
and apply it to complicated, real-world situations.

Some Pointers
Once you’ve finished reading this book, set aside some time to do a thorough review.
You might want to return to the book several times and make use of all the methods
it offers for reviewing the material before sitting the examination.
■ Reread all the Two-Minute Drills or have someone quiz you. You also can

use the drills as a way to do a quick cram before the exam.
■ Reread all the Exam Watch notes. Remember that these notes are based on

the OCA exam. They will draw your attention to what you should expect—and
what you should be on the lookout for.
■ Retake the Self Tests. It is a good idea to take the Self Test right after

you’ve read the chapter because the questions help reinforce what you’ve just
learned, and then take them all again at the end. In the examination, the
questions do not come conveniently grouped by subject: you will have to be
prepared to jump from one topic to another.
■ Complete the Exercises. Did you do the chapter Exercises and the Lab

Questions when you read each chapter? If not, do them! These exercises are
designed to cover exam topics, and there’s no better way to get to know this
material than by practicing. Be sure you understand why you are performing
each step in each exercise. If there is something you are not completely clear
about, reread that section in the chapter.

Introduction

xxiii

Test Structure
The OCA examinations are multiple choice questions, but they are not necessarily
questions where you must pick one answer. Some will ask for two or three answers,
some will say something on the lines of “choose all correct answers.” Most questions
are text based, but some will have an exhibit, which is a diagram or perhaps a screen
shot that illustrates the question.
Read all the questions very carefully. In some cases, when the question asks for one
answer, you may think they are all wrong or that several are correct. Often, when you
reread the question carefully, you will see what the examiners are looking for.
You will have an average of only a minute or two for each question. Go through
them all, fast, answering the ones you know and marking the ones you don’t know
for review. Then go through the marked questions again. That will take up most of
the time. If there are some questions you really don’t know, guess the answer: there
are no marks deducted for incorrect answers.

How to Prepare for the Exam
Study and practice! Go through each chapter of the book, doing all the exercises and
trying out further experiments. Make use of other resources if anything is still not
clear: the Oracle Documentation Library (a free download from Oracle Corporation’s
website) is essential reading. If you have the opportunity to attend an Oracle University
course or have access to the Oracle University Self-paced Online courses, these may
also be of value.
The example questions at the end of each chapter and on the CD are not copies of
real OCA questions (because that would be fraudulent), but they are realistic examples
of the types of question and the format of questions with which you will be faced. They
cover all the examined material. The day before you sit the exam, it makes sense to go
through them all, looking up any for which you do not know the answer.
This book is not intended to be just an exam crammer: it is intended to teach you
how to develop applications with SQL. If you know how to do that, you will pass the
exam. Do not memorize answers to questions—learn the techniques, principles, and
syntax that will let you work out the answers to any question.

xxiv

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

Exam 1Z0-051
OCA Official Objective

Chapter

Page

Data Retrieval Using the SQL SELECT Statement
List the capabilities of SQL SELECT statements

2

52

Execute a basic SELECT statement

2

59

Limit the rows that are retrieved by a query

3

104

Sort the rows that are retrieved by a query

3

136

Use ampersand substitution to restrict and sort output at runtime

3

142

Describe various types of functions available in SQL

4

170

Use character, number, and date functions in SELECT statements

4

177

Describe various types of conversion functions that are
available in SQL

5

228

Use the TO_CHAR, TO_NUMBER, and TO_DATE conversion
functions

5

231

Apply conditional expressions in a SELECT statement

5

245

Identify the available group functions

6

279

Describe the use of group functions

6

279

Group data by using the GROUP BY clause

6

287

Include or exclude grouped rows by using the HAVING clause

6

294

Write SELECT statements to access data from more than one table
using equijoins and nonequijoins

7

310

Join a table to itself by using a self-join

7

331

View data that generally does not meet a join condition by using
outer joins

7

334

Generate a Cartesian product of all rows from two or more tables

7

342

Restricting and Sorting Data

Single-Row Functions

Using Conversion Functions and Conditional Expressions

Reporting Aggregated Data Using the Group Functions

Displaying Data from Multiple Tables

Introduction

OCA Official Objective

Chapter

xxv

Page

Using Subqueries to Solve Problems
Define subqueries

8

358

Describe the types of problems that the subqueries can solve

8

360

List the types of subqueries

8

364

Write single-row and multiple-row subqueries

8

369

Describe set operators

9

382

Use a set operator to combine multiple queries into a single query

9

386

Control the order of rows returned

9

393

Describe each data manipulation language (DML) statement

10

404

Insert rows into a table

10

413

Update rows in a table

10

419

Delete rows from a table

10

423

Control transactions

10

427

Categorize the main database objects

11

450

Review the table structure

11

456

List the data types that are available for columns

11

457

Create a simple table

11

462

Explain how constraints are created at the time of table creation

11

469

Create simple and complex views

12

488

Retrieve data from views

12

497

Create, maintain, and use sequences

12

501

Create and maintain indexes

12

509

Create private and public synonyms

12

498

Using the Set Operators

Manipulating Data

Using DDL Statements to Create and Manage Tables

Creating Other Schema Objects

This page intentionally left blank

1
Oracle Server
Technologies and the
Relational Paradigm

CERTIFICATION OBJECTIVES
1.01

Position the Server Technologies

1.05

1.02

Understand Relational Structures

✓

1.03

Summarize the SQL Language

1.04

Use the Client Tools

Q&A

Create the Demonstration Schemas
Two-Minute Drill
Self Test

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

2

Chapter 1:

Oracle Server Technologies and the Relational Paradigm

T

he content of this chapter is not directly tested by the OCP examination, but it is
vital to understanding the purpose of SQL and what it is meant to achieve. This is
considered to be prerequisite knowledge that every student should have, beginning
with an appreciation of how the Oracle server technologies fit together and the relative position
of each product.
The Oracle server technologies product set is more than a database. There is also
the Oracle Application Server and the Oracle Enterprise Manager. Taken together,
these are the server technologies that make up the Grid. Grid computing is an
emerging environment for managing the complete IT environment and providing
resources to users on demand.
The relational paradigm for database information management was first
formalized in the late ‘60s and has been continually refined since. A paradigm is
a set of standards agreed upon by all those involved that specifies how problems
should be understood and addressed. There are other paradigms within the data
processing world. The Oracle database is an implementation of a Relational
Database Management System (RDBMS) that conforms to the relational paradigm,
but it then goes beyond it to include some aspects of other paradigms such as
hierarchical and object-oriented models.
Structured Query Language (SQL, pronounced “sequel”) is an international
standard for managing data stored in relational databases. Oracle Database 11g offers
an implementation of SQL that is generally compliant with the current standard,
which is SQL-2003. Full details of the compliancy are in Appendix B of the SQL
Language Reference, which is part of the Oracle Database Documentation Library.
As a rule, compliancy can be assumed.
Throughout this book, two tools are used extensively for exercises: SQL*Plus and
SQL Developer. These are tools that developers use every day in their work. The
exercises and many of the examples are based on two demonstration sets of data,
known as the HR and OE schemas. There are instructions on how to launch the
tools and create the demonstration schemas, though you may need assistance from
your local database administrator to get started.
This chapter consists of summarized descriptions of the Oracle Server
Technologies; the concepts behind the relational paradigm and normalizing of data
into relational structures; the SQL language; the client tools; and the demonstration
schemas.

Position the Server Technologies

3

CERTIFICATION OBJECTIVE 1.01

Position the Server Technologies
There is a family of products that makes up the Oracle server technologies:
■ The Oracle Database
■ The Oracle Application Server
■ The Oracle Enterprise Manager
■ Various application development tools and languages

These products each have a position in the Oracle product set. The database
is the repository for data and the engine that manages access to it. The Oracle
Application Server runs software on behalf of end users: it generates the user
interface in the form of windows displayed in users’ browsers and submits calls for
data retrieval and modification to the database for execution. The Oracle Enterprise
Manager is a comprehensive administration tool for monitoring, managing, and
tuning the Oracle processes and also (through plug-ins) third-party products. Lastly,
there are tools and languages for developing applications; either applications that
run on end users’ machines in the client-server model, or applications that run
centrally on application servers.
The combination of the server technologies and the development tools make
up a platform for application development and delivery that enables the Grid. The
Grid is an approach to the delivery of IT services that maximizes the cost efficiency
of the whole environment by delivering computing power from a pool of available
resources to wherever it is needed, on demand.

The Oracle Server Architecture
An Oracle database is a set of files on disk. It exists until these files are deliberately
deleted. There are no practical limits to the size and number of these files, and
therefore no practical limits to the size of a database. Access to the database
is through the Oracle instance. The instance is a set of processes and memory
structures: it exists on the CPU(s) and in the memory of the server node, and
this existence is temporary. An instance can be started and stopped. Users of the
database establish sessions against the instance, and the instance then manages all

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Chapter 1:

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access to the database. It is absolutely impossible in the Oracle environment for any
user to have direct contact with the database. An Oracle instance with an Oracle
database makes up an Oracle server.
The processing model implemented by the Oracle server is that of client-server
processing, often referred to as two-tier. In the client-server model, the generation
of the user interface and much of the application logic is separated from the
management of the data. For an application developed using SQL (as all relational
database applications will be), this means that the client tier generates the SQL
commands, and the server tier executes them. This is the basic client-server split,
with (as a general rule) a local area network between the two sides. The network
communications protocol used between the user process and the server process is
Oracle’s proprietary protocol, Oracle Net.
The client tier consists of two components: the users and the user processes. The
server tier has three components: the server processes that execute the SQL, the
instance, and the database itself. Each user interacts with a user process. Each user
process interacts with a server process, usually across a local area network. The server
processes interact with the instance, and the instance with the database. Figure 1-1
shows this relationship diagrammatically. A session is a user process in communication
with a server process. There will usually be one user process per user and one server
process per user process. The user and server processes that make up sessions are
launched on demand by users and terminated when no longer required; this is the
log-on and log-off cycle. The instance processes and memory structures are launched
by the database administrator and persist until the administrator deliberately
terminates them; this is the database start-up and shut-down cycle.
The user process can be any client-side software that is capable of connecting
to an Oracle server process. Throughout this book, two user processes will be used
extensively: SQL*Plus and SQL Developer. These are simple processes provided by
Oracle for establishing sessions against an Oracle server and issuing ad hoc SQL.
FIGURE 1-1

The indirect
connection
between a user
and a database

Server-side components

Client-side components

Instance
User

User process

Server process

Session components
Database

Position the Server Technologies

5

A widely used alternative is TOAD (the Tool for Application Developers) from
Quest Software, though this is licensed software and before using it one must always
ensure that the license is legal. End-user applications will need to be written with
something more sophisticated than these tools, something capable of managing
windows, menus, proper onscreen dialogs, and so on. Such an application could be
written with the Oracle Developer Suite products; with Microsoft Access linked to
the Oracle ODBC drivers; with any third-generation language (such as C or Java)
for which Oracle has provided a library of function calls that will let it interact with
the server; or with any number of Oracle-compatible third-party tools. What the
user process actually is does not matter to the Oracle server at all. When an end
user fills in a form and clicks a Submit button, the user process will be generating an
INSERT statement (detailed in Chapter 11) and sending it to a server process for
execution against the instance and the database. As far as the server is concerned,
the INSERT statement might just as well have been typed into SQL*Plus as what is
known as ad hoc SQL.
Never forget that all communication with an Oracle server follows this clientserver model. The separation of user code from server code dates back to the
earliest releases of the database and is unavoidable. Even if the user process is
running on the same machine as the server (as is the case if, for example, one is
running a database on one’s own laptop PC for development or training purposes),
the client-server split is still enforced, and network protocols are still used for the
communications between the two processes. Applications running in an application
server environment (described in the next section) also follow the client-server
model for their database access.
The simplest form of the database server is one instance connected to one
database, but in a more complex environment one database can be opened by
many instances. This is known as a RAC (Real Application Cluster). RAC can
bring many potential benefits, which may include scalability, performance, and zero
downtime. The ability to add dynamically more instances running on more nodes to
a database is a major part of the database’s contribution to the Grid.

The Oracle Application Server
With the emergence of the Web as the standard communications platform for
delivering applications to end users has come the need for application servers. An
application server replaces the client-side software traditionally installed on enduser terminals; it runs applications centrally, presenting them to users in windows
displayed locally in web browsers. The applications make use of data stored in one or
more database servers.

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The Oracle Application Server is a platform for developing, deploying, and
managing web applications. A web application can be defined as any application with
which users communicate with HTTP. Web applications usually run in at least three
tiers: a database tier manages access to the data, the client tier (often implemented
as a web browser) handles the local window management for communications with
the users, and an application tier in the middle executes the program logic that
generates the user interface and the SQL calls to the database.
Web applications can be developed with a number of technologies, predominant
among which is Java. Applications written in Java should conform to the J2EE
(Java 2 Enterprise Edition) standard, which defines how such applications should
be packaged and deployed. J2EE and related standards are controlled by Sun
Microsystems and accepted by virtually all software developers. Oracle Application
Server is a J2EE-compliant application server. Oracle’s implementation of the
standards allows for automatic load balancing and fault tolerance across multiple
application servers on multiple machines though J2EE clustering. Clustering
virtualizes the provision of the application service; users ask for an application that
might be available from a number of locations, and the cluster works out from where
any one session or request can best be serviced. If one location fails, others will take
up the load, and more resources can be made available to an application as necessary.
The ability to separate the request for a service from the location of its provision
and to add or remove J2EE servers from a cluster dynamically is a major part of the
Oracle Application Server’s contribution to the Grid.
It is important to note that Oracle’s commitment to international standards is
very strong. Applications running in the Oracle Application Server environment
can connect to any database for which there are Java-compliant drivers; it is not
necessary to use an Oracle database. Applications developed with the Oracle
Application Server toolkits can be deployed to a third-party J2EE-compliant
application server.
The simplest processing model of web applications is three tier: a client tier that
manages the user interface; a middle tier that generates the interface and issues SQL
statements to the data tier; and a data tier that manages the data itself. In the Oracle
environment, the client tier will be a browser (such as Mozilla or Microsoft Internet
Explorer) that handles local window management, controls the keyboard, and tracks
mouse movements. The middle tier will be an Oracle Application Server running the
software (probably written in Java) that generates the windows sent to the client tier
for display and the SQL statements sent to the data tier for execution. The data tier
will be an Oracle server: an instance and a database. In this three-tier environment,
there are two types of sessions: end-user sessions from the client tier to the middle tier, and
database sessions from the middle tier to the data tier. The end-user sessions will

Position the Server Technologies

7

be established with HTTP. The database sessions are client-server sessions consisting
of a user process and a server process, as described in the previous section.
It is possible for an application to use a one-for-one mapping of end-user session
to database session: each user, from their browser, will establish a session against the
application server, and the application server will then establish a session against
the database server on the user’s behalf. However, this model has been proven to be
very inefficient when compared to the connection pooling model. With connection
pooling, the application server establishes a relatively small number of persistent
database sessions and makes them available on demand (queuing requests if
necessary) to a relatively large number of end-user sessions against the application
server. Figure 1-2 illustrates the three-tier architecture using connection pooling.
From the point of view of the database, it makes no difference whether a SQL
statement comes from a client-side process such as SQL*Plus or Microsoft Access or
from a pooled session to an application server. In the former case, the user process
all happens on one machine; in the latter, the user process has been divided into
two tiers: an applications tier that generates the user interface and a client tier that
displays it.

Oracle Enterprise Manager
The increasing size and complexity of IT installations makes management a challenging
task. This is hardly surprising: no one ever said that managing a powerful environment
should necessarily be simple. However, management tools can make the task easier and
the management staff more productive.
FIGURE 1-2

Stateful persistent sessions
over Oracle Net

Browser

The connection
pooling model
Browser

Browser

Application
server

Browser
Stateless nonpersistent
sessions over HTTP

Database
server

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Chapter 1:

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Oracle Enterprise Manager comes in three forms:
■ Database Control
■ Application Server Control
■ Grid Control

Oracle Enterprise Manager Database Control is a graphical tool for managing
one database, which may be a RAC clustered database. It consists of a Java process
running on the database server machine. Administrators connect to Database
Control from a browser, and Database Control then connects to the database server.
Database Control has facilities for real-time management and monitoring, running
scheduled jobs, and reporting alert conditions interactively and through e-mail.
Oracle Enterprise Manager Application Server Control is a graphical tool for
managing one application server instance or a group of instances. The grouping
technology is dependent on the version. Up to Oracle Application Server 10g
release 2, multiple application servers were managed as a “farm,” with a metadata
repository (typically residing in an Oracle database) as the central management
point. From release 3 onward, the technology is based on J2EE clustering, which is
not proprietary to Oracle.
Oracle Enterprise Manager Grid Control globalizes the management environment. A
management repository (residing in an Oracle database) and one or more management
servers manage the complete environment: all the databases and application servers,
wherever they may be. Grid Control can also manage the nodes, or machines, on which
the servers run, as well as (through plug-ins) a wide range of third-party products. Each
managed node runs an agent process, which is responsible for monitoring the managed
target on the node: executing jobs against them and reporting status, activity levels, and
alert conditions back to the management server(s).
Grid Control gives a holistic view of the environment and, if well configured,
makes administration staff far more productive than they are without it. It becomes
possible for one administrator to manage effectively hundreds of targets.

Grid Computing
Critical to the concept of Grid computing is virtualization. This means that at all
levels there is a layer of abstraction between what is requested and what is provided.
End users ask for an application service and let the Grid work out which clustered
J2EE application server can best provide it. Application servers ask for a database
service and let the Grid work out from which RAC node the data can best be served.
Within the Grid there is a mapping of possible services to available service providers,
and there are algorithms for assigning the workload and resources appropriately.

Position the Server Technologies

9

The result is that end users have neither the need nor the capacity to know from
where their computing resources are actually being provided. The analogy often
drawn is with delivery of domestic electricity: it is supplied on demand, and the home
owner has no way of telling which power station is currently supplying him.
The Grid is not exclusive to Oracle. At the physical level, some operating system
and hardware vendors are providing Grid-like capabilities. These include the ability
to partition servers into virtual machines and dynamically add or remove CPU(s)
and RAM from the virtual machines according to demand. This is conceptually
similar to Oracle’s approach of dynamically assigning application server and database
server resources to logical services. There is no reason why the two approaches
cannot be combined. Both are working toward the same goal and can work together.
The result should be an environment where adequate resources are always available
on demand, without facing the issues of excess capacity at some times and underperformance at others. It should also be possible to design a Grid environment with
no single point of failure, thus achieving the goal of 100 percent uptime that is being
demanded by many users.
The SQL application developer need not know how the Grid has been
implemented. The SQL will be invoked from an application server and executed
by an instance against a database: the Grid will take care of making sure that at any
moment pools of application servers and instances sized appropriately for the current
workload are available.

EXERCISE 1-1
Investigate Your Database and Application Environment
This is a paper-based exercise, with no specific solution.
Attempt to identify the user processes, application servers, and database servers
used in your environment. Try to work out where the SQL is being generated and
where it is being executed. Bear in mind that usually the user processes used by
end users will be graphical and will frequently go through application servers; the
database administration and development staff will often prefer to use client-server
tools that connect to the database server directly.

Development Tools and Languages
The Oracle server technologies include various facilities for developing applications,
some existing within the database, others external to it.

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Oracle Server Technologies and the Relational Paradigm

Within the database, it is possible to use three languages. The one that is
unavoidable, and the subject of this book, is SQL. SQL is used for data access, but
it cannot be used for developing complete applications. It has no real facilities
for developing user interfaces, and it also lacks the procedural structures needed
for manipulating rows individually. The other two languages available within the
database fill these gaps. They are PL/SQL and Java. PL/SQL is a third-generation
language (3GL) proprietary to Oracle. It has the usual procedural constructs (such as
if-then-else and looping) and facilities for user interface design. In the PL/SQL code,
one can embed calls to SQL. Thus, a PL/SQL application might use SQL to retrieve
one or more rows from the database, then perform various actions based on their
content, and then issue more SQL to write rows back to the database. Java offers a
similar capability to embed SQL calls within the Java code. This is industry standard
technology: any Java programmer should be able write code that will work with an
Oracle database (or indeed with any other Java-compliant database.)
Other languages are available for developing client-server applications that run
externally to the database. The most commonly used are C and Java, but it is possible
to use most of the mainstream 3GLs. For all these languages, Oracle Corporation
provides OCI (Oracle Call Interface) libraries that let code written in these languages
establish sessions against an Oracle database and invoke SQL commands.
Many organizations will not want to use a 3GL to develop database applications.
Oracle Corporation provides rapid application development tools as part of the
Oracle Developer Suite, and there are many third-party products. These can make
programmers far more productive than if they were working with a 3GL. Like the
languages, all these application development tools end up doing the same thing:
constructing SQL statements that are sent to the database server for execution.
All developers and administrators working in the Oracle environment must know
PL/SQL. C and Java are not necessary, unless the project specifically uses them.

CERTIFICATION OBJECTIVE 1.02

Understand Relational Structures
Critical to an understanding of SQL is an understanding of the relational paradigm
and the ability to normalize data into relational structures. Normalization is the work
of systems analysts, as they model business data into a form suitable for storing in

Understand Relational Structures

11

relational tables. It is a science that can be studied for years, and there are many
schools of thought that have developed their own methods and notations.

Rows and Tables
The relational paradigm models data as two-dimensional tables. A table consists
of a number of rows, each consisting of a set of columns. Within a table, all the
rows have the same column structure, though it is possible that in some rows
some columns may have nothing in them. An example of a table would be a list
of one’s employees, each employee being represented by one row. The columns
might be employee number, name, and a code for the department in which the
employee works. Any employees not currently assigned to a department would have
that column blank. Another table could represent the departments: one row per
department, with columns for the department’s code and the department’s name.
A note on terminology: what Oracle refers to as a table may also be called a
relation or an entity. Rows are sometimes called records or tuples, and columns may
be called attributes or fields. The number of “rows in the table” is the “cardinality of
the tuples.”
Relational tables conform to certain rules that constrain and define the data.
At the column level, each column must be of a certain data type, such as numeric,
date-time, or character. The character data type is the most general, in that it can
accept any type of data. At the row level, usually each row must have some uniquely
identifying characteristic: this could be the value of one column, such as the
employee number and department number in the preceding examples, which cannot
be repeated in different rows. There may also be rules that define links between the
tables, such as a rule that every employee must be assigned a department code that
can be matched to a row in the departments table. Following are examples of the
tabulated data definitions:
Departments table:
Column Name

Description

Data Type

Length

DEPTNO

Department number

Numeric

2

DNAME

Department name

Character

14

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Chapter 1:

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Employees table:
Column Name

Description

Data Type

Length

EMPNO

Employee number

Numeric

4

ENAME

Employee name

Character

10

DEPTNO

Department number

Numeric

2

The tables could contain these rows:
Departments:
DEPTNO

DNAME

10

ACCOUNTING

20

RESEARCH

30

SALES

40

OPERATIONS

Employees:
EMPNO

ENAME

DEPTNO

7369

SMITH

20

7499

ALLEN

30

7521

WARD

30

7566

JONES

20

7654

MARTIN

30

7698

BLAKE

30

7782

CLARK

10

7788

SCOTT

20

Looking at the tables, the two-dimensional structure is clear. Each row is of fixed
length, each column is of fixed length (padded with spaces when necessary), and the
rows are delimited with a new line. The rows have been stored in code order, but
this is a matter of chance, not design: relational tables do not impose any particular
ordering on their rows. Department number 10 has one employee, and department

Understand Relational Structures

13

number 40 has none. Changes to data are usually very efficient with the relational
model. New employees can be appended to the employees table, or they can be
moved from one department to another simply by changing the DEPTNO value in
their row.
Consider an alternative structure, where the data is stored according to the
hierarchical paradigm. The hierarchical model was developed before the relational
model, for technology reasons. In the early days of computing, storage devices lacked
the capability for maintaining the many separate files that were needed for the
many relational tables. Note that this problem is avoided in the Oracle database by
abstracting the physical storage (files) from the logical storage (tables): there is no
direct connection between tables and files and certainly not a one-to-one mapping.
In effect, many tables can be stored in a very few files.
A hierarchical structure stores all related data in one unit. For example, the
record for a department would include all that department’s employees. The
hierarchical paradigm can be very fast and very space efficient. One file access may
be all that is needed to retrieve all the data needed to satisfy a query. The employees
and departments listed previously could be stored hierarchically as follows:
10,ACCOUNTING,7782,CLARK
20,RESEARCH,7369,SMITH,7566,JONES,7788,SCOTT
30,SALES,7499,ALLEN,7521,WARD,7654,MARTIN,7698,BLAKE
40,OPERATIONS

In this example layout, the rows and columns are of variable length. Columns
are delimited with a comma, rows with a new line. Data retrieval is typically
very efficient if the query can navigate the hierarchy: if one knows an employee’s
department, the employee can be found quickly. If one doesn’t, the retrieval may
be slow. Changes to data can be a problem if the change necessitates movement.
For example, to move employee 7566, JONES from RESEARCH to SALES would
involve considerable effort on the part of the database because the move has to be
implemented as a removal from one line and an insertion into another. Note that
in this example, while it is possible to have a department with no employees (the
OPERATIONS department) it is absolutely impossible to have an employee without
a department: there is nowhere to put him or her. This is excellent if there is a
business rule stating that all employees must be in a department but not so good if
that is not the case.

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The relational paradigm is highly efficient in many respects for many types of
data, but it is not appropriate for all applications. As a general rule, a relational
analysis should be the first approach taken when modeling a system. Only if it proves
inappropriate should one resort to nonrelational structures. Applications where the
relational model has proven highly effective include virtually all Online Transaction
Processing (OLTP) systems and Decision Support Systems (DSS). The relational
paradigm can be demanding in its hardware requirements and in the skill needed
to develop applications around it, but if the data fits, it has proved to be the most
versatile model. There can be, for example, problems caused by the need to maintain
the indexes that maintain the links between tables and the space requirements of
maintaining multiple copies of the indexed data in the indexes themselves and in
the tables in which the columns reside. Nonetheless, relational design is in most
circumstances the optimal model.
A number of software publishers have produced database management systems
that conform (with varying degrees of accuracy) to the relational paradigm; Oracle
is only one. IBM was perhaps the first company to commit major resources to it,
but their product (which later developed into DB2) was not ported to non-IBM
platforms for many years. Microsoft’s SQL Server is another relational database that
has been limited by the platforms on which it runs. Oracle databases, by contrast,
have always been ported to every major platform from the first release. It may be this
that gave Oracle the edge in the RDBMS market place.
A note on terminology: confusion can arise when discussing relational databases
with people used to working with Microsoft products. SQL is a language and SQL
Server is a database, but in the Microsoft world, the term SQL is often used to refer
to either.

Data Normalization
The process of modeling data into relational tables is known as normalization and
can be studied at university level for years. There are commonly said to be three
levels of normalization: the first, second, and third normal forms. There are higher
levels of normalization: fourth and fifth normal forms are well defined, but any
normal data analyst (and certainly any normal human being) will not need to be
concerned with them. It is possible for a SQL application to address un-normalized
data, but this will usually be inefficient as that is not what the language is designed
to do. In most cases, data stored in a relational database and accessed with SQL
should be normalized to the third normal form.

Understand Relational Structures

15

SCENARIO & SOLUTION
Your organization is designing a new
application. Who should be involved?

Everyone! The project team must involve business analysts
(who model the business processes), systems analysts
(who model the data), system designers (who decide how
to implement the models), developers (you), database
administrators, system administrators, and (most importantly)
end users.

It is possible that relational structures may
not be suitable for a particular application.
How can this be determined, and what
should be done next? Can Oracle help?

Attempt to normalize the data into two-dimensional tables,
linked with one-to-many relationships. If this really cannot be
done, consider other paradigms. Oracle may well be able to help.
For instance, maps and other geographical data really don’t work
relationally. Neither does text data (such as word processing
documents). But the Spatial and Text database options can be
used for these purposes. There is also the possibility of using
user-defined objects to store nontabular data.

There are often several possible normalized models for an application. It
is important to use the most appropriate—if the systems analyst gets this
wrong, the implications can be serious for performance, storage needs, and
development effort.
As an example of normalization, consider an un-normalized table called BOOKS
that stores details of books, authors, and publishers, using the ISBN number as the
primary key. A primary key is the one attribute (or attributes) that can uniquely
identify a record. These are two entries:
ISBN

Title

Authors

Publisher

12345

Oracle 11g OCP SQL
Fundamentals 1 Exam Guide

John Watson,
Roopesh Ramklass

McGraw-Hill, Spear
Street, San Francisco,
CA 94105

67890

Oracle 11g New Features
Exam Guide

Sam Alapati

McGraw-Hill, Spear
Street, San Francisco,
CA 94105

Storing the data in this table gives rise to several anomalies. First, here is the
insertion anomaly: it is impossible to enter details of authors who are not yet

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published, because there will be no ISBN number under which to store them.
Second, a book cannot be deleted without losing the details of the publisher: a
deletion anomaly. Third, if a publisher’s address changes, it will be necessary to
update the rows for every book he has published: an update anomaly. Furthermore,
it will be very difficult to identify every book written by one author. The fact that a
book may have several authors means that the “author” field must be multivalued,
and a search will have to search all the values. Related to this is the problem of
having to restructure the table of a book that comes along with more authors than
the original design can handle. Also, the storage is very inefficient due to replication
of address details across rows, and the possibility of error as this data is repeatedly
entered is high. Normalization should solve all these issues.
The first normal form is to remove the repeating groups, in this case, the multiple
authors: pull them out into a separate table called AUTHORS. The data structures
will now look like the following.
Two rows in the BOOKS table:
ISBN

TITLE

PUBLISHER

12345

Oracle 11g OCP SQL Fundamentals
1 Exam Guide

McGraw-Hill, Spear Street,
San Francisco, California

67890

Oracle 11g New Features Exam Guide

McGraw-Hill, Spear Street,
San Francisco, California

And three rows in the AUTHOR table:
NAME

ISBN

John Watson

12345

Roopesh Ramklass

12345

Sam Alapati

67890

The one row in the BOOKS table is now linked to two rows in the AUTHORS
table. This solves the insertion anomaly (there is no reason not to insert as many
unpublished authors as necessary), the retrieval problem of identifying all the books
by one author (one can search the AUTHORS table on just one name) and the
problem of a fixed maximum number of authors for any one book (simply insert as
many or as few AUTHORS as are needed).

Understand Relational Structures

17

This is the first normal form: no repeating groups.
The second normal form removes columns from the table that are not dependent
on the primary key. In this example, that is the publisher’s address details: these
are dependent on the publisher, not the ISBN. The BOOKS table and a new
PUBLISHERS table will then look like this:
BOOKS
ISBN

TITLE

PUBLISHER

12345

Oracle 11g OCP SQL Fundamentals 1 Exam Guide

McGraw-Hill

67890

Oracle 11g New Features Exam Guide

McGraw-Hill

PUBLISHERS
PUBLISHER

STREET

CITY

STATE

McGraw-Hill

Spear Street

San Francisco

California

All the books published by one publisher will now point to a single record in
PUBLISHERS. This solves the problem of storing the address many times, and also
solves the consequent update anomalies and the data consistency errors caused by
inaccurate multiple entries.
Third normal form removes all columns that are interdependent. In the
PUBLISHERS table, this means the address columns: the street exists in only one
city, and the city can be in only one state; one column should do, not three. This
could be achieved by adding an address code, pointing to a separate address table:
PUBLISHERS
PUBLISHER

ADDRESS CODE

McGraw-Hill

123

ADDRESSES
ADDRESS CODE

STREET

CITY

STATE

123

Spear Street

San Francisco

California

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One characteristic of normalized data that should be emphasized now is the use
of primary keys and foreign keys. A primary key is the unique identifier of a row
in a table, either one column or a concatenation of several columns (known as a
composite key). Every table should have a primary key defined. This is a requirement
of the relational paradigm. Note that the Oracle database deviates from this
standard: it is possible to define tables without a primary key—though it is usually
not a good idea, and some other RDBMSs do not permit this.
A foreign key is a column (or a concatenation of several columns) that can be
used to identify a related row in another table. A foreign key in one table will match
a primary key in another table. This is the basis of the many-to-one relationship. A
many-to-one relationship is a connection between two tables, where many rows in
one table refer to a single row in another table. This is sometimes called a parentchild relationship: one parent can have many children. In the BOOKS example so
far, the keys are as follows:
TABLE

KEYS

BOOKS

Primary key: ISBN
Foreign key: Publisher

AUTHORS

Primary key: Name + ISBN
Foreign key: ISBN

PUBLISHERS

Primary key: Publisher
Foreign key: Address code

ADDRESSES

Primary key: Address code

These keys define relationships such as that one book can have several authors.
There are various standards for documenting normalized data structures,
developed by different organizations as structured formal methods. Generally
speaking, it really doesn’t matter which method one uses as long as everyone reading
the documents understands it. Part of the documentation will always include a
listing of the attributes that make up each entity (also known as the columns that
make up each table) and an entity-relationship diagram representing graphically the
foreign to primary key connections. A widely used standard is as follows:
■ Primary key columns identified with a hash (#)
■ Foreign key columns identified with a back slash (\)
■ Mandatory columns (those that cannot be left empty) with an asterisk (*)
■ Optional columns with a lowercase “o”

Understand Relational Structures

19

The BOOKS tables can now be described as follows:
Table BOOKS
#*
o
\*

Primary key, required
Optional
Foreign key, link to the PUBLISHERS table

ISBN
Title
Publisher

Table AUTHORS
#*
#\o

Name
ISBN

Together with the ISBN, the primary key
Part of the primary key, and a foreign key to the BOOKS table.
Optional, because some authors may not yet be published.

Table PUBLISHERS
#*
\o

Publisher
Address code

Primary key
Foreign key, link to the ADDRESSES table

Table ADDRESSES
#*
o
o
o

Address code
Street
City
State

Primary key

The second necessary part of documenting the normalized data model is the
entity-relationship diagram. This represents the connections between the tables
graphically. There are different standards for these; Figure 1-3 shows the entityrelationship diagram for the BOOKS example using a very simple notation limited
to showing the direction of the one-to-many relationships, using what are often
called crow’s feet to indicate which sides of the relationship are the many and
the one. It can be seen that one BOOK can have multiple AUTHORS, one
PUBLISHER can publish many books. Note that the diagram also states that both
AUTHORS and PUBLISHERS have exactly one ADDRESS. More complex
notations can be used to show whether the link is required or optional, information
which will match that given in the table columns listed previously.

FIGURE 1-3

An entityrelationship
diagram

ADDRESSES

AUTHORS

BOOKS

PUBLISHERS

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This is a very simple example of normalization, and is not in fact complete. If
one author were to write several books, this would require multiple values in the
ISBN column of the AUTHORS table. That would be a repeating group, which
would have to be removed because repeating groups break the rule for first normal
form. A major exercise with data normalization is ensuring that the structures can
handle all possibilities.
A table in a real-world application may have hundreds of columns and dozens
of foreign keys. The standards for notation vary across organizations—the example
given is very basic. Entity-relationship diagrams for applications with hundreds or
thousands of entities can be challenging to interpret.

EXERCISE 1-2
Perform an Extended Relational Analysis
This is a paper-based exercise, with no specific solution.
Consider the situation where one author can write many books, and one book can
have many authors. This is a many-to-many relationship, which cannot be fit into
the relational model. Sketch out data structures that demonstrate the problem, and
develop another structure that would solve it. Following is a possible solution.
The un-normalized table of books with many authors could look like this:
BOOKS
#*
\*

Title
Authors

There could be two rows in this table:
Title

Authors

11g SQL Fundamentals Exam Guide

John Watson, Roopesh Ramklass

10g DBA Exam Guide

John Watson, Damir Bersinic

And that of authors could look like this:
AUTHORS
#*
\*

Name
Books

Understand Relational Structures

21

There could be three rows in this table:
Name

Books

John Watson

11g SQL Fundamentals Exam Guide, 10g DBA Exam Guide

Roopesh Ramklass

11g SQL Fundamentals Exam Guide

Damir Bersinic

10g DBA Exam Guide

This many-to-many relationship needs to be resolved into many-to-one
relationships by taking the repeating groups out of the two tables and storing them
in a separate books-per-author table. It will also become necessary to introduce
some codes, such as ISBNs to identify books and social security numbers to identify
authors. This is a possible normalized structure:
BOOKS
#*
o

ISBN
Title

AUTHORS
#*
o

SSNO
Name

BOOKAUTHORS
#\*
#\*

ISBN
SSNO

Part of the primary key and a foreign key to BOOKS
Part of the primary key and a foreign key to AUTHORS

The rows in these normalized tables would be as follows:
BOOKS
ISBN

Title

12345

11g SQL Fundamentals Exam Guide

67890

DBA Exam Guide

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Oracle Server Technologies and the Relational Paradigm

AUTHORS
SSNO

Name

11111

John Watson

22222

Damir Bersinic

33333

Roopesh Ramklass

BOOKAUTHORS
ISBN

SSNO

12345

11111

12345

22222

67890

11111

67890

33333

Figure 1-4 shows the entity-relationship diagram for the original un-normalized
structure, followed by the normalized structure.
As a further exercise, consider the possibility that one publisher could have
offices at several addresses, and one address could have offices for several companies.
Authors will also have addresses, and this connection too needs to be defined. These
enhancements can be added to the example worked through previously.

FIGURE 1-4

Un-normalized
and normalized
data models

First, an un-normalized many-to-many relationship:
BOOKS

AUTHORS

The many-to-many relationship resolved, by interposing another entity:
BOOKS

BOOKAUTHORS

AUTHORS

Summarize the SQL Language

23

CERTIFICATION OBJECTIVE 1.03

Summarize the SQL Language
SQL is defined, developed, and controlled by international bodies. Oracle Corporation
does not have to conform to the SQL standard but chooses to do so. The language itself
can be thought as being very simple (there are only 16 commands), but in practice SQL
coding can be phenomenally complicated. That is why a whole book is needed to cover
the bare fundamentals.

SQL Standards
Structured Query Language (SQL) was first invented by an IBM research group in
the ’70s, but in fact Oracle Corporation (then trading as Relational Software, Inc.)
claims to have beaten IBM to market by a few weeks with the first commercial
implementation: Oracle 2, released in 1979. Since then the language has evolved
enormously and is no longer driven by any one organization. SQL is now an
international standard. It is managed by committees from ISO and ANSI. ISO is
the Organisation Internationale de Normalisation, based in Geneva; ANSI is the
American National Standards Institute, based in Washington, DC. The two bodies
cooperate, and their SQL standards are identical.
Earlier releases of the Oracle database used an implementation of SQL that had
some significant deviations from the standard. This was not because Oracle was
being deliberately different: it was usually because Oracle implemented features
that were ahead of the standard, and when the standard caught up, it used different
syntax. An example is the outer join (detailed in Chapter 8), which Oracle
implemented long before standard SQL; when standard SQL introduced an outer
join, Oracle added support for the new join syntax while retaining support for its
own proprietary syntax. Oracle Corporation ensures future compliance by inserting
personnel onto the various ISO and ANSI committees and is now assisting with
driving the SQL standard forward.

SQL Commands
These are the 16 SQL commands, separated into commonly used groups:

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The Data Manipulation Language (DML) commands:
■ SELECT
■ INSERT
■ UPDATE
■ DELETE
■ MERGE

The Data Definition Language (DDL) commands:
■ CREATE
■ ALTER
■ DROP
■ RENAME
■ TRUNCATE
■ COMMENT

The Data Control Language (DCL) commands:
■ GRANT
■ REVOKE

The Transaction Control Language (TCL) commands:
■ COMMIT
■ ROLLBACK
■ SAVEPOINT

The first command, SELECT, is the main subject of Chapters 2 through 9.
The remaining DML commands are covered in Chapter 10, along with the TCL
commands. DDL is detailed in Chapters 11 and 12. DCL, which has to do with
security, is only briefly mentioned: it falls more into the domain of the database
administrator than the developers.
According to all the docs, SELECT is a DML statement. In practice, no one
includes it when they refer to DML—they talk about it as though it were
a language in its own right (it almost is) and use DML to mean only the
commands that change data.

Summarize the SQL Language

25

A Set-oriented Language
Most 3GLs are procedural languages. Programmers working in procedural languages
specify what to do with data, one row at a time. Programmers working in a setoriented language say what they want to do to a group (a “set”) of rows and let the
database work out how to do it to however many rows are in the set.
Procedural languages are usually less efficient than set-oriented languages at
managing data, as regards both development and execution. A procedural routine
for looping through a group of rows and updating them one by one will involve
many lines of code, where SQL might do the whole operation with one command:
programmers’ productivity increases. During program execution, procedural code
gives the database no options; it must run the code as it has been written. With
SQL, the programmer states what he or she wants to do but not how to do it: the
database has the freedom to work out how best to carry out the operation. This will
usually give better results.
Where SQL fails to provide a complete solution is that it is purely a data access
language. Most applications will need procedural constructs, such as flow control:
conditional branching and iteration. They will also usually need screen control,
user interface facilities, and variables. SQL has none of these. SQL is a set-oriented
language capable of nothing other than data access. For application development,
one will therefore need a procedural language that can invoke SQL calls. It is
therefore necessary for SQL to work with a procedural language.
Consider an application that prompts a user for a name, retrieves all the people
with that name from a table, prompts the user to choose one of them, and then
deletes the chosen person. The procedural language will draw a screen and generate
a prompt for a name. The user will enter the name. The procedural language will
construct a SQL SELECT statement using the name and submit the statement
through a database session to the database server for execution. The server will
return a set of rows (all the people with that name) to the procedural language,
which will format the set for display to the user and prompt him to choose one (or
more) of them. The identifier for the chosen person (or people) will then be used to
construct a SQL DELETE statement for the server to execute. If the identifier is a
unique identifier (the primary key) then the set of rows to be deleted will be a set of
just one row; if the identifier is nonunique, then the set selected for deletion would
be larger. The procedural code will know nothing about the likely size of the sets
retrieved or deleted.

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CERTIFICATION OBJECTIVE 1.04

Use the Client Tools
There are numerous tools that can be used to connect to an Oracle database. Two
of the most basic are SQL*Plus and SQL Developer. These are provided by Oracle
Corporation and are perfectly adequate for much of the work that a developer or
a database administrator needs to do. The choice between them is partly a matter
of personal preference, partly to do with the environment, and partly to do with
functionality. SQL Developer undoubtedly offers far more functionality that
SQL*Plus, but it is more demanding in that it needs a graphical terminal, whereas
SQL*Plus can be used on character mode devices.
Oracle Corporation has a history of producing simple tools for interacting with a
database that users disliked and which therefore fell into disuse. Most recently, these
include the iSQL*Plus tool which was introduced with release 9i and dropped with
release 11g. The tool that has lasted longest is SQL*Plus, and even though Oracle
Corporation is promoting SQL Developer very strongly as a replacement, all people
working in the Oracle environment will be well advised to become familiar with it.
Many experienced developers and database administrators (perhaps
including the authors of this book) treat the newer tools with a certain
degree of skepticism—though this may be nothing more than an indication
that these people are somewhat old-fashioned. Throughout this book both
tools will be used.

SQL*Plus
SQL*Plus is a client-server tool for connecting to a database and issuing ad hoc SQL
commands. It can also be used for creating PL/SQL code and has facilities for formatting
results. It is available on all platforms to which the database has been ported—the
sections that follow give some detail on using SQL*Plus on Linux and Windows. There
are no significant differences with using SQL*Plus on any other platform.
In terms of architecture, SQL*Plus is a user process written in C. It establishes
a session against an instance and a database over the Oracle Net protocol. The
platforms for the client and the server can be different. For example, there is no
reason not to use SQL*Plus on a Windows PC to connect to a database running on
a mainframe (or the other way round) provided that Oracle Net has been configured
to make the connection.

Use the Client Tools

27

SQL*Plus on Linux
The SQL*Plus executable file on a Linux installation is sqlplus. The location of
this file will be installation specific but will typically be something like:
/u01/app/oracle/product/db_1/bin/sqlplus

Your Linux account should be set up appropriately to run SQL*Plus. There are
some environment variables that will need to be set. These are
ORACLE_HOME
PATH
LD_LIBRARY_PATH
The ORACLE_HOME variable points to the Oracle Home. An Oracle Home is the
Oracle software installation: the set of files and directories containing the executable
code and some of the configuration files. The PATH must include the bin directory in
the Oracle Home. The LD_LIBRARY_PATH should include the lib directory in the
Oracle Home, but in practice you may get away without setting this. Figure 1-5 shows
a Linux terminal window and some tests to see if the environment is correct.
In Figure 1-5, first the echo command checks whether the three variables have
been set up correctly: there is an ORACLE_HOME, and the bin and lib directories
in it have been set as the first element of the PATH and LD_LIBRARY_PATH
variables. Then which confirms that the SQL*Plus executable file really is available,

FIGURE 1-5

Checking the
Linux session
setup

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in the PATH. Finally, SQL*Plus is launched with a username, a password, and
a connect identifier passed to it on the command line. If the tests do not return
acceptable results and SQL*Plus fails to launch, this should be discussed with your
system administrator and your database administrator. Some common errors with the
logon itself are described in the section “Creating and Testing a Database Connection”
later in this chapter.
The format of the login string is the database username followed by a forward slash
character as a delimiter, then a password followed by an @ symbol as a delimiter, and
finally an Oracle Net connect identifier. In this example, the username is system,
whose password is oracle, and the database is identified by orcl.
Following the logon, the next lines of text display the version of SQL*Plus being
used, which is 11.1.0.6.0, the version of the database to which the connection has
been made (which happens to be the same as the version of the SQL*Plus tool),
and which options have been installed within the database. The last line is the
prompt to the user, SQL> , at which point the user can enter any SQL*Plus or
SQL command. If the login does not succeed with whatever username (probably
not system) you have been allocated, this should be discussed with your database
administrator.

SQL*Plus on Windows
Historically, there were always two versions of SQL*Plus for Microsoft Windows: the
character version and the graphical version. The character version is the executable
file sqlplus.exe, and the graphical version was sqlplusw.exe. With the
current release the graphical version no longer exists, but many developers will
prefer to use it and the versions shipped with earlier releases are perfectly good tools
for working with an 11g database. There are no problems with mixing versions: an
11g SQL*Plus client can connect to a 10g database, and a 10g SQL*Plus client can
connect to an 11g database. Following a default installation of either the Oracle
database or just the Oracle client on Windows, SQL*Plus will be available as a
shortcut on the Windows Start menu. The navigation path will be as follows:
1. Start
2. Programs
3. Oracle—OraDB11g_home1
4. Application Development
5. SQL Plus

Use the Client Tools

29

Note that the third part of the navigation path may vary depending on the
installation. The location of the executable file launched by the shortcut will,
typically, be something like the following:
D:\oracle\app\product\11.1.0\db_2\BIN\sqlplus.exe

However, the exact path will be installation specific. Figure 1-6 shows a logon to
a database with SQL*Plus, launched from the shortcut. The first line of text shows
the version of SQL*Plus, which is the 11.1.0.4.0 beta release, and the time the
program was launched. The third line of text is a logon prompt:
Enter user-name:

followed by the logon string entered manually, which was
system/oracle@orcl

A change some people like to make to the shortcut that launches SQL*Plus is to
prevent it from immediately presenting a login prompt. To do this, add the NOLOG
switch to the end of the command:
sqlplus /nolog

There is no reason not to launch SQL*Plus from an operating system prompt
rather than from the Start menu shortcut: simply open a command window and run
it. The program will immediately prompt for a logon, unless you invoke it with the
NOLOG switch described above.

FIGURE 1-6

A database logon
with SQL*Plus for
Windows

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The tests of the environment and the need to set the variables if they are not
correct, previously described for a Linux installation, are not usually necessary on a
Windows installation. This is because the variables are set in the Windows Registry
by the Oracle Universal Installer when the software is installed. If SQL*Plus does
not launch successfully, check the Registry variables. Figure 1-7 shows the relevant
section of the Registry, viewed with the Windows regedit.exe Registry Editor
utility. Within the Registry Editor, navigate to the key:
HKEY_LOCAL_MACHINE
SOFTWARE
ORACLE
KEY_OraDb11g_home1
The final element of this navigation path will have a different name if there have
been several 11g installations on the machine.
Note the values of the Registry variables ORACLE_HOME and ORACLE_
HOME_NAME. These will relate to location of the sqlplus.exe executable and
the Start menu navigation path to reach the shortcut that will launch it.

Creating and Testing a Database Connection
SQL*Plus does not have any way of storing database connection details. Each time
a user wishes to connect to a database, the user must tell SQL*Plus who they are
and where the database is. There are variations depending on site-specific security
FIGURE 1-7

The Oracle
Registry variables

Use the Client Tools

31

facilities, but the most common means of identifying oneself to the database is
by presenting a username and a password. There are two commonly used forms
of connect identifier for identifying the database: either by giving an alias that is
resolved into the full connect details, or by entering the full details.
From an operating system prompt, these commands will launch SQL*Plus and
connect as database user SCOTT whose password is TIGER using each technique:
sqlplus scott/tiger@orcl
sqlplus scott/tiger@linsrv1.bplc.co.za:1521/orcl.bplc.com

The first example uses an alias, orcl, to identify the database. This must be resolved
into the full connect details. This resolution can be done in a number of ways, but
one way or another it must be accomplished. The usual techniques for this are to
use a locally stored text file called the tnsnames.ora file, or to contact an LDAP
directory such as Microsoft’s Active Directory or Oracle’s Oracle Internet Directory.
The second example provides all the connect details in line. The connect
details needed are the hostname of the computer on which the database instance is
running; the TCP port on which the Oracle Net database listener can be contacted;
and the database service to which the user wishes the database listener to connect
him. The first technique, where the user need only enter an alias, requires the
database administrator to configure a name resolution mechanism; the second
technique can only work if the user knows the details himself.
There are a number of circumstances that will cause a SQL*Plus connection
attempt to fail. Figure 1-8 illustrates some of the more common problems.
FIGURE 1-8

Some common
logon problems

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First, the user launches SQL*Plus from a Windows operating system prompt, using
the NOLOG switch to prevent the immediate login prompt. No problem so far.
Second, from the SQL> prompt, the user issues a connection request, which fails
with a well known error:
ORA-1254: TNS: could not resolve the connect identifier specified

This error is because the connect identifier given, wrongalias, cannot be resolved
into database connection details by the TNS (Transparent Network Substrate—not an
acronym particularly worth remembering) layer of Oracle Net. The name resolution
method to be used and its configuration is a matter for the database administrator. In
this case, the error is obvious: the user entered the wrong connect identifier.
The second connect attempt gives the correct identifier, orcl. This fails with
ORA-12541: TNS:no listener

This indicates that the connect identifier has resolved correctly into the address
of a database listener, but that the listener is not actually running. Note that another
possibility would be that the address resolution is faulty and is sending SQL*Plus
to the wrong address. Following this error, the user should contact the database
administrator and ask him or her to start the listener. Then try again.
The third connect request fails with
ORA-12514: TNS:listener does not currently know of service
requested in connect descriptor

This error is generated by the database listener. SQL*Plus has found the listener
with no problems, but the listener cannot make the onward connection to the database
service. The most likely reason for this is that the database instance has not been
started, so the user should ask the database administrator to start it and then try again.
The fourth connect request fails with
ORA-01017: invalid username/password; logon denied

To receive this message, the user must have contacted the database. The user has
got through all the possible network problems, the database instance is running, and
the database itself has been opened by the instance. The user just has the password
or username wrong. Note that the message does not state whether it is the password
or the username that is wrong: if it were to do so, it would be giving out information
to the effect that the other one was right.
Finally, the fifth connect attempt succeeds.
The preceding example demonstrates a problem-solving technique you will
use frequently. If something fails, work through what it is doing step by step.
Read every error message.

Use the Client Tools

33

SQL Developer
SQL Developer is a tool for connecting to an Oracle database (or, in fact, some
non-Oracle databases too) and issuing ad hoc SQL commands. It can also manage
PL/SQL objects. Unlike SQL*Plus, it is a graphical tool with wizards for commonly
needed actions. SQL Developer is written in Java and requires a Java Runtime
Environment (JRE) to run.
Being written in Java, SQL Developer is available on all platforms that support the
appropriate version of the JRE. There are no significant differences between platforms.

Installing and Launching SQL Developer
SQL Developer is not installed with the Oracle Universal Installer, which is used
to install all other Oracle products. It does not exist in an Oracle Home but is a
completely self contained product. The latest version can be downloaded from
Oracle Corporation’s website.
An installation of the 11g database will include a copy of SQL Developer, but
it will not be the current version. Even if you happen to have an installation
of the database, you will usually want to install the current version of SQQ
Developer as well.
To install SQL Developer, unzip the ZIP file. That’s all. It does require JDK1.5,
the Java Runtime Environment release 1.5, to be available: this comes from
Sun Microsystems. But if JDK1.5 (or a later release) is not already available on
the machine being used, there are downloadable versions of SQL Developer for
Windows that include it. For platforms other than Windows, the JDK1.5 must be
preinstalled. Download it from Sun Microsystem’s website and install according to
the platform-specific directions. To check that the JDK is available with the correct
version, from an operating system prompt run the following command:
java –version

This should return something like the following:
java version "1.5.0_13"
Java(TM) 2 Runtime Environment, Standard Edition (build 1.5.0_13-b05)
Java HotSpot(TM) Client VM (build 1.5.0_13-b05, mixed mode, sharing)

If it does not, using which java may help identify the problem: the search path
could be locating an incorrect version.
Once SQL Developer has been unzipped, change your current directory to the
directory in which SQL Developer was unzipped and launch it. On Windows, the

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executable file is sqldeveloper.exe. On Linux, it is the sqldeveloper.sh
shell script. Remember to check that the DISPLAY environment variable has been
set to a suitable value (such as 127.0.0.1:0.0, if SQL Developer is being run on the
system console) before running the shell script.
Any problems with installing the JRE and launching SQL Developer should be
referred to your system administrator.

The SQL Developer User Interface
Figure 1-9 shows the SQL Developer User Interface after connecting to a database.
The general layout of the SQL Developer window is a left pane for navigation
around objects, and a right pane to display and enter information.
In the figure, the left-hand pane shows that a connection has been made to a
database. The connection is called orcl_sys. This name is just a label chosen when
the connection was defined, but most developers will use some sort of naming
convention—in this case, the name chosen is the database identifier, which is orcl,
and the name of the user the connection was made as, which was sys. The branches
beneath list all the possible object types that can be managed. Expanding the
branches would list the objects themselves. The right-hand pane has an upper part
prompting the user to enter a SQL statement and a lower part that will display the
FIGURE 1-9

The SQL
Developer User
Interface

Use the Client Tools

35

result of the statement. The layout of the panes and the tabs visible on them are
highly customizable.
The menu buttons across the top menu bar give access to standard facilities:
A normal Windows-like file menu, from which one can save work and
exit from the tool.

■ File

A normal Windows-like edit menu, from which one can undo, redo,
copy, paste, find, and so on.

■ Edit

■ View

The options for customizing the SQL Developer user interface.

Facilities for moving between panes and for moving around code
that is being edited.

■ Navigate

Forces execution of the SQL statements, SQL script, or PL/SQL block
that is being worked on.

■ Run

Rather than running a whole block of code, steps through it line by
line with breakpoints.

■ Debug

Options for use when writing SQL and PL/SQL code, such as
keyword completion and automatic indenting.

■ Source

Links to external programs, including SQL*Plus.

■ Tools

Tools for converting applications designed for third-party databases
(Microsoft Access, SQL Server, and MySQL) to the Oracle environment.

■ Migrate
■ Help

It’s pretty good.

SQL Developer can be a very useful tool, and it is very customizable. Experiment
with it, read the Help, and set up the user interface the way that works best for you.

Creating a Database Connection
Database connections can be created and saved for reuse. Figure 1-10 shows the
window where connections can be defined. To reach this window, click the “+”
symbol visible on the Connections tab shown previously in Figure 1-9.
The name for the connection is arbitrary. In this example, the name chosen is the
name of the database connect identifier (orcl) suffixed with the username (hr) that
will be embedded in the connection.
The username and password must both be supplied, but only the username will be
saved unless the Save Password check box is selected. Saving a password means that
future connections can be made without any password prompt. This is convenient
but highly dangerous if there is any possibility that the computer you are working on
is not secure. In effect, you are delegating the authentication to your local operating
system: if you can log on to that, you can log on to the database.

36

Chapter 1:

Oracle Server Technologies and the Relational Paradigm

FIGURE 1-10

How to define
a new database
connection

Assuming that you are using SQL Developer to connect to an Oracle database
rather than to third-party database, select the Oracle tab.
The Role drop-down box gives you the option to connect as sysdba. A sysdba
connection is required before certain particularly serious operations (such as
database startup and shutdown) can be carried out. It will never be needed for the
exercises covered in this book.
The Connection Type radio buttons let you choose between three options:
This prompts for the machine name of the database server, the port
on which the database listener will accept connection requests, and the
instance (the SID) or the service to which the connection will be made.

■ Basic

If a name resolution method has been configured, then an alias for the
database can be entered, rather than the full details needed by the Basic option.

■ TNS

This allows entry of a full JDBC (Java Database Connectivity)
connect string. This is completely Oracle independent and could be used to
connect to any database that conforms to the JDBC standard.

■ Advanced

Selecting Basic requires the user to know how to connect to the database;
selecting TNS requires some configuration to have been done on the client machine
by the database administrator, in order that the alias can be resolved into the full
connection details.

Create the Demonstration Schemas

37

SCENARIO & SOLUTION
If a connection fails what can you
do? How should you try to fix the
problem?

If a connection attempt fails, there should be some sort of error
message. Read it! If it isn’t immediately self explanatory, look it up in
the Oracle documentation and on Metalink. Try to follow the flow
of a connection: from the user process to the database listener to the
instance to the database, and check each step.

If you can’t fix the problem
yourself, where can you go for help?

The documentation and Metalink will get you a long way. Most
support groups will refuse to talk to you unless you can show that you
have tried to solve the problem yourself. Then talk to your network
administrators, system administrators, and database administrators.
Often, a large part of solving the problem is finding out whose
responsibility it is: be wary of people trying to avoid responsibility but
don’t dump problems on people inappropriately.

After you enter the details, the Test button will force SQL Developer to attempt a
logon. If this returns an error, then either the connection details are wrong, or there
is a problem on the server side. Typical server-side problems are that the database
listener is not running, or that the database has not been started. Whatever the
error is, it will be prefixed with an error number—some of the common errors were
described in the preceding section, which described the use of SQL*Plus.

CERTIFICATION OBJECTIVE 1.05

Create the Demonstration Schemas
Throughout this book, there are hundreds of examples of running SQL code against
tables of data. For the most part, the examples use tables in two demonstration
schemas provided by Oracle: the HR schema, which is sample data that simulates
a simple human resources application, and the OE schema, which simulates a more
complicated order entry application.
These schemas can be created when the database is created; it is an option
presented by the Database Configuration Assistant. If they do not exist, they can be
created later by running some scripts that will exist in the database Oracle Home.
An earlier demonstration schema was SCOTT (password TIGER).This schema
is simpler than HR or OE. Many people with long experience of Oracle will
prefer to use this.The creation script is still supplied, it is utlsampl.sql.

38

Chapter 1:

Oracle Server Technologies and the Relational Paradigm

Users and Schemas
First, two definitions. In Oracle parlance, a database user is a person who can log
on to the database. A database schema is all the objects in the database owned by
one user. The two terms can often be used interchangeably, as there is a one-to-one
relationship between users and schemas. Note that while there is in fact a CREATE
SCHEMA command, this does not actually create a schema—it is only a quick way
of creating objects in a schema. A schema is initially created empty, when a user is
created with the CREATE USER command.
Schemas are used for storing objects. These may be data objects such as tables
or programmatic objects such as PL/SQL stored procedures. User logons are used
to connect to the database and access these objects. By default, users have access
to the objects in their own schema and to no others, but most applications change
this. Typically, one schema may be used for storing data that is accessed by other
users who have been given permission to use the objects, even though they do not
own them. In practice, very few users will ever have objects in their own schema,
or permission to create them: they will have access rights (which will be strictly
controlled) only to objects in another schema. These objects will be used by all
users who run the application whose data that schema stores. Conversely, the users
who own the data-storing schemas may never in fact log on: the only purpose of
their schemas is to contain data used by others.
It is impossible for a data object to exist independently of a schema. Or in other
words, all tables must have an owner. The owner is the user in whose schema the
table resides. The unique identifier for a table (or any other schema object) is the
username, followed by the object name. It follows that it is not possible for two
tables with the same name to exist in the same schema, but that two tables with the
same name (though possibly different structures or contents) can exist in different
schemas. If an object does not exist in one’s own schema, to access it one must
qualify its name with the name of the schema in which it resides. For example,
HR.EMPLOYEES is the table called EMPLOYEES in user HR’s schema. Only a
user connected as HR could get to the table by referring to EMPLOYEES without a
schema name qualifier.

The HR and OE Schemas
The HR demonstration schema consists of seven tables, linked by primary key to
foreign key relationships. Figure 1-11 illustrates the relationships between the tables,
as an entity relationship diagram.

Create the Demonstration Schemas

FIGURE 1-11

39

REGIONS

The HR entity
relationship
diagram

COUNTRIES

LOCATIONS

DEPARTMENTS

JOB_HISTORY

EMPLOYEES

JOBS

Two of the relationships shown in Figure 1-11 may not be immediately
comprehensible. First, there is a many-to-one relationship from EMPLOYEES to
EMPLOYEES. This is what is known as a self-referencing foreign key. This means
that many employees can be connected to one employee, and it’s based on the fact
that many employees may have one manager, but the manager is also an employee.
The relationship is implemented by the column manager_id being a foreign key to
employee_id, which is the table’s primary key.
The second relationship that may require explanation is between DEPARTMENTS
and EMPLOYEES, which is bidirectional. The one department-to-many employees
relationship simply states that there may be many staff members in each department,
based on the EMPLOYEES dept_id column being a foreign key to the DEPARTMENTS
primary key dept_id column. The one employee-to-many departments relationship
shows that one employee could be the manager of several departments and is
implemented by the manager_id column in DEPARTMENTS being a foreign key to the
primary key employee_id column in EMPLOYEES.
Table 1-1 shows the columns of each table in the HR schema, using the notation
described in the earlier section “Data Normalization” to indicate primary keys (#),
foreign keys (\), and whether columns are optional (o) or mandatory (*).

40

Chapter 1:

TABLE 1-1

The tables and
columns on the
HR schema

Oracle Server Technologies and the Relational Paradigm

Table

Columns

REGIONS

#*
o

region_name

COUNTRIES

#*

country_id

o

country_name

\o

region_id

#*

location_id

o

street_address

o

postal_code

*

city

o

state_province

\o

country_id

#*

department_id

*

department_name

\o

manager_id

\o

location_id

#*

employee_id

o

first_name

*

last_name

*

e-mail

o

phone_number

*

hire_date

\*

job_id

o

salary

o

commission_pct

\o

manager_id

\o

department_id

#*

job_id

*

job_title

o

min_salary

o

max_salary

LOCATIONS

DEPARTMENTS

EMPLOYEES

JOBS

region_id

Create the Demonstration Schemas

TABLE 1-1

The tables and
columns on the
HR schema (cntd.)

Table

Columns

JOB_HISTORY

#*

employee_id

#*

start_date

*

end_date

\*

job_id

\o

department_id

41

The tables are:
■ REGIONS has rows for major geographical areas.
■ COUNTRIES has rows for each country, which are optionally assigned to a

region.
■ LOCATIONS includes individual addresses, which are optionally assigned to

a country.
■ DEPARTMENTS has a row for each department, optionally assigned to a

location and optionally with a manager (who must exist as an employee).
■ EMPLOYEES has a row for every employee, each of whom must be assigned

to a job and optionally to a department and to a manager. The managers must
themselves be employees.
■ JOBS lists all possible jobs in the organization. It is possible for many

employees to have the same job.
■ JOB_HISTORY lists previous jobs held by employees, uniquely identified by

employee_id and start_date; it is not possible for an employee to hold two jobs
concurrently. Each job history record will refer to one employee, who will have
had one job at that time and may have been a member of one department.
This HR schema is used for most of the exercises and many of the examples
embedded in the chapters of this book and does need to be available.
There are rows in EMPLOYEES that do not have a matching parent row in
DEPARTMENTS.This could be by design but might well be a design mistake
that is possible because the DEPARTMENT_ID column in EMPLOYEES is not
mandatory.There are similar possible errors in the REGIONS—COUNTRIES—
LOCATIONS hierarchy, which really does not make a lot of sense.

42

Chapter 1:

Oracle Server Technologies and the Relational Paradigm

The OE schema is considerably more complex than the HR schema. The table
structures are much more complicated: they include columns defined as nested
tables, user-defined data types, and XML data types. There are a number of optional
exercises at the end of each chapter that are usually based on the OE schema. The
objects referred to are described as they are used.

Demonstration Schema Creation
If the database you are using was created specifically for studying for the OCP SQL
examination, the demonstration schemas should have been created already. They
are an option presented by the Database Configuration Assistant when it creates a
database. After database creation, the schemas may have to be unlocked and their
passwords set; by default the accounts are locked, which means you cannot log on to
them. These commands, which could be issued from SQL*Plus or SQL Developer,
will make it possible to log on as users HR and OE using the passwords HR and OE:
alter user hr account unlock identified by hr;
alter user oe account unlock identified by oe;

These alter user commands can only be issued when connected to the database
as a user with DBA privileges, such as the user SYSTEM.
If the schemas were not created at database creation time, they can be created
later by running scripts installed into the Oracle Home of the database. These
scripts will need to be run from SQL*Plus or SQL Developer as a user with SYSDBA
privileges. The script will prompt for certain values as it runs. For example, on
Linux, first launch SQL*Plus from an operating system prompt:
sqlplus / as sysdba

There are various options for this connection, but the preceding syntax will usually
work if the database is running on the same machine where you are running
SQL*Plus. Then invoke the script from the SQL> prompt:
SQL> @?/demo/schema/human_resources/hr_main.sql

The “?” character is a variable that SQL*Plus will expand into the path to the
Oracle Home directory. The script will prompt for HR’s password, default tablespace,
and temporary tablespace; the SYS password; and a destination for a logfile of the
script’s running. Typical values for the default tablespace and temporary tablespace
are USERS and TEMP, but these will have to have been created already. After
completion, you will be connected to the database as the new HR user. To verify
this, run this statements.
SQL> show user;

Certification Summary

43

You will see that you are currently connected as HR; then run:
SQL> select table_name from user_tables;

You will see a list of the seven tables in the HR schema.
To create the OE schema, follow the same process, nominating the script:
?/demo/schema/order_entry/oe_main.sql

The process for creating the schemas on Windows is identical, except for the path
delimiters—where most operating systems use forward slashes, Windows uses back
slashes. So the path to the Windows HR creation script becomes:
@?\demo\schema\human_resources\hr_main.sql

Note that running these schema creation scripts will drop the schemas first if
they already exist. Dropping a schema means removing every item in it and then
removing the user. This should not be a problem, unless the schema has been
used for some development work that needs to be kept. The drop is absolutely
nonreversible, so any object in a dropped schema will be irretrievable lost.
If the demonstration creation schema scripts do not exist as just described, this
will be because the Oracle Home installation has not been completed. Installing a
database Oracle Home can be done from one CD, but there is a second CD (called
the companion CD) that includes a number of components that are, strictly speaking,
optional. These include the demonstration schemas. The companion CD should
normally be installed; if this has not been done, this must be discussed with the
database administrator.
The demonstration schemas should not exist in production databases. It is not
good, for security reasons, to have unnecessary schemas in a database that
have well known usernames, capabilities, and (possibly) passwords.

CERTIFICATION SUMMARY
SQL is a language for managing access to normalized data stored in relational
databases. It is not an application development language, but is invoked by such
languages when they need to access data. The Oracle server technologies provide a
platform for developing and deploying such applications. The combination of the
Oracle server technologies and SQL result in an environment conforming to the
relational database paradigm that is an enabling technology for Grid computing.

44

Chapter 1:

Oracle Server Technologies and the Relational Paradigm

Numerous client tools can be used to connect to an Oracle database. Two
provided by Oracle Corporation are SQL*Plus and SQL Developer: SQL*Plus
is installed as part of every Oracle client and Oracle database install, but SQL
Developer can be installed as a separate product. Both tools can be used for
preparing for the OCP examinations, and students should be familiar with both.
The demonstration schemas store example data that is used to illustrate the use of
SQL, and also of more advanced Oracle development facilities.

Two-Minute Drill

✓

45

TWO-MINUTE DRILL
Position the Server Technologies
❑ The Oracle database stores and manages access to user data.
❑ The Oracle Application Server runs applications that connect users to the

database.
❑ Oracle Enterprise Manager is a tool for managing databases, application servers,

and, if desired, the entire computing environment.
❑ Languages built into the database for application development are SQL,

PL/SQL, and Java.

Understand Relational Structures
❑ Data must be normalized into two-dimensional tables.
❑ Tables are linked through primary and foreign keys.
❑ Entity-relationship diagrams represent the tables graphically.

Summarize the SQL Language
❑ The DML commands are SELECT, INSERT, UPDATE, DELETE, and MERGE.
❑ The DDL commands are CREATE, ALTER, DROP, RENAME, TRUNCATE,

and COMMENT.
❑ The DCL commands are GRANT and REVOKE.
❑ The TCL commands are COMMIT, ROLLBACK, and SAVEPOINT.

Use the Client Tools
❑ SQL*Plus is a command-line utility installed into the Oracle Home.
❑ SQL Developer is a graphical tool installed into its own directory.
❑ Both tools require a database connection, consisting of a username, a password,

and a connect identifier.

Create the Demonstration Schemas
❑ The demonstration schemas are provided by Oracle to facilitate learning but

must be created before they can be used.

46

Chapter 1:

Oracle Server Technologies and the Relational Paradigm

SELF TEST
Position the Server Technologies
1. What components of the IT environment can Oracle Enterprise Manager Grid Control
manage? (Choose the best answer.)
A. Oracle databases
B. Oracle application servers
C. Third-party products
D. The server machines
E. All of the above
2. What languages can run within the database? (Choose all that apply.)
A. SQL
B. C
C. PL/SQL
D. Java
E. Any other language linked to the OCI libraries

Understand Relational Structures
3. Data that is modeled into a form suitable for processing in a relational database may be
described as being (Choose the best answer.)
A. First normal form
B. Third normal form
C. Abnormal form
D. Paranormal form
4. An entity-relationship diagram shows data modeled into (Choose the best answer.)
A. Two-dimensional tables
B. Multidimensional tables
C. Hierarchical structures
D. Object-oriented structures

Self Test

47

Summarize the SQL Language
5. SQL is a set-oriented language. Which of these features is a consequence of this? (Choose the
best answer.)
A. Individual rows must have a unique identifier.
B. Sets of users can be managed in groups.
C. SQL statements can be placed within blocks of code in other languages, such as Java and
PL/SQL.
D. One statement can affect multiple rows.
6. Which of these constructs is not part of the SQL language? (Choose all that apply.)
A. Iteration, based on DO..WHILE
B. Iteration, based on FOR..NEXT
C. Branching, based on IF..THEN..ELSE
D. Transaction control, based on COMMIT
E. Transaction control, based on ROLLBACK

Use the Client Tools
7. Which of these statements regarding SQL Developer are correct? (Choose two answers.)
A. SQL Developer cannot connect to databases earlier than release 10g.
B. SQL Developer can be installed outside an Oracle Home.
C. SQL Developer can store passwords.
D. SQL Developer relies on an LDAP directory for name resolution.
8. Which of the following are requirements for using SQL Developer? (Choose two correct
answers.)
A. A Java Runtime Environment
B. The OCI libraries
C. A name resolution method such as LDAP or a TNSNAMES.ORA file
D. The SQL*Plus libraries
E. A graphical terminal

48

Chapter 1:

Oracle Server Technologies and the Relational Paradigm

Create the Demonstration Schemas
9. Where may the demonstration schemas be created? (Choose the best answer.)
A. The demonstration schemas must be created in a demonstration database.
B. The demonstration schemas cannot be created in a production database.
C. The demonstration schemas can be created in any database.
D. The demonstration schemas can be created in any database if the demonstration user is
created first.
10. How can you move a schema from one user to another? (Choose the best answer.)
A. Use the ALTER SCHEMA MOVE… command.
B. You cannot move a schema from one user to another.
C. A schema can only be moved if it is empty (or if all objects within it have been dropped).
D. Attach the new user to the schema, then detach the old user from the schema.

LAB QUESTION
The OE schema includes these tables:
■

CUSTOMERS

■

INVENTORIES

■

ORDERS

■

ORDER_ITEMS

■

PRODUCT_DESCRIPTIONS

■

PRODUCT_INFORMATION

■

WAREHOUSES

A CUSTOMER can place many ORDERS, and an order can have many ORDER_ITEMS. Each
item will be of one product, described by its PRODUCT_INFORMATION, and each product may
have several PRODUCT_DESCRIPTIONS, in different languages. There are a number of WAREHOUSES, each of which can store many products; one product may be stored in many warehouses.
An INVENTORIES entry relates products to warehouses, showing how much of each product is in
each warehouse.
Sketch out this schema as an entity-relationship diagram, showing the many-to-one connections
between the tables and ensuring that there are no many-to-many connections.

Self Test Answers

49

SELF TEST ANSWERS
Position the Server Technologies
✓
1. ®
®
˚
✓
2. ®
®
˚

E. Grid Control can manage the complete environment (according to Oracle Corporation).
A, B, C, D. All of these can be managed by Grid Control.
A, C, D. SQL, PL/SQL, and Java can all run in the database.
B, E. C cannot run inside the database, and OCI is used by external processes to connect to
the database; it does not run within it.

Understand Relational Structures
✓ B. Third normal form is the usual form aimed for by systems analysts when they normalize
3. ®
data into relational structures.
®
˚ A, C, D. A is wrong because first normal form is only the first stage of data normalization.
C and D would be more suitable to the X-Files than to a database.
✓ A. The relational model uses two-dimensional tables.
4. ®
®
˚ B, C, D. B is wrong because two dimensions is the limit for relational structures. C and D
are wrong because they refer to nonrelational structures (though there are facilities within the
Oracle database for simulating them).

Summarize the SQL Language
✓ D. In a set-oriented language, one command can affect many rows (a set), whereas a
5. ®
procedural language processes rows one by one.
®
˚ A, B, C. A is wrong because while rows should have a unique a identifier in a well designed
application, this is not actually a requirement. B is wrong because users cannot be grouped in
the Oracle environment. C is wrong because (even though the statement is correct) it is not
relevant to the question.
✓ A, B, C. These are all procedural constructions, which are not part of a set-oriented
6. ®
language. They are all used in PL/SQL.
®
˚ D, E. These are SQL’s transaction control statements.

Use the Client Tools
✓ B, C. B is correct because SQL Developer can be installed in its own directory. C is correct
7. ®
because passwords can be saved as part of a connection definition (though this may not be a
good idea).
®
˚ A, D. A is wrong because the Oracle Net protocol lets SQL Developer connect to a
number of versions of the database. D is wrong because LDAP is only one of several techniques
for name resolution.

50

Chapter 1:

Oracle Server Technologies and the Relational Paradigm

✓ A, E. A is correct because SQL Developer is written in Java and therefore requires a Java
8. ®
Runtime Environment. E is correct because SQL Developer needs a graphics terminal to display
windows.
®
˚ B, C, D. B is wrong because SQL Developer uses JDBC to connect to databases, not
OCI. C is wrong because, while SQL Developer can use LDAP or a TNSNAMES.ORA file,
it can also use and store the basic connection details. D is wrong because SQL Developer is a
completely independent product.

Create the Demonstration Schemas
✓ C. The demonstration schemas can be created in any database, either at database creation
9. ®
time or by running scripts later.
®
˚ A, B, D. A and B are wrong because, while they may be good practice, they are not a
technical requirement. D is wrong because it fails to understand that a schema can only be (and
always is) created with a user.
✓ B. A schema and a user are inseparable.
10. ®
®
˚ A, C, D. A is wrong because there is no such command. C and D are wrong because they
assume the impossible: that you can separate a user from his or her schema.

LAB ANSWER
Figure 1-12 shows a solution.
FIGURE 1-12

An entityrelationship
diagram
describing the OE
schema

CUSTOMERS

ORDERS

ORDER_ITEMS

WAREHOUSES

INVENTORIES

PRODUCT_
INFORMATION

PRODUCT_
DESCRIPTIONS

2
Data Retrieval Using
the SQL SELECT
Statement

CERTIFICATION OBJECTIVES
2.01

List the Capabilities of SQL SELECT
Statements

2.02

Execute a Basic SELECT Statement

✓
Q&A

Two-Minute Drill
Self Test

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

52

Chapter 2:

Data Retrieval Using the SQL SELECT Statement

T

his chapter explores the concepts of extracting or retrieving data stored in relational
tables using the SELECT statement. The statement is introduced in its basic form
and is progressively built on to extend its core functionality. As you learn the rules
governing this statement, an important point to remember is that the SELECT statement never
alters information stored in the database. Instead, it provides a read-only method of extracting
information. When you ask a question using SQL SELECT, you are guaranteed to extract results,
even from the difficult questions.

CERTIFICATION OBJECTIVE 2.01

List the Capabilities of SQL SELECT Statements
Knowing how to retrieve data in a set format using a query language is the first
step toward understanding the capabilities of SELECT statements. Describing the
relations involved provides a tangible link between the theory of how data is stored
in tables and the practical visualization of the structure of these tables. These topics
form an important precursor to the discussion of the capabilities of the SELECT
statement. The three primary areas explored are as follows:
■ Introducing the SQL SELECT statement
■ The DESCRIBE table command
■ Capabilities of the SELECT statement

Introducing the SQL SELECT Statement
The SELECT statement from Structured Query Language (SQL) has to be the single
most powerful nonspoken language construct. The SELECT statement is an elegant,
flexible, and highly extensible mechanism created to retrieve information from a
database table. A database would serve little purpose if it could not be queried to
answer all sorts of interesting questions. For example, you may have a database that
contains personal financial records like your bank statements, your utility bills, and
your salary statements. You could easily ask the database for a date-ordered list of
your electrical utility bills for the last six months or query your bank statement for
a list of payments made to a certain account over the same period. The beauty of
the SELECT statement is encapsulated in its simple English-like format that allows
questions to be asked of the database in a natural manner.

List the Capabilities of SQL SELECT Statements

53

Tables, also known as relations, consist of rows of information divided by
columns. Consider two of the sample tables introduced in the previous chapter: the
EMPLOYEES table and the DEPARTMENTS table. This sample dataset is based on
the Human Resources (HR) information for some fictitious organization. In Oracle
terminology, each table belongs to a schema (owner): in this case the HR schema.
The EMPLOYEES table stores rows or records of information. These contain
several attributes (columns) that describe each employee in this organization. The
DEPARTMENTS table contains descriptive information about each department
within this organization, stored as rows of data divided into columns.
Assuming a connection to a database containing the sample HR schema is
available, then using either SQL*Plus or SQL Developer you can establish a user
session. Once connected to the database, you are ready to begin your tour of SQL.

The DESCRIBE Table Command
To get the answers one seeks, one must ask the correct questions. An understanding
of the terms of reference, which in this case, are relational tables, is essential for the
formulation of the correct questions. A structural description of a table is useful to
establish what questions can be asked of it. The Oracle server stores information
about all tables in a special set of relational tables called the data dictionary, in
order to manage them. The data dictionary is quite similar to a regular language
dictionary. It stores definitions of database objects in a centralized, ordered, and
structured format.
A clear distinction must be drawn between storing the definition and the
contents of a table. The definition of a table includes information like table name,
table owner, details about the columns that comprise it, and its physical storage size
on disk. This information is also referred to as metadata. The contents of a table are
stored in rows and are referred to as data.
The structural metadata of a table may be obtained by querying the database for
the list of columns that comprise it using the DESCRIBE command. The general
form of the syntax for this command is intuitively:
DESC[RIBE] .tablename
This command shall be systematically unpacked. The DESCRIBE keyword can
be shortened to DESC. All tables belong to a schema or owner. If you are describing
a table that belongs to the schema to which you have connected, the 
portion of the command may be omitted. Figure 2-1 shows how the EMPLOYEES
table is described from SQL*Plus after connecting to the database as the HR user
with the DESCRIBE EMPLOYEES command and how the DEPARTMENTS table

54

Chapter 2:

Data Retrieval Using the SQL SELECT Statement

FIGURE 2-1

Describing
EMPLOYEES,
DEPARTMENTS,
and DUAL tables

is described using the shorthand notation: DESC HR.DEPARTMENTS. The HR.
notational prefix could be omitted since the DEPARTMENTS table belongs to the
HR schema. The HR schema (and every other schema) has access to a special table
called DUAL, which belongs to the SYS schema. This table can be structurally
described with the command: DESCRIBE SYS.DUAL.
Describing tables yields interesting and useful results. You know which columns
of a table can be selected since their names are exposed. You also know the nature
of the data contained in these columns since the column data type is exposed.
Column data types are discussed in detail in Chapter 11. For the current discussion,
it is sufficient to regard the different data type columns as detailed in this section.
Numeric columns are often specified as NUMBER(p,s), where the first parameter
is precision and the second is scale. In Figure 2-1, the SALARY column of the
EMPLOYEES table has a data type of: NUMBER(8,2). This means that the values
stored in this column can have at most 8 digits. Of these 8 digits, 2 must be to the right
of the decimal point. A SALARY value of 999999.99 is acceptable, but a SALARY
value of 9999999.9 is not, even though both these numbers contain 8 digits.
VARCHAR2(length) data type columns store variable length alphanumeric
character data, where length determines the maximum number of characters a column
can contain. The FIRST_NAME column of the EMPLOYEES table has data type
VARCHAR2(20), which means it can store employees’ names of up to 20 characters.
Note that if this column contains no data or its content is less than 20 characters,

List the Capabilities of SQL SELECT Statements

55

it will not necessarily use the same space as it would use to store a name that is
20 characters long. The CHAR(size) column data type specifies fixed-length columns
where row space is preallocated to contain a fixed number of characters regardless
of its contents. CHAR is much less commonly used than VARCHAR2. Unless the
length of the data is predictable and constant, the CHAR data type utilizes storage
inefficiently, padding any unused components with spaces.
DATE and TIMESTAMP column data types store date and time information.
DATE stores a moment in time with precision including day, month, year, hours,
minutes, and seconds. TIMESTAMP(f) stores the same information as DATE but is
also capable of storing fractional seconds.
A variety of data types is available for use as column data types. Many have
a specialized purpose like Binary Large Objects (BLOBs), used for storing
binary data like music or video data.The vast majority of tables, however, use
the primitive column data types of NUMBER, VARCHAR2, and DATE.The
TIMESTAMP data type has become widely used since its introduction in Oracle
9i. Becoming familiar and interacting with these generic primitive data types
prepares you for dealing with a significant range of database-related queries.
Mandatory columns, which are forced to store data for each row, are exposed by
the “Null?” column output from the DESCRIBE command having the value: NOT
NULL. You are guaranteed that any column of data which is restricted by the NOT
NULL constraint when the table is created must contain some data. It is important
to note that NULL has special meaning for the Oracle server. NULL refers to an
absence of data. Blank spaces do not count as NULL since they are present in the
row and have some length even though they are not visible.

EXERCISE 2-1
Describing the Human Resources Schema
The HR schema contains seven tables representing a data model of a fictitious
Human Resources department. The EMPLOYEES table, which stores details of the
staff, and the DEPARTMENTS table, which contains the details of the departments
in the organization, have been described. In this step-by-step exercise, a connection
is made using SQL Developer as the HR user and the remaining five sample tables
are described. They are the JOBS table, which keeps track of the different job types
available in the organization, and the JOB_HISTORY table, which keeps track of
the job details of employees who changed jobs but remained in the organization.
To understand the data model further, the LOCATIONS, COUNTRIES, and

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REGIONS tables, which keep track of the geographical information pertaining to
departments in the organization, will be described.
1. Launch SQL Developer and choose New from the File menu. Choose Database Connection. If this is the first time you are connecting to the database
from SQL Developer, you are required to create a connection. Provide a
descriptive connection name and input HR as the username. The remaining connection details should be obtained from your database administrator.
Once the connection is saved, choose the Connect button.
2. Navigate to the SQL Editor, which is the section titled Enter SQL Statement.
3. Type in the command: DESCRIBE JOBS. Terminating this command with a
semicolon is optional.
4. Execute the DESCRIBE command, either by pressing the F5 key or by
clicking the solid green triangular arrow icon located on the toolbar above
the SQL Editor.
5. The JOBS table description appears in the Results frame as shown in the
following illustration.

6. Steps 3 to 5 can be repeated to describe the remaining JOB_HISTORY,
LOCATIONS, COUNTRIES, and REGIONS tables.

List the Capabilities of SQL SELECT Statements

57

7. SQL Developer provides an alternative to the DESCRIBE command when it
comes to obtaining the structural information of tables.
8. Navigate to the LOCATIONS table using the Tree navigator located on the
left frame underneath the connection name.
9. SQL Developer describes the table automatically on the right side of the tool
as shown in the following illustration.

Capabilities of the SELECT Statement
Relational database tables are built on a strong mathematical foundation called
relational theory. In this theory, relations, or tables, are operated on by a formal
language called relational algebra. SQL is a commercial interpretation of the
relational algebraic constructs. Three concepts from relational theory encompass the
capability of the SELECT statement: projection, selection, and joining.
Projection refers to the restriction of attributes (columns) selected from a relation
or table. When requesting information from a table, you can ask to view all the
columns. For example, in the HR.DEPARTMENTS table, you can retrieve all

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rows and all columns with a simple SELECT statement. This query will return
DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, and LOCATION_
ID information for every department record stored in the table. What if you wanted
a list containing only the DEPARTMENT_NAME and MANAGER_ID columns?
Well, you would request just those two columns from the table. This restriction of
columns is called projection.
Selection refers to the restriction of the tuples or rows selected from a relation
(table). It is often not desirable to retrieve every row from a table. Tables may
contain many rows and, instead of asking for all of them, selection provides a means
to restrict the rows returned. Perhaps you have been asked to identify only the
employees who belong to department 30. With selection it is possible to limit the
results set to those rows of data which have a DEPARTMENT_ID value of 30.
Joining, as a relational concept, refers to the interaction of tables with each other
in a query. Third normal form, as discussed in Chapter1, presented the notion of
separating different types of data into autonomous tables to avoid duplication and
maintenance anomalies and to associate related data using primary and foreign key
relationships. These relationships provide the mechanism to join tables with each
other. Joining is discussed extensively in Chapter7.
Assume there is a need to retrieve the e-mail addresses for all employees who
work in the Sales department. The EMAIL column belongs to the EMPLOYEES
table, while the DEPARTMENT_NAME column belongs to the DEPARTMENTS
table. Projection and selection from the DEPARTMENTS table may be used to obtain
the DEPARTMENT_ID value that corresponds to the Sales department. The
matching rows in the EMPLOYEES table may be joined to the DEPARTMENTS
table based on this common DEPARTMENT_ID value. The EMAIL column may
then be projected from this set of results.
The SQL SELECT statement is mathematically governed by these three tenets.
An unlimited combination of projections, selections, and joins provides the
language to extract the relational data required.

The three concepts
of projection, selection, and joining,
which form the underlying basis for the
capabilities of the SELECT statement,
are always measured in the exam.These
concepts may be presented in a list with

other false concepts and you may be asked
to choose the correct three fundamental
concepts. Or, a list of SQL statements may
be presented, and you may be asked to
choose the statement that demonstrates
one or more of these concepts.

Execute a Basic SELECT Statement

59

CERTIFICATION OBJECTIVE 2.02

Execute a Basic SELECT Statement
The practical capabilities of the SELECT statement are realized in its execution.
The key to executing any query language statement is a thorough understanding
of its syntax and the rules governing its usage. This topic is discussed first. It is
followed by a discussion of the execution of a basic query before expressions and
operators, which exponentially increase the utility of data stored in relational tables,
are introduced. Next, the concept of a null value is demystified, as its pitfalls are
exposed. These topics will be covered in the following four sections:
■ Syntax of the primitive SELECT statement
■ Rules are meant to be followed
■ SQL expressions and operators
■ NULL is nothing

Syntax of the Primitive SELECT Statement
In its most primitive form, the SELECT statement supports the projection of
columns and the creation of arithmetic, character, and date expressions. It also
facilitates the elimination of duplicate values from the results set. The basic
SELECT statement syntax is as follows:
SELECT *|{[DISTINCT] column|expression [alias],…}
FROM table;
The special keywords or reserved words of the SELECT statement syntax appear
in uppercase. When using the commands, however, the case of the reserved words in
your query statement does not matter. The reserved words cannot be used as column
names or other database object names. SELECT, DISTINCT, and FROM are three
keyword elements. A SELECT statement always comprises two or more clauses.
The two mandatory clauses are the SELECT clause and the FROM clause. The pipe
symbol | is used to denote OR. So you can read the first form of the above SELECT
statement as:
SELECT *
FROM table;
In this format, the asterisk symbol (*) is used to denote all columns. SELECT *
is a succinct way of asking the Oracle server to return all possible columns. It is used

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as a shorthand, time-saving symbol instead of typing in SELECT column1, column2,
column3, column4,…,columnX, to select all the columns. The FROM clause specifies
which table to query to fetch the columns requested in the SELECT clause.
You can issue the following SQL command to retrieve all the columns and all the
rows from the REGIONS table in the HR schema:
SELECT * FROM REGIONS;

As shown in Figure 2-2, when this command is executed in SQL*Plus, it returns
all the rows of data and all the columns belonging to this table. Use of the asterisk
in a SELECT statement is sometimes referred to as a “blind” query because the exact
columns to be fetched are not specified.
The second form of the basic SELECT statement has the same FROM clause as
the first form, but the SELECT clause is different:
SELECT {[DISTINCT] column|expression [alias],…}
FROM table;
This SELECT clause can be simplified into two formats:
SELECT column1 (possibly other columns or expressions) [alias optional]
OR
SELECT DISTINCT column1 (possibly other columns or expressions) [alias
optional]
FIGURE 2-2

Projecting all
columns from the
REGIONS table

Execute a Basic SELECT Statement

61

An alias is an alternative name for referencing a column or expression. Aliases
are typically used for displaying output in a user-friendly manner. They also serve as
shorthand when referring to columns or expressions to reduce typing. Aliases will
be discussed in detail later in this chapter. By explicitly listing only the relevant
columns in the SELECT clause you, in effect, project the exact subset of the results
you wish to retrieve. The following statement will return just the REGION_NAME
column subset of the REGIONS table as shown in Figure 2-2:
SELECT REGION_NAME
FROM REGIONS;

You may be asked to obtain all the job roles in the organization that employees
have historically fulfilled. For this you can issue the command: SELECT * FROM
JOB_HISTORY. However, in addition, the SELECT * construct returns the
EMPLOYEE_ID, START_DATE, and END_DATE columns. The uncluttered results
set containing only JOB_ID and DEPARTMENT_ID columns can be obtained with
the following statement executed in SQL*Plus, as shown in Figure 2-3.
Using the DISTINCT keyword allows duplicate rows to be eliminated from the
results set. In numerous situations a unique set of rows is required. It is important
to note that the criterion employed by the Oracle server in determining whether a

FIGURE 2-3

Projecting specific
columns from the
JOB_HISTORY
table

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row is unique or distinct depends entirely on what is specified after the DISTINCT
keyword in the SELECT clause. Selecting distinct JOB_ID values from the JOB_
HISTORY table will return the eight distinct job types as shown in Figure 2-4.
Compare this output to Figure 2-3, where ten rows are returned. Can you see that
there are two occurrences of the AC_ACCOUNT and ST_CLERK JOB_ID values?
These are the two duplicate rows that have been eliminated by looking for distinct
JOB_ID values. Selecting the distinct DEPARTMENT_ID column from the JOB_
HISTORY table returns only six rows, as Figure 2-5 demonstrates. DEPARTMENT_
ID values 50, 80, 90, and 110 each occur twice in the JOB_HISTORY table, and thus
four rows have been eliminated by searching for distinct DEPARTMENT_ID values.
An important feature of the DISTINCT keyword is the elimination of duplicate
values from combinations of columns. There are ten rows in the JOB_HISTORY
table. Eight rows contain distinct JOB_ID values. Six rows contain distinct
DEPARTMENT_ID values. Can you guess how many rows contain distinct
combinations of JOB_ID and DEPARTMENT_ID values? As Figure 2-6 reveals,
FIGURE 2-4

Selecting unique
JOB_IDs from the
JOB_HISTORY
table

Execute a Basic SELECT Statement

63

FIGURE 2-5

Selecting unique
DEPARTMENT_
IDs from the
JOB_HISTORY
table

there are nine rows returned in the results set that contain distinct JOB_ID and
DEPARTMENT_ID combinations, with one row from Figure 2-3 having being
eliminated. This is, of course, the row that contains a JOB_ID value of ST_CLERK
and a DEPARTMENT_ID value of 50.
The ability to project specific columns from a table is very useful. Coupled
with the ability to remove duplicate values or combinations of values
empowers you to assist with basic user reporting requirements. In many
application databases, tables can sometimes store duplicate data. End user
reporting frequently requires this data to be presented as a manageable set
of unique records.This is something you now have the ability to do. Be careful,
though, when using blind queries to select data from large tables. Executing a
SELECT * FROM HUGE_TABLE; statement may cause performance issues if
the table contains millions of rows of data.

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FIGURE 2-6

Data Retrieval Using the SQL SELECT Statement

Unique JOB_ID, DEPARTMENT_ID combinations from JOB_HISTORY

Rules Are Meant to be Followed
SQL is a fairly strict language in terms of syntax rules, but it remains simple and
flexible enough to support a variety of programming styles. This section discusses
some of the basic rules governing SQL statements.

Uppercase or Lowercase
It is a matter of personal taste about the case in which SQL statements are submitted
to the database. The examples used thus far have been written in uppercase. Many
developers, including the authors of this book, prefer to write their SQL statements
in lowercase. There is also a common misconception that SQL reserved words need

Execute a Basic SELECT Statement

65

to be specified in uppercase. Again, this is up to you. Adhering to a consistent and
standardized format is advised. The following three statements are syntactically
equivalent:
SELECT * FROM LOCATIONS;
Select * from locations;
select * from locations;

There is one caveat regarding case sensitivity. When interacting with literal
values, case does matter. Consider the JOB_ID column from the JOB_HISTORY
table. This column contains rows of data which happen to be stored in the database
in uppercase; for example, SA_REP and ST_CLERK. When requesting that the
results set be restricted by a literal column, the case is critical. The Oracle server
treats the request for all the rows in the
JOB_HISTORY table that contain a value of
St_Clerk in the JOB_ID column differently
from the request for all rows which have a value
SQL statements may
of ST_CLERK in JOB_ID column.
be submitted to the database in either
Metadata about different database objects
lowercase or uppercase. You must pay
is stored by default in uppercase in the data
careful attention to case when interacting
dictionary. If you query a database dictionary
with character literal data and aliases.
table to return a list of tables owned by the
Asking for a column called JOB_ID or
HR schema, it is likely that the table names
job_id returns the same column, but
returned are stored in uppercase. This does
asking for rows where the JOB_ID value
not mean that a table cannot be created with
is PRESIDENT is different from asking for
a lowercase name; it can be. It is just more
rows where the JOB_ID value is President.
common and the default behavior of the Oracle
Character literal data should always be
server to create and store tables, columns, and
treated in a case-sensitive manner.
other database object metadata in uppercase in
the database dictionary.

Statement Terminators
Semicolons are generally used as SQL statement terminators. SQL*Plus always
requires a statement terminator, and usually a semicolon is used. A single SQL
statement or even groups of associated statements are often saved as script files
for future use. Individual statements in SQL scripts are commonly terminated by
a line break (or carriage return) and a forward slash on the next line, instead of a
semicolon. You can create a SELECT statement, terminate it with a line break,
include a forward slash to execute the statement, and save it in a script file. The
script file can then be called from within SQL*Plus. Note that SQL Developer does

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not require a statement terminator if only a single statement is present, but it will
not object if one is used. It is good practice to always terminate your SQL statements
with a semicolon. Several examples of SQL*Plus statements follow:
select country_name, country_id, location_id from countries;
select city, location_id,
state_province, country_id
from locations
/

The first example of code demonstrates two important rules. First, the statement
is terminated by a semicolon. Second, the entire statement is written on one line. It
is entirely acceptable for a SQL statement to either be written on one line or to span
multiple lines as long as no words in the statement span multiple lines. The second
sample of code demonstrates a statement that spans three lines that is terminated by
a new line and executed with a forward slash.

Indentation, Readability, and Good
Practice
A common technique
employed by some exam question
designers tests attention to detail. A
single missing punctuation mark like
a semicolon may make the difference
between a correct answer and an
incorrect one. Incorrect spelling of object
names further tests attention to detail.
You may be asked to choose the correct
statement that queries the REGIONS
table. One of the options may appear
correct but references the REGION table.
This misspelling can lead to an incorrect
statement being chosen.

Consider the following query:
select city, location_id,
state_province, country_id
from locations
/

This example highlights the benefits of
indenting your SQL statement to enhance the
readability of your code. The Oracle server does
not object if the entire statement is written
on one line without indentation. It is good
practice to separate different clauses of the
SELECT statement onto different lines. When
an expression in a clause is particularly complex,
it is often better to separate that term of the
statement onto a new line. When developing
SQL to meet your reporting needs, the process
is often iterative. The SQL interpreter is far more useful during development if
complex expressions are isolated on separate lines, since errors are usually thrown in
the format of: “ERROR at line X:” This makes the debugging process much simpler.

Execute a Basic SELECT Statement

67

SCENARIO & SOLUTION
You want to construct and execute queries against
tables stored in an Oracle database. Are you confined
to using SQL*Plus or SQL Developer?

No. Oracle provides SQL*Plus and SQL Developer
as free tools to create and execute queries. There are
numerous tools available from Oracle (for example,
Discoverer, Forms, and JDeveloper) and other thirdparty vendors that provide an interface to the tables
stored in an Oracle database.

To explore your database environment further, you
would like a list of tables, owned by your current
schema, available for you to query. How do you
interrogate the database dictionary to provide this
metadata?

The data dictionary is a set of tables and views
of other tables that can be queried via SQL. The
statement SELECT TABLE_NAME from USER_
TABLES; queries the database dictionary for a list of
table names that belong to the current user.

When querying the JOBS table for every row
containing just the JOB_ID and MAX_SALARY
columns, is a projection, selection, or join being
performed?

A projection is performed since the columns in the
JOBS table have been restricted to the JOB_ID and
MAX_SALARY columns.

EXERCISE 2-2
Answering Our First Questions with SQL
In this step-by-step exercise, a connection is made using SQL*Plus as the HR user to
answer two questions using the SELECT statement.
Question 1: How many unique departments have employees currently working in
them?
1. Start SQL*Plus and connect to the HR schema.
2. You may initially be tempted to find the answer in the DEPARTMENTS
table. A careful examination reveals that the question asks for information
about employees. This information is contained in the EMPLOYEES table.
3. The word “unique” should guide you to use the DISTINCT keyword.

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4. Combining steps 2 and 3, you can construct the following SQL statement:
select distinct department_id
from employees;

5. As shown in the following illustration, this query returns 12 rows. Notice that
the third row is empty. This is a null value in the DEPARTMENT_ID column.

6. The answer to the first question is therefore: Eleven unique departments have
employees working in them, but at least one employee has not been assigned
to a department.
Question 2: How many countries are there in the Europe region?
1. This question comprises two parts. Consider the REGIONS table, which
contains four regions each uniquely identified by a REGION_ID value, and
the COUNTRIES table, which has a REGION_ID column indicating which
region a country belongs to.
2. The first query needs to identify the REGION_ID of the Europe region. This
is accomplished by the SQL statement:
select * from regions;

3. The following illustration shows that the Europe region has a REGION_ID
value of 1:

Execute a Basic SELECT Statement

69

4. To identify which countries have 1 as their REGION_ID, you need to execute the following SQL query
select region_id, country_name from countries;

5. Manually counting the country rows with a REGION_ID of 1 in the following
illustration helps answer the second question:

6. The answer to the second question is therefore: There are eight countries in
the Europe region as far as the HR data model is concerned.

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SQL Expressions and Operators
The general form of the SELECT statement introduced the notion that columns
and expressions are selectable. An expression is usually made up of an operation
being performed on one or more column values. The operators that can act upon
column values to form an expression depend on the data type of the column. They
are the four cardinal arithmetic operators (addition, subtraction, multiplication,
and division) for numeric columns; the concatenation operator for character or
string columns; and the addition and subtraction operators for date and timestamp
columns. As in regular arithmetic, there is a predefined order of evaluation (operator
precedence) when more than one operator occurs in an expression. Round brackets
have the highest precedence. Division and multiplication operations are next in
the hierarchy and are evaluated before addition and subtraction, which have lowest
precedence. These precedence levels are shown in Table 2-1.
Operations with the same level of precedence are evaluated from left to right.
Round brackets may therefore be used to enforce nondefault operator precedence.
Using brackets generously when constructing complex expressions is good practice
and is encouraged. It leads to readable code that is less prone to error. Expressions
open up a large number of useful data manipulation possibilities.

Arithmetic Operators
Consider the example of the JOB_HISTORY table, which stores the start date and
end date of an employee’s term in a previous job role. It may be useful for tax or
pension purposes, for example, to calculate how long an employee worked in that
role. This information can be obtained using an arithmetic expression. There are a
few interesting elements of both the SQL statement and the results returned from
Figure 2-7 that warrant further discussion.
Five elements have been specified in the SELECT clause. The first four are regular
columns of the JOB_HISTORY table namely: EMPLOYEE_ID, JOB_ID, START_
DATE, and END_DATE. The latter two terms provide the source information required
to calculate the number of days that an employee filled a particular position. Consider
employee number 176 on the ninth row of output. This employee started as a Sales
TABLE 2-1

Precedence
of Arithmetic
Operators

Precedence Level

Operator Symbol

Operation

Highest

()

Brackets or parentheses

Medium

/

Division

Medium

∗

Multiplication

Lowest

−

Subtraction

Lowest

+

Addition

Execute a Basic SELECT Statement

FIGURE 2-7

71

Arithmetic expression to calculate number of days worked

Manager on January 1, 1999 and ended employment on December 31, 1999. Therefore,
this employee worked for exactly one year, which, in 1999, consisted of 365 days.
The number of days for which an employee was employed can be calculated
by using the fifth element in the SELECT clause, which is an expression. This
expression demonstrates that arithmetic performed on columns containing date
information returns numeric values which represent a certain number of days.
To enforce operator precedence of the subtraction operation, the subexpression
end_date-start_date is enclosed in round brackets. One day was added to compensate
for the arithmetic loss of a day arising from the subtraction operation. Suppose an
employee started work on January 1 and quit later that day. One day must be added
to the formula, otherwise the subexpression end_date-start_date would incorrectly
return zero days worked.

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A hypothetical formula for predicting the probability of a meteor shower in a
particular geographic region has been devised. The two expressions listed in Figure 2-8
are identical except for the Meteor Shower Probability % expression. However, as the
results in the following table demonstrate, a different calculation is being made by each
expression. Notice that the two expressions differ very slightly. Expression 2 has a pair
of parentheses at the very end, enclosing (10 − 5 ). Consider how the expressions are
evaluated for the Asia region where REGION_ID is 3 as shown in the following table:
Step

Expression 1

Expression 2

1.

region_id * 100/5 + 20 / 10 − 5

region_id * 100/5 + 20 / (10 − 5)

2.

Substitute region_id with value:
3 * 100 /5 + 20 / 10 − 5

Substitute region_id with value:
3 * 100 / 5 + 20 / (10 − 5)

3.

The operators with the highest
precedence are the two division
and one multiplication operators.
These must be evaluated first. If
more than one operator with the
same level of precedence is present
in an expression, then these will
be evaluated from left to right.
Therefore, the first subexpression
to be evaluated is: 3*100:
300 / 5 + 20 / 10 − 5

The operator with the highest precedence
is the pair of parentheses and these must
be evaluated first. Therefore, the first
subexpression to be evaluated is:
(10 − 5):

The next subexpression to be
evaluated is: 300/5:

The next operators in the expression
with the highest precedence are the two
division and one multiplication operators.
If more than one operator with the same
level of precedence is present in an
expression, then these will be evaluated
from left to right. Therefore, the next
subexpression to be evaluated is: 3*100:
300/5 + 20/5

4.

60 + 20/10 − 5

3*100/5 + 20/5

5.

The next subexpression to be
evaluated is: 20/10:
60 + 2 − 5

The next subexpression to be evaluated is:
300/5:
60 + 20/5

6.

The remaining operators are one
addition and one subtraction
operator which share the same
level of precedence. These will
therefore be evaluated from left to
right. The next subexpression to
be evaluated is: 60+2:
62 − 5=57

The next subexpression to be evaluated
is: 20/5:

60 + 4 = 64

Execute a Basic SELECT Statement

FIGURE 2-8

73

Use of the concatenation and arithmetic operators

Expressions offer endless possibilities and are one of the fundamental
constructs in SELECT statements. As you practice SQL on your test database
environment, you may encounter two infamous Oracle errors: “ORA-00923:
FROM keyword not found where expected” and “ORA-00942: table or view
does not exist”.These are indicative of spelling or punctuation errors, such as
missing enclosing quotes around character literals. Do not be perturbed by
these messages. Remember, you cannot cause damage to the database if all
you are doing is selecting data. It is a read-only operation and the worst you
can do is execute a nonperformant query.

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Expression and Column Aliasing
Figure 2-7 introduced a new concept called column aliasing. Notice how the
expression column has a meaningful heading named Days Employed. This heading
is an alias. An alias is an alternate name for a column or an expression. If this
expression did not make use of an alias, the column heading would be: (END_
DATE-START_DATE)+1, which is unattractive and not very descriptive. Aliases
are especially useful with expressions or calculations and may be implemented in
several ways. There are a few rules governing the use of column aliases in SELECT
statements. In Figure 2-7, the alias given for the calculated expression called
“Days Employed” was specified by leaving a space and entering the alias in double
quotation marks. These quotation marks are necessary for two reasons. First, this
alias is made up of more than one word. Second, case preservation of an alias is only
possible if the alias is double quoted. As Figure 2-9 shows, an “ORA-00923: FROM
FIGURE 2-9

Use of column and expression aliases

Execute a Basic SELECT Statement

75

keyword not found where expected” error is returned when a multiworded alias is
not double quoted.
The ORA-00923 error is not randomly generated by the server. The Oracle
interpreter tries to process the statement and finds a problem with it. As it processes
this particular statement, it finds a problem with line 2. An asterisk symbol is
inserted at the starting point of the problem: the word Employed. Line 2 was
processed and the expression was aliased with the word Days. The space after Days
indicates to the Oracle interpreter that, since there is no additional comma to
indicate another term belonging to the SELECT clause, it is complete. Therefore, it
expects to find the FROM clause next. Instead it finds the word Employed and yields
this error. Error messages from the Oracle server are informative and you should read
them carefully to resolve problems. This error is avoided by enclosing an alias that
contains a space or other special characters, such as # and $, in double quotation
marks as shown around the alias “Days Employed” in Figure 2-7.
The second example in Figure 2-9 illustrates another interesting characteristic
of column aliasing. Double quotation marks have once again been dispensed with
and an underscore character is substituted for the space between the words to avoid
an error being returned. The Oracle interpreter processes the statement, finds no
problem, and executes it. Notice that although the alias was specified as Date_
Employed, with only the title letters of the alias being capitalized, the expression
heading was returned as DATE_EMPLOYED: all letters were automatically
converted to uppercase. Thus, to preserve the case of the alias, it must be enclosed in
double quotation marks.
The aliases encountered so far have been specified by leaving a space after a
column or expression and inserting the alias. SQL offers a more formalized way of
inserting aliases. The AS keyword is inserted between the column or expression
and the alias. Figure 2-10 illustrates the mixed use of the different types of column
aliasing. Both the EMPLOYEE_ID and JOB_ID columns are aliased using the AS
keyword, while the “Days Employed” expression is aliased using a space. The AS
keyword is optional since it is also possible to use a space before specifying an alias,
as discussed earlier. Use of the AS keyword does, however, improve the readability of
SQL statements, and the authors believe it is a good SQL coding habit to form.

Character and String Concatenation Operator
The double pipe symbols || represent the character concatenation operator. This
operator is used to join character expressions or columns together to create a larger
character expression. Columns of a table may be linked to each other or to strings of
literal characters to create one resultant character expression.

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FIGURE 2-10

Use of the AS
keyword to
specify column
aliases

Figure 2-8 shows that the concatenation operator is flexible enough to be used
multiple times and almost anywhere in a character expression. Here, the character
literal “The” is concatenated to the data contents of the REGION_NAME column.
This new string of characters is further concatenated to the character literal “region
is on Planet Earth,” and the whole expression is aliased with the friendly column
heading “Planetary Location.” Notice how each row in the results set is constructed
by the systematic application of the expression to every row value from the table.
Consider the first data row from the “Planetary Location” expression column.
It returns “The Europe region is on Planet Earth.” A legible sentence for the rows
of data has been created by concatenating literal strings of characters and spaces
to either side of each row’s REGION_NAME column value. The REGION_ID
column has been aliased to show that regular columns as well as expressions may
be aliased. Further, column headings are by default displayed in uppercase but can
be overridden using an alias like “Region Id.” The data types of the columns being
queried determine how SQL*Plus and SQL Developer present their default data
outputs. If the data type is numeric then the column data is formatted to be right
aligned. If the data type is character or date, then the column data is formatted to be
left aligned.

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77

Literals and the DUAL Table
Literal values in expressions are a common occurrence. These values refer to
numeric, character, or date and time values found in SELECT clauses that do not
originate from any database object. Concatenating character literals to existing
column data can be useful, as introduced in Figure 2-8. What about processing
literals that have nothing to do with existing column data? To ensure relational
consistency, Oracle offers a clever solution to the problem of using the database to
evaluate expressions that have nothing to do with any tables or columns. To get the
database to evaluate an expression, a syntactically legal SELECT statement must
be submitted. What if you wanted to know the sum of two numbers or two numeric
literals? These questions can only be answered by interacting with the database in
a relational manner. Oracle solves the problem of relational interaction with the
database operating on literal expressions by offering a special table called DUAL.
Recall the DUAL table described in Figure 2-1. It contains one column called
DUMMY of character data type. You can execute the query: SELECT * FROM
DUAL, and the data value “X” is returned as the contents of the DUMMY column.
The DUAL table allows literal expressions to be selected from it for processing
and returns the expression results in its single row. It is exceptionally useful since it
enables a variety of different processing requests to be made from the database. You
may want to know how many seconds there are in a year. Figure 2-11 demonstrates
an arithmetic expression executed against the DUAL table. Testing complex
expressions during development, by querying the dual table, is an effective method
to evaluate whether these expressions are working correctly. Literal expressions can
be queried from any table, but remember that the expression will be processed for
every row in the table.
select 'literal'||'processing using the REGIONS table'
from regions;

The preceding statement will return four lines in the results set, since there are
four rows of data in the REGIONS table.

Two Single Quotes or the Alternative Quote Operator
The literal character strings concatenated so far have been singular words prepended
and appended to column expressions. These character literals are specified using
single quotation marks. For example:
select 'I am a character literal string'

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FIGURE 2-11

Data Retrieval Using the SQL SELECT Statement

Using the DUAL table

What about character literals that contain single quotation marks? Plurals pose a
particular problem for character literal processing. Consider the following statement:
select 'Plural's have one quote too many' from dual;

As the example in Figure 2-12 shows, executing this statement causes an ORA00923 Oracle error to be generated. It might seem like an odd error, but upon closer
examination, the Oracle interpreter successfully processes the SELECT statement
until position 16, at which point it expects a “FROM” clause. Position 1 to position
16 is:
select 'Plural's

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79

FIGURE 2-12

Error while
dealing with
literals with
implicit quotes

The Oracle server processes this segment to mean that the character literal
‘Plural’ is aliased as column “s”. At this point, the interpreter expects a “FROM”
clause, but instead finds the word “have.” It then generates an error.
So, how are words that contain single quotation marks dealt with? There are
essentially two mechanisms available. The most popular of these is to add an
additional single quotation mark next to each naturally occurring single quotation
mark in the character string. Figure 2-13 demonstrates how the previous error is
avoided by replacing the character literal 'Plural's with the literal 'Plural''s.
The second example in Figure 2-13 shows that using two single quotes to
handle each naturally occurring single quote in a character literal can become
messy and error prone as the number of affected literals increases. Oracle offers a
neat way to deal with this type of character literal in the form of the alternative
quote (q) operator. Notice that the problem is that Oracle chose the single quote
characters as the special pair of symbols that enclose or wrap any other character
literal. These character-enclosing symbols could have been anything other than
single quotation marks.
Bearing this in mind, consider the alternative quote (q) operator. The q operator
enables you to choose from a set of possible pairs of wrapping symbols for character
literals as alternatives to the single quote symbols. The options are any single-byte

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FIGURE 2-13

Data Retrieval Using the SQL SELECT Statement

Use of two single quotes with literals with implicit quotes

or multibyte character or the four brackets: (round brackets), {curly braces}, [square
brackets], or . Using the q operator, the character delimiter can
effectively be changed from a single quotation mark to any other character, as shown
in Figure 2-14.
The syntax of the alternative quote operator is as follows:
q'delimiter'character literal which may include the single quotes delimiter'
where delimiter can be any character or bracket. The first and second examples in
Figure 2-14 show the use of angle and square brackets as character delimiters, while
the third example demonstrates how an uppercase “X” has been used as the special
character delimiter symbol through the alternative quote operator.

Execute a Basic SELECT Statement

FIGURE 2-14

81

The alternate quote (q) operator

NULL Is Nothing
The concept of a null value was introduced in the earlier discussion of the
DESCRIBE command. Both the number zero and a blank space are different from
null since they occupy space. Null refers to an absence of data. A row that contains
a null value lacks data for that column. Null is formally defined as a value that is
unavailable, unassigned, unknown, or inapplicable. In other words, the rules of
engaging with null values need careful examination. Failure to heed the special
treatment that null values require will almost certainly lead to an error, or worse, an
inaccurate answer.
Null values may be a tricky concept to come to grips with. The problem stems
from the absence of null on a number line. It is not a real, tangible value that can be
related to the physical world. Null is a placeholder in a nonmandatory column until

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INSIDE THE EXAM
There are two certification objectives in this
chapter. The capabilities of the SELECT
statement introduce the three fundamental
theoretical concepts of projection, selection,
and joining. Practical examples that illustrate
selection include building the SELECT clause
and using the DISTINCT keyword to limit
the rows returned. Projection is demonstrated
in examples where columns and expressions
are restricted for retrieval. The second objective of executing a SQL statement measures
your understanding of the basic form of the
SELECT statement. The exam measures two
aspects. First, syntax is measured: you are
required to spot syntax errors. SQL syntax
errors are raised when the Oracle interpreter
does not understand a statement. These errors

could take the form of statements missing
terminators such as a missing semicolon, not
enclosing character literals in appropriate
quote operators, or statements making use of
invalid reserved words.
Second, the meaning of a statement is measured. You will be presented with a syntactically legitimate statement and asked to choose
between accurate and inaccurate descriptions of that statement. The exam measures
knowledge around the certification objectives
using multiple choice format questions. Your
understanding of column aliasing, arithmetic
and concatenation operators, character literal
quoting, the alternative quote operator, SQL
statement syntax, and basic column data types
will be tested.

some real data is stored in its place. Until then, beware of conducting arithmetic
with null columns.
This section focuses on interacting with null column data with the SELECT
statement and its impact on expressions.

Not Null and Nullable Columns
Tables store rows of data that are divided into one or more columns. These columns
have names and data types associated with them. Some of them are constrained by
database rules to be mandatory columns. It is compulsory for some data to be stored
in the NOT NULL columns in each row. When columns of a table, however, are not
compelled by the database constraints to hold data for a row, these columns run the
risk of being empty.
In Figure 2-15, the EMPLOYEES table is described, and a few columns are
selected from it. There are five NOT NULL columns and six NULLABLE columns.
Nullable is a term sometimes used to describe a column that is allowed to store
null values. One of the nullable columns is the COMMISSION_PCT column.

Execute a Basic SELECT Statement

FIGURE 2-15

83

Null values in the Commission_Pct column

Figure 2-15 shows the first two rows of data from the EMPLOYEES table. This is
sufficient to illustrate that both these employee records have null values in their
COMMISSION_PCT columns.
SQL Developer makes it simple to observe null values in columns, as displayed
in Figure 2-16. Here, the word (null) is output when a null value is encountered, as
with the COMMISSION_PCT column. SQL Developer supports customizing this
default description of null column data.
The column aliased as “Null Arithmetic” is an expression made up of
COMMISSION_PCT + EMPLOYEE_ID + 10. Instead of returning a numeric value,
this column returns null. There is an important reason for this:
Any arithmetic calculation with a NULL value always returns NULL.

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FIGURE 2-16

Data Retrieval Using the SQL SELECT Statement

Null arithmetic always returns a null value.

Oracle offers a mechanism for interacting arithmetically with NULL values using
the general functions discussed in Chapter 5. As the column expression aliased as
“Division by Null” illustrates, even division by a null value results in null, unlike
division by zero, which results in an error. Finally, notice the impact of the null
keyword when used with the character concatenation operator. Null is concatenated
between the FIRST_NAME and LAST_NAME columns, yet it has no impact. The
character concatenation operators ignore null, whilst the arithmetic operations
involving null values always result in null.

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85

Foreign Keys and Nullable Columns
Data model design sometimes leads to problematic situations when tables are related
to each other via a primary and foreign key relationship, but the column that the
foreign key is based on is nullable.
The DEPARTMENTS table has, as its primary key, the DEPARTMENT_ID
column. The EMPLOYEES table has a DEPARTMENT_ID column that is
constrained by its foreign key relationship to the DEPARTMENT_ID column in
the DEPARTMENTS table. This means that no record in the EMPLOYEES table
is allowed to have in its DEPARTMENT_ID column a value that is not in the
DEPARTMENTS table. This referential integrity forms the basis for third normal
form and is critical to overall database integrity.
But what about NULL values? Can the DEPARTMENT_ID column in the
DEPARTMENTS table contain nulls? The answer is no. Oracle insists that any
column that is a primary key is implicitly constrained to be mandatory. But what
about implicit constraints on foreign key columns? This is a quandary for Oracle,
since in order to remain flexible and cater to the widest audience, it cannot insist
that columns related through referential integrity constraints must be mandatory.
Further, not all situations demand this functionality.

SCENARIO & SOLUTION
You are constructing an arithmetic expression that
calculates taxable income based on an employee’s
SALARY and COMMISSION_PCT columns, both
of which are nullable. Is it possible to convert the
null values in either column to zero to always return
a numeric taxable income?

Yes, but not with the information you have covered
so far. Null values require special handling. In
Chapter 5, we discuss the NVL function, which
provides a mechanism to convert null values into
more arithmetic-friendly data values.

An alias provides a mechanism to rename a column
or an expression. Under what conditions should you
enclose an alias in double quotes?

If an alias contains more than one word or if the
case of an alias must be preserved, then it should
be enclosed in double quotation marks. Failure to
double quote a multiworded alias will raise an Oracle
error. Failure to double quote a single-word alias will
result in the alias being returned in uppercase.

When working with character literal values that
include single quotation marks, how should you
specify these literals in the SELECT clause without
raising an error?

There are two mechanisms available. The more
common approach is to replace each naturally
occurring single quote with two single quotes. The
other approach is to make use of the alternate quote
operator to specify an alternate pair of characters
with which to enclose character literals.

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The DEPARTMENT_ID column in the EMPLOYEES table is actually nullable.
Therefore, the risk exists that there are records with null DEPARTMENT_ID values
present in this table. In fact, there are such records in the EMPLOYEES table. The
HR data model allows employees, correctly or not, to belong to no department.
When performing relational joins between tables, it is entirely possible to miss or
exclude certain records that contain nulls in the join column. Chapter 7 explores
ways to deal with this challenge.

EXERCISE 2-3
Experimenting with Expressions and the DUAL Table
In this step-by-step exercise a connection is made using SQL Developer as the HR
user. Use expressions and operators to answer three questions related to the SELECT
statement:
Question 1: It was demonstrated earlier how the number of days for which staff were
employed in a job could be calculated. For how many years were staff employed
while fulfilling these job roles and what were their EMPLOYEE_ID, JOB_ID,
START_DATE, and END_DATE values? Alias the expression column in your query
with the alias Years Employed. Assume that a year consists of 365.25 days.
1. Start SQL Developer and connect to the HR schema.
2. The projection of columns required includes EMPLOYEE_ID, JOB_ID,
START_DATE, END_DATE, and an expression called Years Employed from
the JOB_HISTORY table.
3. The expression can be calculated by dividing one plus the difference between
END_DATE and START_DATE by 365.25 days, as shown next:
select employee_id, job_id, start_date, end_date,
((end_date-start_date) + 1)/365.25 "Years Employed"
from job_history;

4. Executing the preceding SELECT statement yields the results displayed in
the following illustration:

Execute a Basic SELECT Statement

87

Question 2: Query the JOBS table and return a single expression of the form The Job
Id for the  job is: . Take note that the job_title should have an
apostrophe and an “s” appended to it to read more naturally. A sample of this output
for the organization president is: “The Job Id for the President’s job is: AD_PRES.”
Alias this column expression as “Job Description” using the AS keyword.
1. There are multiple solutions to this problem. The approach chosen here is
to handle the naturally occurring single quotation marks with an additional
single quote.
2. A single expression aliased as Job Description is required and may be constructed by dissecting the requirement into the literal “The Job Id for the”
being concatenated to the JOB_TITLE column. This string is then concatenated to the literal “'s job is:,” which is further concatenated to the JOB_ID
column. An additional single quotation mark is added to yield the SELECT
statement that follows:
select 'The Job Id for the '||job_title||'''s job is: '||job_id
AS "Job Description"
from jobs;

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3. The results of this SQL query are shown in the following illustration:

Question 3: Using the DUAL table, calculate the area of a circle with radius 6000
units, with pi being approximately 22/7. Use the formula: Area = pi × radius ×
radius. Alias the result as “Area.”
1. Working with the DUAL table may initially seem curious. You get used to
it as its functionality becomes more apparent. This question involves selecting a literal arithmetic expression from the DUAL table to yield a single row
calculated answer that is not based on the column values in any table.
2. The expression may be calculated using the following SQL statement; note
the use of brackets for precedence.
select (22/7) * (6000 * 6000) Area
from dual

3. The results returned show the approximate area of the circle as 113142857.14
square units.

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89

CERTIFICATION SUMMARY
The SELECT statement construct forms the basis for the majority of interactions
that occur with an Oracle database. These interactions may take the form of queries
issued from SQL Developer or SQL*Plus or any number of Oracle and other thirdparty client tools. At their core, these tools translate requests for information into
SELECT statements, which are then executed by the database.
The structure of a table has been described. Rows of data have been retrieved
and the set-oriented format of the results was revealed. The results were refined
by projection. In other words, your queries can include only the columns you are
interested in retrieving and exclude the remaining columns in a table.
SELECT syntax rules are basic and flexible, and language errors should be rare
due to its English-like grammar. Statement termination using semicolons, regard
for character literal case-sensitivity, and awareness of null values should assist with
avoiding errors.
Expressions expose a vista of data manipulation possibilities through the
interaction of arithmetic and character operators with column or literal data, or a
combination of the two.
The general form of the SELECT statement was explored and the foundation for
the expansion of this statement was constructed.
The Self Test exercises are made up of two components. The first component
is comprised of questions that give you an idea about what you may be asked
during the exam. The second component enables you to practice the language
skills discussed in this chapter in a lab format. The solutions to both categories of
questions are discussed in detail in the solutions section.

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✓

Data Retrieval Using the SQL SELECT Statement

TWO-MINUTE DRILL
List the Capabilities of SQL SELECT Statements
❑ The three fundamental operations that SELECT statements are capable of

are projection, selection, and joining.
❑ Projection refers to the restriction of columns selected from a table. Using

projection, you retrieve only the columns of interest and not every possible
column.
❑ Selection refers to the extraction of rows from a table. Selection includes the

further restriction of the extracted rows based on various criteria or conditions. This allows you to retrieve only the rows that are of interest and not
every row in the table.
❑ Joining involves linking two or more tables based on common attributes.

Joining allows data to be stored in third normal form in discrete tables, instead of in one large table.
❑ An unlimited combination of projections, selections, and joins provides the

language to extract the relational data required.
❑ A structural definition of a table can be obtained using the DESCRIBE

command.
❑ Columns in tables store different types of data using various data types, the most

common of which are NUMBER, VARCHAR2, DATE, and TIMESTAMP.
❑ The data type NUMBER(x,y) implies that numeric information stored in this

column can have at most x digits, but at least y of these digits must appear on
the right hand side of the decimal point.
❑ The DESCRIBE command lists the names, data types, and nullable status of

all columns in a table.
❑ Mandatory columns are also referred to as NOT NULL columns

Execute a Basic SELECT Statement
❑ The syntax of the primitive SELECT clause is as follows:

SELECT *|{[DISTINCT] column|expression [alias],…}
❑ The SELECT statement is also referred to as a SELECT query and comprises

at least two clauses, namely the SELECT clause and the FROM clause.

Two-Minute Drill

91

❑ The SELECT clause determines the projection of columns. In other words, the

SELECT clause specifies which columns are included in the results returned.
❑ The asterisk (*) operator is used as a wildcard symbol to indicate all columns.

So, the statement SELECT * FROM ACCOUNTS returns all the columns
available in the ACCOUNTS table.
❑ The FROM clause specifies the source table or tables from which items are

selected.
❑ The DISTINCT keyword preceding items in the SELECT clause causes

duplicate combinations of these items to be excluded from the returned
results set.
❑ SQL statements should be terminated with a semicolon. As an alternative,

a new line can be added after a statement and a forward slash can be used to
execute the statement.
❑ SQL statements can be written and executed in lowercase or uppercase. Be

careful when interacting with character literals since these are case-sensitive.
❑ Arithmetic operators and the string concatenation operator acting on col-

umn and literal data form the basis of SQL expressions.
❑ Expressions and regular columns may be aliased using the AS keyword or by

leaving a space between the column or expression and the alias.
❑ If an alias contains multiple words or the case of the alias is important, it

must be enclosed in double quotation marks.
❑ Naturally occurring single quotes in a character literal can be selected by

making use of either an additional single quote per naturally occurring quote
or the alternative quote operator.
❑ The DUAL table is a single column and single row table that is often used to

evaluate expressions that do not refer to specific columns or tables.
❑ Columns which are not governed by a NOT NULL constraint have the

potential to store null values and are sometimes referred to as nullable columns.
❑ NULL values are not the same as a blank space or zero. NULL values refer to

an absence of data. Null is defined as a value that is unavailable, unassigned,
unknown, or inapplicable.
❑ Caution must be exercised when working with null values since arithmetic

with a null value always yields a null result.

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SELF TEST
The following questions will help you measure your understanding of the material presented in this
chapter. Read all the choices carefully because there might be more than one correct answer. Choose
all the correct answers for each question.
The following test is typical of the questions and format of the OCP 11g examination for the topic
“Retrieving Data using the SQL SELECT Statement.” These questions often make use of the Human
Resources schema.

List the Capabilities of SQL SELECT Statements
1. Which query creates a projection of the DEPARTMENT_NAME and LOCATION_ID
columns from the DEPARTMENTS table? (Choose the best answer.)
A. SELECT DISTINCT DEPARTMENT_NAME, LOCATION_ID
FROM DEPARTMENTS;
B. SELECT DEPARTMENT_NAME, LOCATION_ID
FROM DEPARTMENTS;
C. SELECT DEPT_NAME, LOC_ID
FROM DEPT;
D. SELECT DEPARTMENT_NAME AS “LOCATION_ID”
FROM DEPARTMENTS;
2. After describing the EMPLOYEES table, you discover that the SALARY column has a data
type of NUMBER(8,2). Which SALARY value(s) will not be permitted in this column?
(Choose all that apply.)
A. SALARY=12345678
B. SALARY=123456.78
C. SALARY=12345.678
D. SALARY=123456
E. SALARY=12.34
3. After describing the JOB_HISTORY table, you discover that the START_DATE and END_
DATE columns have a data type of DATE. Consider the expression END_DATE-START_
DATE. (Choose two correct statements.)
A. A value of DATE data type is returned.
B. A value of type NUMBER is returned.
C. A value of type VARCHAR2 is returned.

Self Test

93

D. The expression is invalid since arithmetic cannot be performed on columns with DATE
data types.
E. The expression represents the days between the END_DATE and START_DATE less one
day.
4. The DEPARTMENTS table contains a DEPARTMENT_NAME column with data type
VARCHAR2(30). (Choose two true statements about this column.)
A. This column can store character data up to a maximum of 30 characters.
B. This column must store character data that is at least 30 characters long.
C. The VARCHAR2 data type is replaced by the CHAR data type.
D. This column can store data in a column with data type VARCHAR2(50) provided that the
contents are at most 30 characters long.

Execute a Basic SELECT Statement
5. Which statement reports on unique JOB_ID values from the EMPLOYEES table? (Choose all
that apply.)
A. SELECT JOB_ID FROM EMPLOYEES;
B. SELECT UNIQUE JOB_ID FROM EMPLOYEES;
C. SELECT DISTINCT JOB_ID, EMPLOYEE_ID FROM EMPLOYEES;
D. SELECT DISTINCT JOB_ID FROM EMPLOYEES;
6. Choose the two illegal statements. The two correct statements produce identical results. The
two illegal statements will cause an error to be raised:
A. SELECT DEPARTMENT_ID|| ' represents the '||
DEPARTMENT_NAME||' Department' as "Department Info"
FROM DEPARTMENTS;
B. SELECT DEPARTMENT_ID|| ' represents the ||
DEPARTMENT_NAME||' Department' as "Department Info"
FROM DEPARTMENTS;
C. select department_id|| ' represents the '||department_name||
' Department' "Department Info"
from departments;
D. SELECT DEPARTMENT_ID represents the DEPARTMENT_NAME Department as
"Department Info"
FROM DEPARTMENTS;

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7. Which expressions do not return NULL values? (Choose all that apply.)
A. select ((10 + 20) * 50) + null from dual;
B. select 'this is a '||null||'test with nulls' from dual;
C. select null/0 from dual;
D. select null||'test'||null as “Test” from dual;
8. Choose the correct syntax to return all columns and rows of data from the EMPLOYEES table.
A. select all from employees;
B. select employee_id, first_name, last_name, first_name, department_id
from employees;
C. select % from employees;
D. select * from employees;
E. select *.* from employees;
9. The following character literal expression is selected from the DUAL table:
SELECT 'Coda''s favorite fetch toy is his orange ring' FROM DUAL;
(Choose the result that is returned.)
A. An error would be returned due to the presence of two adjacent quotes
B. Coda's favorite fetch toy is his orange ring
C. Coda''s favorite fetch toy is his orange ring
D. 'Coda''s favorite fetch toy is his orange ring'
10. There are four rows of data in the REGIONS table. Consider the following SQL statement:
SELECT '6 * 6' “Area” FROM REGIONS;
How many rows of results are returned and what value is returned by the Area column?
(Choose the best answer.)
A. 1 row returned, Area column contains value 36
B. 4 rows returned, Area column contains value 36 for all 4 rows
C. 1 row returned, Area column contains value 6 * 6
D. 4 rows returned, Area column contains value 6 * 6 for all 4 rows
E. A syntax error is returned.

LAB QUESTION
In this chapter you worked through examples in the Human Resources schema. Oracle provides a
number of example schemas for you to experiment with and to learn different concepts from. For the
practical exercises, you will be using the Order Entry, or OE, schema. The solutions for these exercises

Lab Question

95

will be provided later using SQL Developer. Using SQL Developer or SQL*Plus, connect to the OE
schema and complete the following tasks.
1. Obtain structural information for the PRODUCT_INFORMATION and ORDERS tables.
2. Select the unique SALES_REP_ID values from the ORDERS table. How many different
sales representatives have been assigned to orders in the ORDERS table?
3. Create a results set based on the ORDERS table that includes the ORDER_ID, ORDER_
DATE, and ORDER_TOTAL columns. Notice how the ORDER_DATE output is formatted
differently from the START_DATE and END_DATE columns in the HR.JOB_ID table.
4. The PRODUCT_INFORMATION table stores data regarding the products available for sale
in a fictitious IT hardware store. Produce a set of results that will be useful for a sales person.
Extract product information in the format  with code:  has status of: . Alias the expression as “Product.” The results
should provide the LIST_PRICE, the MIN_PRICE, the difference between LIST_PRICE,
and MIN_PRICE aliased as “Max Actual Savings,” along with an additional expression that
takes the difference between LIST_PRICE and MIN_PRICE and divides it by the LIST_
PRICE and then multiplies the total by 100. This last expression should be aliased as “Max
Discount %.”
5. Calculate the surface area of the Earth using the DUAL table. Alias this expression as
“Earth's Area.” The formula for calculating the area of a sphere is: 4πr2. Assume, for this
example, that the earth is a simple sphere with a radius of 3,958.759 miles and that π is 22/7.

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SELF TEST ANSWERS
List the Capabilities of SQL SELECT Statements
✓ B. A projection is an intentional restriction of the columns returned from a table.
1. ®
®
˚ A is eliminated since the question has nothing to do with duplicates, distinctiveness,
or uniqueness of data. C incorrectly selects nonexistent columns called DEPT_NAME and
LOC_ID from a nonexistent table called DEPT. D returns just one of the requested columns:
DEPARTMENT_NAME. Instead of additionally projecting the LOCATION_ID column
from the DEPARTMENTS table, it attempts to alias the DEPARTMENT_NAME column as
LOCATION_ID.
✓ A and C. Columns with NUMBER(8,2) data type can store, at most, eight digits; of which,
2. ®
at most, two of those digits are to the right of the decimal point. Although A and C are the
correct answers, note that since the question is phrased in the negative, these values are NOT
allowed to be stored in such a column. A is not allowed because it contains eight whole number
digits, but the data type is constrained to store six whole number digits and two fractional digits.
C is not allowed since it has three fractional digits and the data type allows a maximum of two
fractional digits.
®
˚ B, D, and E can legitimately be stored in this data type and, therefore, are the incorrect
answers to this question. D shows that numbers with no fractional part are legitimate values for
this column, as long as the number of digits in the whole number portion does not exceed six
digits.
✓ B and E. The result of arithmetic between two date values represents a certain number of days.
3. ®
®
˚ A, C, and D are incorrect. It is a common mistake to expect the result of arithmetic
between two date values to be a date as well, so A may seem plausible, but it is false.
✓ A and D. The scale of the VARCHAR2 data type, specified in brackets, determines its
4. ®
maximum capacity for storing character data as mentioned by A. If a data value that is at most
30 characters long is stored in any data type, it can also be stored in this column as stated by D.
®
˚ B is incorrect because it is possible to store character data of any length up to 30 characters
in this column. C is false, since the CHAR data type exists in parallel with the VARCHAR2
data type.

Execute a Basic SELECT Statement
✓ D. Unique JOB_ID values are projected from the EMPLOYEES table by applying the
5. ®
DISTINCT keyword to just the JOB_ID column.
®
˚ A, B, and C are eliminated since A returns an unrestricted list of JOB_ID values including
duplicates; B makes use of the UNIQUE keyword in the incorrect context; and C selects the

Self Test Answers

97

distinct combination of JOB_ID and EMPLOYEE_ID values. This has the effect of returning
all the rows from the EMPLOYEES table since the EMPLOYEE_ID column contains unique
values for each employee record. Additionally, C returns two columns, which is not what was
originally requested.
✓ B and D represent the two illegal statements that will return syntax errors if they are
6. ®
executed. This is a tricky question because it asks for the illegal statements and not the legal
statements. B is illegal because it is missing a single quote enclosing the character literal
“represents the.” D is illegal because it does not make use of single quotes to enclose its
character literals.
®
˚ A and C are the legal statements and, therefore, in the context of the question, are the
incorrect answers. A and C appear to be different since the case of the SQL statements are
different and A uses the alias keyword AS, whereas C just leaves a space between the expression
and the alias. Yet both A and C produce identical results.
✓ B and D do not return null values since character expressions are not affected in the same
7. ®
way by null values as arithmetic expressions. B and D ignore the presence of null values in their
expressions and return the remaining character literals.
®
˚ A and C return null values because any arithmetic expression that involves a null will
return a null.
✓ D. An asterisk is the SQL operator that implies that all columns must be selected from a
8. ®
table.
®
˚ A, B, C, and E are incorrect. A uses the ALL reserved word but is missing any column
specification and will, therefore, generate an error. B selects some columns but not all columns
and, therefore, does not answer the question. C and E make use of illegal selection operators.
✓ B. The key to identifying the correct result lies in understanding the role of the single
9. ®
quotation marks. The entire literal is enclosed by a pair of quotes to avoid the generation of an
error. The two adjacent quotes are necessary to delimit the single quote that appears in literal B.
®
˚ A, C, and D are incorrect. A is eliminated since no error is returned. C inaccurately returns
two adjacent quotes in the literal expression and D returns a literal with all the quotes still
present. The Oracle server removes the quotes used as character delimiters after processing the
literal.
✓ D. The literal expression '6 * 6' is selected once for each row of data in the REGIONS
10. ®
table.
®
˚ A, B, C, and E are incorrect. A returns one row instead of four and calculates the product
6 * 6. The enclosing quote operators render 6 * 6 a character literal and not a numeric literal
that can be calculated. B correctly returns four rows but incorrectly evaluates the character
literal as a numeric literal. C incorrectly returns one row instead of four and E is incorrect,
because the given SQL statement can be executed.

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LAB ANSWER
The assumption is made that an Oracle database is available for you to practice on. The database
administrator (DBA) in your organization may assist you with installing and setting this up. In order
for any client tool such as SQL*Plus or SQL Developer to connect to the database, a listener process
should be running and the database must be opened. Additionally, you may have to request that the
HR and OE schema accounts be unlocked and that the passwords be reset. If these sample schemas
are not present, it is a simple matter to get the DBA to run the scripts, which are installed when
the database is installed, to create them. Connect to the OE schema using either SQL*Plus or SQL
Developer.
1. The DESCRIBE command gives us the structural description of a table. The following
illustration shows these two tables being described:

Lab Answer

99

2. The request for unique values usually involves using the DISTINCT keyword as part of your
SELECT statement. The two components of the statement involve the SELECT clause and the
FROM clause. You were asked for unique SALES_REP_ID values FROM the ORDERS table. It
is simple to translate this request into the following SELECT statement:
select distinct sales_rep_id
from orders;

From the results in the illustration, you can answer the original question: There are nine
different sales representatives responsible for orders listed in the ORDERS table, but there is
one order that contains null values in their SALES_REP_ID fields.

3. When asked to create a results set, it translates to SELECT one or more columns from a table.
In this case, your SELECT clause is constructed from the three columns requested. There is no

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request for unique values, so there is no need to consider the DISTINCT keyword. The FROM
clause need only include the ORDERS table to build the following SELECT statement:
select order_id, order_date, order_total
from orders;

Consider the output in the following illustration, specifically the ORDER_DATE column. This
column contains the day, month, year, hours, minutes, seconds, and fractional seconds up to six
decimal places or accurate up to a millionth of a second. The description of the ORDERS table
exposes ORDER_DATE as a TIMESTAMP(6) with LOCAL TIMEZONE column. This means that
the data in this column can be stored with fractional precision up to six decimal places and that the
data is time zone-aware. Basically, data may be worked on by people in different time zones. So Oracle
provides a data type that normalizes the local time to the database time zone to avoid confusion.
Compared to the START_DATE and END_DATE columns in the HR.JOB_ID table, the ORDER_
DATE column data type is far more sophisticated. Essentially, though, both these data types store date
and time information but to differing degrees of precision.

Lab Answer

101

4. The SELECT clause to answer this question should contain an expression aliased as “Product”
made up of concatenations of character literals with the PRODUCT_NAME, PRODUCT_ID,
and PRODUCT_STATUS columns. Additionally, the SELECT clause must contain the
LIST_PRICE and MIN_PRICE columns and two further arithmetic expressions aliased as “Max
Actual Savings” and “Max Discount %.” The FROM clause need only include the PRODUCT_
INFORMATION table. Proceed by constructing each of the three expressions in turn and
put them all together. The “Product” expression could be derived with the following SELECT
statement:
select product_name||' with code: '||product_id'||' has status of:'||order_
status AS Product

The “Max Actual Savings” expression could be derived with the following SELECT statement:
select list_price – min_price AS "Max Actual Savings"

The “Max Discount %” expression takes the calculation for “Max Actual Savings”, divides this
amount by the LIST_PRICE, and multiplies it by 100. It could be derived with the following
SELECT statement:
Select ((list_price-min_price)/list_price) * 100 AS "Max Discount %"

These three expressions, along with the two regular columns, form the SELECT clause executed
against the PRODUCT_INFORMATION table as shown next:

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5. The versatile DUAL table clearly forms the FROM clause. The SELECT clause is more
interesting, since no actual columns are being selected, just an arithmetic expression. A possible
SELECT statement to derive this calculation could be:
select (4 * (22/7) * (3958.759 * 3958.759)) AS "Earth's Area"
from dual;

This calculation approximates that planet Earth’s surface area is 197016572.595304 square miles.

3
Restricting and
Sorting Data

CERTIFICATION OBJECTIVES
3.01

Limit the Rows Retrieved by a Query

✓

3.02

Sort the Rows Retrieved by a Query

Q&A

3.03

Ampersand Substitution

Two-Minute Drill
Self Test

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

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Chapter 3:

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L

imiting the columns retrieved by a SELECT statement is known as projection and
was introduced in Chapter 2. Restricting the rows returned is known as selection.
This chapter discusses the WHERE clause, which is an enhancement to the selection
functionality of the SELECT statement. The WHERE clause specifies one or more conditions that
the Oracle server evaluates to restrict the rows returned by the statement. A further language
enhancement is introduced by the ORDER BY clause, which provides data sorting capabilities.
Ampersand substitution introduces a way to reuse the same statement to execute different
queries by substituting query elements at runtime. This area of runtime binding in SQL statements
is thoroughly explored.

CERTIFICATION OBJECTIVE 3.01

Limit the Rows Retrieved by a Query
One of the cornerstone principles in relational theory is selection. Selection is
actualized using the WHERE clause of the SELECT statement. Conditions that
restrict the dataset returned take many forms and operate on columns as well
as expressions. Only those rows in a table that conform to these conditions are
returned. Conditions restrict rows using comparison operators in conjunction
with columns and literal values. Boolean operators provide a mechanism to
specify multiple conditions to restrict the rows returned. Boolean, conditional,
concatenation, and arithmetic operators are discussed to establish their order of
precedence when they are encountered in a SELECT statement. The following four
areas are investigated:
■ The WHERE clause
■ Comparison operators
■ Boolean operators
■ Precedence rules

The WHERE clause
The WHERE clause extends the SELECT statement by providing the language to
restrict rows returned based on one or more conditions. Querying a table with just
the SELECT and FROM clauses results in every row of data stored in the table

Limit the Rows Retrieved by a Query

105

being returned. Using the DISTINCT keyword, duplicate values are excluded, and
the resultant rows are restricted to some degree. What if very specific information is
required from a table, for example, only the data where a column contains a specific
value? How would you retrieve the countries that belong to the Europe region from
the COUNTRIES table? What about retrieving just those employees who work
as sales representatives? These questions are answered using the WHERE clause
to specify exactly which rows must be returned. The format of the SQL SELECT
statement which includes the WHERE clause is:
SELECT *|{[DISTINCT] column|expression [alias],…}
FROM table
[WHERE condition(s)];
The SELECT and FROM clauses were examined in Chapter 2. The WHERE
clause always follows the FROM clause. The square brackets indicate that the
WHERE clause is optional. One or more conditions may be simultaneously applied
to restrict the result set. A condition is specified by comparing two terms using a
conditional operator. These terms may be column values, literals, or expressions.
The equality operator is most commonly used to restrict result sets. Two examples of
WHERE clauses are shown next:
select country_name
from countries
where region_id=3;
select last_name, first_name from employees
where job_id='SA_REP';

The first example projects the COUNTRY_NAME column from the COUNTRIES
table. Instead of selecting every row, the WHERE clause restricts the rows returned
to only those which contain a 3 in the REGION_ID column. The second example
projects two columns, LAST_NAME and FIRST_NAME from the EMPLOYEES
table. The rows returned are restricted to those which contain the value SA_REP in
their JOB_ID columns.

Numeric-Based Conditions
Conditions must be formulated appropriately for different column data types. The
conditions restricting rows based on numeric columns can be specified in several
different ways. Consider the SALARY column in the EMPLOYEES table. This
column has a data type of NUMBER(8,2). Figure 3-1 shows two different ways in
which the SALARY column has been restricted. The first and second examples

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Chapter 3:

Restricting and Sorting Data

FIGURE 3-1

Two ways to
select numeric
values in a
WHERE clause

retrieve the LAST_NAME and SALARY values of the employees who earn $ 10,000.
Notice the difference in the WHERE clauses of the following queries. The first query
specifies the number 10000, while the second encloses the number within single
quotes like a character literal. Both formats are acceptable to Oracle since an implicit
data type conversion is performed when necessary.
select last_name, salary from employees
where salary = 10000;
select last_name, salary from employees
where salary = '10000';

A numeric column can be compared to another numeric column in the same row
to construct a WHERE clause condition, as the following query demonstrates:
select last_name, salary from employees
where salary = department_id;

The first example in Figure 3-2 shows how the WHERE clause is too restrictive
and results in no rows being selected. This is because the range of SALARY values
is 2100 to 999999.99, and the range of DEPARTMENT_ID values is 10 to 110.
Since there is no overlap in the range of DEPARTMENT_ID and SALARY values,
there are no rows that satisfy this condition and therefore nothing is returned. The
example also illustrates how a WHERE clause condition compares one numeric
column to another.

Limit the Rows Retrieved by a Query

107

FIGURE 3-2

Using the
WHERE clause
with numeric
expressions

The second example in Figure 3-2 demonstrates extending the WHERE clause
condition to compare a numeric column, SALARY, to the numeric expression:
DEPARTMENT_ID*100. For each row, the value in the SALARY column is
compared to the product of the DEPARTMENT_ID value and 100. The WHERE
clause also permits expressions on either side of the comparison operator. You could
issue the following statement to yield identical results:
select last_name, salary from employees
where salary/10 = department_id*10;

As in regular algebra, the expression (SALARY = DEPARTMENT_ID * 100)
is equivalent to (SALARY/10 = DEPARTMENT_ID * 10). The notable feature
about this example is that the terms on either side of the comparison operator are
expressions.

Character-Based Conditions
Conditions determining which rows are selected based on character data, are specified
by enclosing character literals in the conditional clause, within single quotes. The
JOB_ID column in the EMPLOYEES table has a data type of VARCHAR2(10).
Suppose you wanted a report consisting of the LAST_NAME values of those
employees currently employed as sales representatives. The JOB_ID value for a sales
representative is SA_REP. The following statement produces such a report.
select last_name
from employees
where job_id='SA_REP';

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If you tried specifying the character literal without the quotes, an Oracle error
would be raised. Remember that character literal data is case sensitive, so the
following WHERE clauses are not equivalent.
Clause 1: where job_id=SA_REP
Clause 2: where job_id='Sa_Rep'
Clause 3: where job_id='sa_rep'

Clause 1 generates an “ORA-00904: “SA_REP”: invalid identifier” error since
the literal SA_REP is not wrapped in single quotes. Clause 2 and Clause 3 are
syntactically correct but not equivalent. Further, neither of these clauses yields any
data since there are no rows in the EMPLOYEES table which have JOB_ID column
values that are either Sa_Rep or sa_rep, as shown in Figure 3-3.
Character-based conditions are not limited to comparing column values
with literals. They may also be specified using other character columns and
expressions. The LAST_NAME and FIRST_NAME columns are both specified as
VARCHAR2(25) data typed columns. Consider the query:
select employee_id, job_id
from employees
where last_name=first_name;

Both the LAST_NAME and FIRST_NAME columns appear on either side of the
equality operator in the WHERE clause. No literal values are present; therefore no

FIGURE 3-3

Using the WHERE
clause with
character data

Limit the Rows Retrieved by a Query

109

single quote characters are necessary to delimit them. This condition stipulates that
only rows which contain the same data value (an exact case-sensitive match) in the
LAST_NAME and FIRST_NAME columns will be returned. This condition is too
restrictive and, as Figure 3-4 shows, no rows are returned.
Character-based expressions form either one or both parts of a condition
separated by a conditional operator. These expressions can be formed by
concatenating literal values with one or more character columns. The following four
clauses demonstrate some of the options for character-based conditions:
Clause
Clause
Clause
Clause

FIGURE 3-4

Character
column-based
WHERE clause

1:
2:
3:
4:

where
where
where
where

'A '||last_name||first_name = 'A King'
first_name||' '||last_name = last_name||' '||first_name
'SA_REP'||'King' = job_id||last_name
job_id||last_name ='SA_REP'||'King'

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Clause 1 concatenates the string literal “A” to the LAST_NAME and FIRST_
NAME columns. This expression is compared to the literal “A King,” and any
row that fulfils this condition is returned. Clause 2 demonstrates that character
expressions may be placed on both sides of the conditional operator. Clause 3
illustrates that literal expressions may also be placed on the left of the conditional
operator. It is logically equivalent to clause 4, which has swapped the operands in
clause 3 around. Both clauses 3 and 4 result in the same row of data being returned,
as shown in Figure 3-5.

Date-Based Conditions
DATE columns are useful when storing date and time information. Date literals
must be enclosed in single quotation marks just like character data; otherwise an
error is raised. When used in conditional WHERE clauses, DATE columns are
FIGURE 3-5

Equivalence
of conditional
expressions

Limit the Rows Retrieved by a Query

111

compared to other DATE columns or to date literals. The literals are automatically
converted into DATE values based on the default date format, which is DD-MONRR. If a literal occurs in an expression involving a DATE column, it is automatically
converted into a date value using the default format mask. DD represents days,
MON represents the first three letters of a month, and RR represents a Year 2000–
compliant year (that is, if RR is between 50 and 99, then the Oracle server returns
the previous century, else it returns the current century). The full four-digit year,
YYYY, can also be specified. Consider the following four SQL statements:
Statement 1:
select employee_id from job_history
where start_date = end_date;
Statement 2:
select employee_id from job_history
where start_date = '01-JAN-2001';
Statement 3:
select employee_id from job_history
where start_date = '01-JAN-01';
Statement 4:
select employee_id from job_history
where start_date = '01-JAN-99';

The first statement tests equality between two DATE columns. Rows that contain
the same values in their START_DATE and END_DATE columns will be returned.
Note, however, that DATE values are only equal to each other if there is an exact
match between all their components including day, month, year, hours, minutes, and
seconds. Chapter 4 discusses the details of storing DATE values. Until then, don’t
worry about the hours, minutes, and seconds components.
In the WHERE clause of the second statement, the START_DATE column
is compared to the character literal: '01-JAN-2001'. The entire four-digit year
component (YYYY) has been specified. This is acceptable to the Oracle server, and
all rows in the JOB_HISTORY table with START_DATE column values equal to
the first of January 2001 will be returned.
The third statement is equivalent to the second since the literal '01-JAN-01' is
converted to the date value 01-JAN-2001. This is due to the RR component being
less than 50, so the current (twenty-first) century, 20, is prefixed to the year RR
component to provide a century value. All rows in the JOB_HISTORY table with
START_DATE column values = 01-JAN-2001 will be returned.

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The century component for the literal '01-JAN-99' becomes the previous
(twentieth) century, 19, yields a date value of 01-JAN-1999 for the fourth statement,
since the RR component, 99, is greater than 50. Rows in the JOB_HISTORY table
with START_DATE column values = 01-JAN-1999 will be returned.
Arithmetic using the addition and subtraction operators is supported in
expressions involving DATE values. An expression like: END_DATE – START_
DATE returns a numeric value representing the number of days between START_
DATE and END_DATE. An expression like: START_DATE + 30 returns a DATE
value that is 30 days later than START_DATE. So the following expression is
legitimate, as shown in Figure 3-6:
select employee_id from job_history
where start_date + 30 = '31-JAN-99';

This query returns rows from the JOB_HISTORY table containing a START_
DATE value equal to 30 days before 31-JAN-1999. Therefore, only rows with a
value of 01-JAN-1999 in the START_DATE column will be retrieved.
FIGURE 3-6

Using the
WHERE clause
with numeric
expressions

Limit the Rows Retrieved by a Query

Conditional clauses
compare two terms using comparison
operators. It is important to understand
the data types of the terms involved so
they can be enclosed in single quotes, if
necessary. A common mistake is to assume
that a WHERE clause is syntactically
correct, when in fact, it is missing quotation

113

marks that delimit character or date
literals. Another common oversight is not
being aware that the terms to the left
and right of the comparison operator in
a conditional clause may be expressions,
columns, or literal values. Both these
concepts may be tested in the exam.

Comparison Operators
The equality operator is used extensively to illustrate the concept of restricting rows
using a WHERE clause. There are several alternative operators that may also be
used. The inequality operators like “less than” or “greater than or equal to” may be
used to return rows conforming to inequality conditions. The BETWEEN operator
facilitates range-based comparison to test whether a column value lies between two
values. The IN operator tests set membership, so a row is returned if the column
value tested in the condition is a member of a set of literals. The pattern matching
comparison operator LIKE is extremely powerful, allowing components of character
column data to be matched to literals conforming to a specific pattern. The last
comparison operator discussed in this section is the IS NULL operator, which returns
rows where the column value contains a null value. These operators may be used in
any combination in the WHERE clause and will be discussed next.

Equality and Inequality
Limiting the rows returned by a query involves specifying a suitable WHERE clause.
If the clause is too restrictive, then few or no rows are returned. If the conditional
clause is too broadly specified, then more rows than are required are returned.
Exploring the different available operators should equip you with the language to
request exactly those rows you are interested in. Testing for equality in a condition
is both natural and intuitive. Such a condition is formed using the “is equal to” (=)
operator. A row is returned if the equality condition is true for that row. Consider
the following query:
select last_name, salary
from employees
where job_id='SA_REP';

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The JOB_ID column of every row in the EMPLOYEES table is tested for equality
with the character literal SA_REP. For character information to be equal, there must
be an exact case-sensitive match. When such a match is encountered, the values
for the projected columns, LAST_NAME and SALARY, are returned for that row,
as shown in Figure 3-7. Note that although the conditional clause is based on the
JOB_ID column, it is not necessary for this column to be projected by the query.
Inequality-based conditions enhance the WHERE clause specification. Range and
pattern matching comparisons are possible using inequality and equality operators,
but it is often preferable to use the BETWEEN and LIKE operators for these
comparisons. The inequality operators are described in Table 3-1.
Inequality operators allow range-based queries to be fulfilled. You may be required
to provide a set of results where a column value is greater than another value.

FIGURE 3-7

Conditions based
on the equality
operator

Limit the Rows Retrieved by a Query

TABLE 3-1

Inequality
Operators

Operator

Description

<

Less than

>

Greater than

<=

Less than or equal to

>=

Greater than or equal to

<>

Not equal to

!=

Not equal to

115

For example, the following query may be issued to obtain a list of LAST_NAME and
SALARY values for employees who earn more that $5000:
select last_name, salary from employees
where salary > 5000;

Similarly, to obtain a list of employees who earn less than $3000, the following
query may be submitted:
select last_name, salary from employees
where salary < 3000;

The composite inequality operators (made up of more than one symbol) are utilized
in the following four clauses:
Clause
Clause
Clause
Clause

1:
2:
3:
4:

where
where
where
where

salary
salary
salary
salary

<=
>=
<>
!=

3000;
5000;
department_id;
4000+department_id;

Clause 1 returns those rows which contain a SALARY value that is less than or
equal to 3000. Clause 2 obtains data where the SALARY value is greater than or
equal to 5000, whilst clauses 3 and 4 demonstrate the two forms of the “not equal to”
operators. Clause 3 returns the rows which have SALARY column values that are
not equal to the DEPARTMENT_ID values. The alternate “not equal to” operator
in clause 4 illustrates that columns, literals, and expressions may all be compared
using inequality operators. Clause 4 returns those rows which contain a SALARY
value that is not equal to the sum of the DEPARTMENT_ID for that row and 4000.
Numeric inequality is naturally intuitive. The comparison of character and date
terms, however, is more complex. Testing character inequality is interesting since
the strings being compared on either side of the inequality operator are converted

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Chapter 3:

Restricting and Sorting Data

to a numeric representation of its characters. Based on the database character set
and NLS (National Language Support) settings, each character string is assigned
a numeric value. These numeric values form the basis for the evaluation of the
inequality comparison. Consider the following statement:
select last_name from employees
where last_name < 'King';

The character literal 'King' is converted to a numeric representation. Assuming
a US7ASCII database character set with AMERICAN NLS settings, the literal
'King' is converted into a sum of its ordinal character values: K + i + n + g =
(75+105+110+103=393). For each row in the EMPLOYEES table, the LAST_
NAME column is similarly converted to a numeric value. If this value is less than
393, then the row is selected. The same process for comparing numeric data using
the inequality operators applies to character data. The only difference is that
character data is converted implicitly by the Oracle server to a numeric value based
on certain database settings.
Inequality comparisons operating on date values follow a similar process to
character data. The Oracle server stores dates in an internal numeric format, and
these values are compared within the conditions. The second of June of a certain
year occurs earlier than the third of June of the same year. Therefore, the numeric
value of the date 02-JUN-2008 is less than the numeric value of the date 03-JUN2008. Consider the following query:
select last_name from employees
where hire_date < '01-JAN-2000';

This query retrieves each employee record containing a HIRE_DATE value that
is earlier than ‘01-JAN-2000’. Rows with employee HIRE_DATE=31-DEC-1999
will be returned, while rows with employee HIRE_DATE values later than the first
of January 2000 will not be returned, as shown in Figure 3-8.
The WHERE clause is a fundamental extension to the SELECT statement
and forms part of most queries. Although many comparison operators exist,
the majority of conditions are based on comparing two terms using both the
equality and the inequality operators.

Range Comparison with the BETWEEN Operator
The BETWEEN operator tests whether a column or expression value falls within
a range of two boundary values. The item must be at least the same as the lower
boundary value, or at most the same as the higher boundary value, or fall within the
range, for the condition to be true.

Limit the Rows Retrieved by a Query

117

FIGURE 3-8

Conditions based
on the inequality
operators

Suppose you want the last names of employees who earn a salary in the range of
$3400 and $4000. A possible solution using the BETWEEN operator is as follows:
select last_name from employees
where salary between 3400 and 4000;

This operator allows the WHERE condition to read in a natural English manner.
The last names of all the employees earning from $3400 to $4000 will be returned.
Boolean operators like AND, OR, and NOT are discussed later in this chapter, but
they are introduced here to enhance the description of the BETWEEN operator.
The AND operator is used to specify multiple WHERE conditions, all of which
must be satisfied for a row to be returned. Using the AND operator, the BETWEEN
operator is equivalent to two conditions using the “greater than or equal to” and

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“less than or equal to” operators, respectively. The preceding SQL statement is
equivalent to the following statement, as shown in Figure 3-9.
select last_name from employees
where salary >= 3400
and salary <= 4000;

The SALARY value for a row is tested first if it is greater than or equal to 3400
and second if it is less than or equal to 4000. If both conditions are satisfied, the
LAST_NAME value from the row forms part of the results set. If only one or neither
of the conditions is satisfied, the row is not selected.
Conditions specified with the BETWEEN operator can therefore be equivalently
denoted using two inequality-based conditions, but it is shorter and simpler to

FIGURE 3-9

The BETWEEN
operator

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119

specify the range condition using the BETWEEN operator. The implication of this
equivalence is that the mechanism utilized to evaluate numeric, character, and date
operands by the inequality operators is the same for the BETWEEN operator. The
following query tests whether the HIRE_DATE column value is later than 24-JUL1994 but earlier than 07-JUN-1996:
select first_name, hire_date from employees
where hire_date between '24-JUL-1994' and '07-JUN-1996';

You are not restricted to specifying literal values as the operands to the
BETWEEN operator, since these may be column values and expressions such as the
following:
select first_name, hire_date from employees
where '24-JUL-1994' between hire_date+30 and '07-JUN-1996';

For a row to be returned by this query, the date literal 24-JUL-1994 must fall
between the row’s HIRE_DATE column value plus 30 days and the date literal
07-JUN-1996.

Set Comparison with the IN Operator
The IN operator tests whether an item is a member of a set of literal values. The set
is specified by a comma separating the literals and enclosing them in round brackets.
If the literals are character or date values, then these must be delimited using
single quotes. You may include as many literals in the set as you wish. Consider the
following example:
select last_name from employees
where salary in (1000,4000,6000);

The SALARY value in each row is compared for equality to the literals specified
in the set. If the SALARY value equals 1000, 4000, or 6000, the LAST_NAME
value for that row is returned. The Boolean OR operator, discussed later in this
chapter, is used to specify multiple WHERE conditions, at least one of which must
be satisfied for a row to be returned. The IN operator is therefore equivalent to
a series of OR conditions. The preceding SQL statement may be written using
multiple OR condition clauses, for example:
select last_name from employees
where salary = 1000
OR salary = 4000
OR salary = 6000;

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This statement will return an employee’s LAST_NAME if at least one of the
WHERE clause conditions is true; it has the same meaning as the previous statement
that uses the IN operator, as shown in Figure 3-10.
Testing set membership using the IN operator is more succinct than using
multiple OR conditions, especially as the number of members in the set increases.
The following two statements demonstrate use of the IN operator with DATE and
CHARACTER data.
select last_name from employees
where last_name in ('King','Garbharran','Ramklass');
select last_name from employees
where hire_date in ('01-JAN-1998','01-DEC-1999');

Pattern Comparison with the LIKE Operator
To review, the BETWEEN operator provides a concise way to specify range-based
conditions, and the IN operator provides an optimal method to test set membership.
Now we introduce the LIKE operator, which is designed exclusively for character
data and provides a powerful mechanism for searching for letters or words.
LIKE is accompanied by two wildcard characters: the percentage symbol (%) and
the underscore character (_). The percentage symbol is used to specify zero or more
wildcard characters, while the underscore character specifies one wildcard character.
A wildcard may represent any character.

FIGURE 3-10

The IN operator

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121

You may be requested to provide a list of employees whose first names begin with
the letter “A.” The following query can be used to provide this set of results:
select first_name from employees
where first_name like 'A%';

The character literal that the FIRST_NAME column is compared to is enclosed in
single quotes like a regular character literal. In addition, it has a percentage symbol,
which has a special meaning in the context of the LIKE operator. The percentage
symbol substitutes zero or more characters appended to the letter A. Employee records
with FIRST_NAME values beginning with the letter A are returned.
The wildcard characters can appear at the beginning, middle or at the end of the
character literal. They can even appear alone as in:
where first_name like '%';

In this case, every row containing a FIRST_NAME value that is not null will be
returned. Wildcard symbols are not mandatory when using the LIKE operator. In
such cases, LIKE behaves as an equality operator testing for exact character matches;
so the following two WHERE clauses are equivalent:
where last_name like 'King';
where last_name = 'King';

The underscore wildcard symbol substitutes exactly one other character in a
literal. Consider searching for employees whose last names are four letters long,
begin with a “K,” have an unknown second letter, and end with an ng. You may issue
the following statement:
where last_name like 'K_ng';

Depending on the dataset, you may retrieve employees named King, Kong, and
Kung. An alternate way to perform pattern matching is to use an interminable
series of OR conditions, but to achieve the preceding results without using the
LIKE operator is prohibitively complex. For example, it may be achieved with the
following series of OR conditions:
where last_name = 'Kang'
OR last_name ='Kbng'
OR last_name ='Kcng'
…
OR last_name ='Kzng'

This example is incomplete, since it is not feasible to list every possible character
that could be substituted. This example demonstrates the sheer effort required to
substitute a single character without using the LIKE operator and the underscore

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wildcard symbol. For an unknown number (zero or more) of character substitutions,
the possibilities are exponentially larger than for single-character substitution. It is
not practically possible to perform character pattern matching without the use of the
LIKE operator and the wildcard symbols.
As Figure 3-11 shows, the two wildcard symbols can be used independently,
together, or even multiple times in a single WHERE condition. The first query
retrieves those records where COUNTRY_NAME begins with the letter “I”
followed by zero or more characters, one of which must be a lowercase “a.”
The second query retrieves those countries whose names contain the letter “i”
as its fifth character. The length of the COUNTRY_NAME values and the letter
they begin with are unimportant. The four underscore wildcard symbols preceding
the lowercase “i” in the WHERE clause represent exactly four characters (which
could be any characters). The fifth letter must be an “i,” and the percentage symbol
FIGURE 3-11

The wildcard
symbols of the
LIKE operator

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123

specifies that the COUNTRY_NAME can have zero or more characters from the
sixth character onward.
What about the scenario when you are searching for a literal that contains a
percentage or underscore character? Oracle provides a way to temporarily disable
their special meaning and regard them as regular characters using the ESCAPE
identifier. The JOBS table contains JOB_ID values that are literally specified with
an underscore character, such as SA_MAN, AD_VP, MK_REP, and SA_REP.
Assume there is, in addition, a row in the JOBS table with a JOB_ID of SA%MAN.
Notice there is no underscore character in this JOB_ID. How can values be retrieved
from the JOBS table if you are looking for JOB_ID values beginning with the
characters SA_? Consider the following SQL statement:
select * from jobs
where job_id like 'SA_%';

This query will return the rows SA_REP, SA_MAN, and SA%MAN. The
requirement in this example is not met since an additional row, SA%MAN, not
conforming to the criterion that it begins with the characters SA_, is also returned,
as the first example in Figure 3-12 shows.
A naturally occurring underscore character may be escaped (or treated as a
regular nonspecial symbol) using the ESCAPE identifier in conjunction with an
ESCAPE character. The second example in Figure 3-12 shows the SQL statement
that retrieves the JOBS table records with JOB_ID values equal to SA_MAN and
SA_REP and which conforms to the original requirement:
select job_id from jobs
where job_id like 'SA\_%' escape '\';

FIGURE 3-12

The ESCAPE
identifier and the
LIKE operator

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The ESCAPE identifier instructs the Oracle server to treat any character found
after the backslash character as a regular nonspecial symbol with no wildcard
meaning. In the preceding WHERE clause, any JOB_ID values that begin with the
three characters “SA_” will be returned. Traditionally, the ESCAPE character is the
backslash symbol, but it does not have to be. The following statement is equivalent
to the previous one but uses a dollar symbol as the ESCAPE character instead.
select job_id from jobs
where job_id like 'SA$_%' escape '$';

The percentage symbol may be similarly escaped when it occurs naturally
as character data. Suppose, there is a requirement to retrieve the row with the
hypothetical JOB_ID: SA%MAN introduced earlier. Querying the JOBS table for
JOB_ID values such as SA%MAN using the following code results in the records
SA_MAN and SA%MAN being returned.
select job_id from jobs
where job_id like 'SA%MAN';

The Oracle server interprets the percentage symbol in the WHERE clause as a
wildcard symbol when used with the LIKE operator. To obtain the row with JOB_ID:
SA%MAN using the LIKE operator, the percentage symbol may be escaped using
the following statement:
select job_id from jobs
where job_id like 'SA\%MAN' ESCAPE '\';

The backslash is defined as the ESCAPE character that instructs the Oracle
server to ignore the wildcard properties of the symbol occurring immediately after
the backslash. In this way, both wildcard symbols can be used as either specialized or
regular characters in different segments of the same character string.

EXERCISE 3-1
Using the LIKE Operator
Retrieve a list of DEPARTMENT_NAME values that end with the three letters
“ing” from the DEPARTMENTS table.
1. Start SQL*Plus and connect to the HR schema.
2. The SELECT clause is
SELECT DEPARTMENT_NAME

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125

3. The FROM clause is
FROM DEPARTMENTS
4. The WHERE clause must perform a comparison between the DEPARTMENT_
NAME column values and a pattern of characters beginning with zero or more
characters but ending with three specific characters, “ing.”
5. The operator enabling character pattern matching is the LIKE operator. The
pattern the DEPARTMENT_NAME column must conform to is '%ing'.
The percentage wildcard symbol indicates that zero or more characters may
precede the “ing” string of characters.
6. Thus, the WHERE clause is
WHERE DEPARTMENT_NAME LIKE '%ing'
7. Executing this statement returns the set of results matching this pattern as
shown in the following illustration:

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Null Comparison with the IS NULL Operator
NULL values inevitably find their way into database tables. It is often required that
only those records that contain a NULL value in a specific column are sought. The
IS NULL operator selects only the rows where a specific column value is NULL.
Testing column values for equality to NULL is performed using the IS NULL
operator instead of the “is equal to” operator (=).
Consider the following query that fetches the LAST_NAME column from
the EMPLOYEES table for those rows which have NULL values stored in the
COMMISSION_PCT column:
select last_name from employees
where commission_pct is null;

This WHERE clause reads naturally and retrieves only the records which contain
NULL COMMISSION_PCT values. As Figure 3-13 shows, the query using the “is equal
to” operator does not return any rows, while the query using the IS NULL operator does.

Boolean Operators
Data is restricted using a WHERE clause with a single condition. Boolean or logical
operators enable multiple conditions to be specified in the WHERE clause of the
SELECT statement. This facilitates a more refined data extraction capability.
Consider isolating those employee records with FIRST_NAME values that begin
with the letter “J” and who earn a COMMISSION_PCT greater than 10 percent.
First, the data in the EMPLOYEES table must be restricted to FIRST_NAME values
like “J%”, and second, the COMMISSION_PCT values for the records must be
tested to ascertain if they are larger than 10 percent. These two separate conditions
may be associated using the Boolean AND operator and are applied consecutively in
a WHERE clause. A result set conforming to any or all conditions or to the negation
of one or more conditions may be specified using Boolean operators.

The AND Operator
The AND operator merges conditions into one larger condition to which a row must
conform to be included in the results set. Boolean operators are defined using truth
tables. Table 3-2, the AND operator truth table, summarizes its functionality.
If two conditions specified in a WHERE clause are joined with an AND operator,
then a row is tested consecutively for conformance to both conditions before being
retrieved. If it conforms to neither or only one of the conditions, the row is excluded
since the result is FALSE. If the row contains a NULL value that causes one of the
conditions to evaluate to NULL, then that row is excluded. A row will only be

Limit the Rows Retrieved by a Query

FIGURE 3-13

Using the IS
NULL operator

TABLE 3-2

AND Operator
Truth Table

Condition X

Condition Y

Result

FALSE

FALSE

FALSE

TRUE

FALSE

FALSE

FALSE

TRUE

FALSE

TRUE

TRUE

TRUE

TRUE

NULL

NULL

NULL

TRUE

NULL

FALSE

NULL

FALSE

NULL

FALSE

FALSE

NULL

NULL

NULL

127

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returned if every condition joined with an AND operator evaluates to TRUE. In a
scenario with more than two conditions joined with the AND operator, only data
conforming to every condition will be returned.
Employee records with FIRST_NAME values beginning with the letter “J” and
COMMISSION_PCT greater than 10 percent can be retrieved using the following
query:
select first_name, last_name, commission_pct, hire_date
from employees
where first_name like 'J%'
and commission_pct > 0.1;

Notice that the WHERE clause now has two conditions, but only one WHERE
keyword. The AND operator separates the two conditions. To specify further
mandatory conditions, simply add them and ensure that they are separated by
additional AND operators. You can specify as many conditions as you wish.
Remember, though, the more AND conditions specified, the more restrictive the
query becomes. Figure 3-14 shows the preceding query followed by two additional
restrictions. The HIRE_DATE value must be larger than 01-JUN-1996, and the
LAST_NAME must contain the letter “o.” The first query returns four rows. Notice
that the additional AND conditions in the second query are satisfied by only two rows.

The OR Operator
The OR operator separates multiple conditions, at least one of which must be
satisfied by the row selected to warrant inclusion in the results set. Table 3-3, the
OR operator truth table, summarizes its functionality.

TABLE 3-3

OR Operator
Truth Table

Condition X

Condition Y

Result

FALSE

FALSE

FALSE

TRUE

FALSE

TRUE

FALSE

TRUE

TRUE

TRUE

TRUE

TRUE

TRUE

NULL

TRUE

NULL

TRUE

TRUE

FALSE

NULL

NULL

NULL

FALSE

NULL

NULL

NULL

NULL

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129

FIGURE 3-14

Using the AND
operator

If two conditions specified in a WHERE clause are joined with an OR operator
then a row is tested consecutively for conformance to either or both conditions
before being retrieved. Conforming to just one of the OR conditions is sufficient
for the record to be returned. If it conforms to none of the conditions, the row is
excluded since the result is FALSE. A row will only be returned if at least one of the
conditions associated with an OR operator evaluates to TRUE.
Retrieving employee records having FIRST_NAME values beginning with
the letter B or those with a COMMISSION_PCT greater than 35 percent can be
written as:
select first_name, last_name, commission_pct, hire_date
from employees
where first_name like 'B%'
or commission_pct > 0.35;

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Notice that the two conditions are separated by the OR keyword. All employee
records with FIRST_NAME values beginning with an uppercase “B” will be returned
regardless of their COMMISSION_PCT values, even if they are NULL. All those
records having COMMISSION_PCT values greater that 35 percent, regardless of
what letter their FIRST_NAME values begin with, are also returned.
Further OR conditions may be specified by separating them with an OR operator.
The more OR conditions you specify, the less restrictive your query becomes. Figure 3-15
shows the preceding query with two additional OR conditions. The HIRE_DATE value
must be larger than 01-JAN-2000 or the LAST_NAME must begin with the letter “B.”
The first query returns fewer rows than the second query since more rows meet the less
restrictive conditions in the second query than in the first.

FIGURE 3-15

Using the OR
operator

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131

SCENARIO & SOLUTION
You have a complex query with multiple conditions.
Is there a restriction on the number of conditions you
can specify in the WHERE clause? Is there a limit to
the number of comparison operators you can use in a
single query?

No. You may specify any number of conditions in the
WHERE clause separated by the Boolean operators.
There is no limit when using the comparison
operators, and they may be specified multiple times if
necessary in a single query.

You have been tasked to locate rows in the
EMPLOYEES table where the SALARY values
contain the numbers 8 and 0 adjacent to each other.
The SALARY column has a NUMBER data type. Is
it possible to use the LIKE comparison operator with
numeric data?

Yes. Oracle automatically casts the data into the
required data type, if possible. In this case, the
numeric SALARY values are momentarily “changed”
into character data allowing the use of the LIKE
operator to locate matching patterns. The following
query locates the required rows:
SELECT * FROM EMPLOYEES
WHERE SALARY LIKE '%80%';”

By restricting the rows returned from the JOBS table
to those which contain the value SA_REP in the
JOB_ID column, is a projection, selection or join
performed?

A selection is performed since rows are restricted.

The NOT Operator
The NOT operator negates conditional operators. A selected row must conform to the
logical opposite of the condition in order to be included in the results set. Table 3-4,
the NOT operator truth table, summarizes its functionality.
Conditional operators may be negated by the NOT operator as shown by the
WHERE clauses listed in Table 3-5.
As the examples in Table 3-5 suggest, the NOT operator can be very useful. It is
important to understand that the NOT operator negates the comparison operator in
a condition, whether it’s an equality, inequality, range based, pattern matching, set
membership, or null testing operator.

TABLE 3-4

The NOT
Operator Truth
Table

Condition X

NOT Condition X

FALSE

TRUE

TRUE

FALSE

NULL

NULL

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TABLE 3-5

Conditions
Negated by the
NOT Operator

Restricting and Sorting Data

Positive

Negative

where last_name='King'

where NOT (last_name='King')

where first_name LIKE 'R%'

where first_name NOT LIKE 'R%'

where department_id IN (10,20,30)

where department_id NOT IN (10,20,30)

where salary BETWEEN 1 and 3000

where salary NOT BETWEEN 1 and 3000

where commission_pct IS NULL

where commission_pct IS NOT NULL

Retrieving employee records with FIRST_NAME values that do NOT begin with
the letter “B” or those that do NOT comply with a COMMISSION_PCT greater
than 35 percent can be written as:
select first_name, last_name, commission_pct, hire_date
from employees
where first_name not like 'B%'
or not (commission_pct > 0.35);

Notice that the two conditions are still separated by the OR operator and the
NOT operator has just been added to them.
AND and OR are Boolean operators that enable multiple WHERE clause
conditions to be specified in a single query. All conditions separated by an
AND operator must evaluate to true after testing a row’s values to prevent
it from being excluded from the final results set. However, only one of the
conditions separated by an OR operator must evaluate to true to avoid its
exclusion from the final results set. If five conditions, A, B, C, D, and E, occur
in a WHERE clause as WHERE A and B or C or D and E, then a row will be
returned if both conditions A and B are fulfilled, or only condition C is met, or
only condition D is met, or both conditions D and E are fulfilled.

Precedence Rules
Arithmetic, character, comparison, and Boolean expressions were examined in the
context of the WHERE clause. But how do these operators interact with each other?
Arithmetic operators subscribe to a precedence hierarchy. Bracketed expressions are
evaluated before multiplication and division operators, which are evaluated before
subtraction and addition operators. Similarly, there is a precedence hierarchy for the
previously mentioned operators as shown in Table 3-6.
Operators at the same level of precedence are evaluated from left to right if they
are encountered together in an expression. When the NOT operator modifies the

Limit the Rows Retrieved by a Query

TABLE 3-6

Operator
Precedence
Hierarchy

Precedence Level

133

Operator Symbol

Operation

1

()

Parentheses or brackets

2

/,*

Division and multiplication

3

+,−

Addition and subtraction

4

||

Concatenation

5

=,<,>,<=,>=

Equality and inequality
comparison

6

[NOT] LIKE, IS [NOT]
NULL, [NOT] IN

Pattern, null, and set
comparison

7

[NOT] BETWEEN

Range comparison

8

!=,<>

Not equal to

9

NOT

NOT logical condition

10

AND

AND logical condition

11

OR

OR logical condition

LIKE, IS NULL, and IN comparison operators, their precedence level remains the
same as the positive form of these operators.
Consider the following SELECT statement that demonstrates the interaction of
various different operators:
select last_name,salary,department_id,job_id,commission_pct
from employees
where last_name like '%a%' and salary > department_id * 200
or
job_id in ('MK_REP','MK_MAN') and commission_pct is not null

The LAST_NAME, SALARY, DEPARTMENT_ID, JOB_ID, and
COMMISSION_PCT columns are projected from the EMPLOYEES table based
on two discrete conditions. The first condition retrieves the records containing
the character “a” in the LAST_NAME field AND with a SALARY value greater
than 200 times the DEPARTMENT_ID value. The product of DEPARTMENT_
ID and 200 is processed before the inequality operator since the precedence of
multiplication is higher than inequality comparison.
The second condition fetches those rows with JOB_ID values of either MK_
MAN or MK_REP in which COMMISSION_PCT values are not null. For a row to
be returned by this query, either the first OR second conditions need to be fulfilled.
Figure 3-16 illustrates three queries. Query 1 returns four rows. Query 2 is based
on the first condition just discussed and returns four rows. Query 3 is based on the
second condition and returns zero rows.

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FIGURE 3-16

Operator
precedence in the
WHERE clause

Changing the order of the conditions in the WHERE clause changes its meaning
due to the different precedence of the operators. Consider the following code sample:
select last_name,salary,department_id,job_id,commission_pct
from employees
where last_name like '%a%' and salary > department_id * 100 and
commission_pct is not null
or
job_id = 'MK_MAN'

There are two composite conditions in this query. The first condition retrieves
the records with the character “a” in the LAST_NAME field AND a SALARY
value greater than 100 times the DEPARTMENT_ID value AND where the
COMMISSION_PCT value is not null. The second condition fetches those rows
with JOB_ID values of MK_MAN. A row is returned by this query, if it conforms to
either condition one OR condition two, but not necessarily to both.

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135

FIGURE 3-17

Effect of
condition clause
ordering due to
precedence rules

As Figure 3-17 illustrates, this query returns six rows. It further shows the division
of the query into two queries based on its two composite conditions. The first
condition results in five rows being returned while the second results in the retrieval
of just one row with a JOB_ID value of MK_MAN.

Boolean operators OR
and AND allow multiple WHERE clause
conditions to be specified.The Boolean
NOT operator negates a conditional
operator and may be used several times
within the same condition.The equality,
inequality, BETWEEN, IN, and LIKE

comparison operators test two terms within
a single condition. Only one comparison
operator is used per conditional clause.
The distinction between Boolean and
comparison operators is important and
forms the basis for many questions related
to this chapter in the exam.

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CERTIFICATION OBJECTIVE 3.02

Sort the Rows Retrieved by a Query
Regular language dictionaries sort words in alphabetical order. Pages printed in a
book are sorted numerically in ascending order from beginning to end. The practical
implementations of selection and projection have been covered thus far. The
usability of the retrieved datasets may be significantly enhanced with a mechanism
to order or sort the information. Information may be sorted alphabetically,
numerically, from earliest to latest, or in ascending or descending order. Further, the
data may be sorted by one or more columns, including columns that are not listed in
the SELECT clause. Sorting is performed once the results of a SELECT statement
have been fetched. The sorting parameters do not influence the records returned by
a query, just the presentation of the results. Exactly the same rows are returned by
a statement including a sort clause as are returned by a statement excluding a sort
clause. Only the ordering of the output may differ. Sorting the results of a query is
accomplished using the ORDER BY clause.

The ORDER BY Clause
When tables are created, they are initially empty and contain no rows. As rows
are inserted, updated, and deleted by one or more users or application systems, the
original ordering of the stored rows is lost. The Oracle server cannot and does not
guarantee that rows are stored sequentially. This is not a problem since a mechanism
to sort the retrieved dataset is available in the form of the ORDER BY clause.
This clause is responsible for transforming the output of a query into more
practical, user-friendly sorted data. The ORDER BY clause is always the last clause
in a SELECT statement. As the full syntax of the SELECT statement is progressively
exposed, you will observe new clauses added but none of them will be positioned
after the ORDER BY clause. The format of the ORDER BY clause in the context of
the SQL SELECT statement is as follows:
SELECT *|{[DISTINCT] column|expression [alias],…}
FROM table
[WHERE condition(s)]
[ORDER BY {col(s)|expr|numeric_pos} [ASC|DESC] [NULLS FIRST|LAST]];

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137

Ascending and Descending Sorting
Ascending sort order is natural for most types of data and is therefore the default sort
order used whenever the ORDER BY clause is specified. An ascending sort order for
numbers is lowest to highest, while it is earliest to latest for dates and alphabetically
for characters. The first form of the ORDER BY clause shows that results of a query
may be sorted by one or more columns or expressions:
ORDER BY col(s)|expr;
Suppose that a report is requested that must contain an employee’s LAST_
NAME, HIRE_DATE, and SALARY information, sorted alphabetically by the
LAST_NAME column for all sales representatives and marketing managers. This
report could be extracted with the following SELECT statement:
select last_name, hire_date, salary
from employees
where job_id in ('SA_REP','MK_MAN')
order by last_name;

The data selected may be ordered by any of the columns from the tables in the
FROM clause, including those that do not appear in the SELECT list. The results
from the preceding query may be sorted by the COMMISSION_PCT column, as
shown in Figure 3-18.
The second example in Figure 3-18 shows that by appending the keyword DESC
to the ORDER BY clause, the rows are returned sorted in descending order based on
the COMMISSION_PCT column. The third example demonstrates the optional
NULLS LAST keywords, which specify that if the sort column contains null values,
then these rows are to be listed last after sorting the remaining rows based on their
NOT NULL values. To specify that rows with null values in the sort column should
be displayed first, append the NULLS FIRST keywords to the ORDER BY clause.
The following example sorts a dataset based on an expression. This expression
calculates an employee’s value to a company based on their HIRE_DATE and
SALARY values. This formula takes the HIRE_DATE value and subtracts a
specified number of days to return an earlier date. The number of days subtracted
is calculated by dividing the SALARY value by 10. The expression is aliased as
EMP_VALUE as follows:
select last_name, salary, hire_date, hire_date-(salary/10) emp_value
from employees
where job_id in ('SA_REP','MK_MAN')
order by emp_value;

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FIGURE 3-18

Sorting data using
the ORDER BY
clause

The EMP_VALUE expression is initialized with the HIRE_DATE value and is
offset further into the past based on the SALARY field. The earliest EMP_VALUE
date appears first in the result set output since the ORDER BY clause specifies that
the results will be sorted by the expression alias. Note that the results could be sorted
by the explicit expression and the alias could be omitted as in ORDER BY HIRE_
DATE-(SALARY/10), but using aliases renders the query easier to read.
Several implicit default options are selected when you use the ORDER BY
clause. The most important of these is that unless DESC is specified, the sort order
is assumed to be ascending. If null values occur in the sort column, the default sort
order is assumed to be NULLS LAST for ascending sorts and NULLS FIRST for
descending sorts. If no ORDER BY clause is specified, the same query executed
at different times may return the same set of results in different row order, so no
assumptions should be made regarding the default row order.

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139

Positional Sorting
Oracle offers an alternate and shorter way to specify the sort column or expression.
Instead of specifying the column name, the position of the column as it occurs in the
SELECT list is appended to the ORDER BY clause. Consider the following example:
select last_name, hire_date, salary
from employees
where job_id in ('SA_REP','MK_MAN')
order by 2;

The ORDER BY clause specifies the numeric literal two. This is equivalent to
specifying ORDER BY HIRE_DATE, since the HIRE_DATE column is the second
column selected in the SELECT clause.
Positional sorting applies only to columns in the SELECT list that have a numeric
position associated with them. Modifying the preceding query to sort the results by
the JOB_ID column is not possible using positional sorting since this column does
not occur in the SELECT list.

Composite Sorting
Results of a query may be sorted by more than one column using composite sorting.
Two or more columns may be specified (either literally or positionally) as the
composite sort key by commas separating them in the ORDER BY clause. Consider
the requirement to fetch the JOB_ID, LAST_NAME, SALARY, and HIRE_DATE
values from the EMPLOYEES table. The further requirements are that the results
must be sorted in reverse alphabetical order by JOB_ID first, then in ascending
alphabetical order by LAST_NAME, and finally in numerically descending order
based on the SALARY column. The following SELECT statement fulfils these
requirements:
select job_id, last_name, salary, hire_date
from employees
where job_id in ('SA_REP','MK_MAN')
order by job_id desc, last_name, 3 desc;

Each column involved in the sort is listed left to right in order of importance
separated by commas in the ORDER BY clause, including the modifier DESC,
which occurs twice in this clause. This example also demonstrates mixing literal and
positional column specifications. As Figure 3-19 shows, there are several rows with
the same JOB_ID value, for example, SA_REP. For these rows, the data is sorted
alphabetically by the secondary sort key, which is the LAST_NAME column.

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FIGURE 3-19

Composite
sorting using
the ORDER BY
clause

For the rows with the same JOB_ID and same LAST_NAME column values such as
SA_REP and Smith, these rows are sorted in numeric descending order by the third
sort column, SALARY.

The concept of sorting data
is usually thoroughly tested.The syntax of
the ORDER BY clause is straightforward,
but multiple sort terms like expressions,
columns, and positional specifiers, coupled
with descending sort orders for some

terms and ascending sort orders for others,
provide a powerful data sorting mechanism
that comes with a corresponding increase
in complexity.This complexity is often
tested, so ensure that you have a solid
understanding of the ORDER BY clause.

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141

EXERCISE 3-2
Sorting Data Using the ORDER BY Clause
The JOBS table contains descriptions of different types of jobs an employee in the
organization may occupy. It contains the JOB_ID, JOB_TITLE, MIN_SALARY,
and MAX_SALARY columns. You are required to write a query that extracts
the JOB_TITLE, MIN_SALARY, and MAX_SALARY columns, as well as an
expression called VARIANCE, which is the difference between the MAX_SALARY
and MIN_SALARY values, for each row. The results must include only JOB_TITLE
values that contain either the word “President” or “Manager.” Sort the list in
descending order based on the VARIANCE expression. If more than one row has
the same VARIANCE value, then, in addition, sort these rows by JOB_TITLE in
reverse alphabetic order.
1. Start SQL Developer and connect to the HR schema.
2. The SELECT clause is
SELECT JOB_TITLE, MIN_SALARY, MAX_SALARY, (MAX_SALARY –
MIN_SALARY) VARIANCE
3. The FROM clause is
FROM JOBS
4. The WHERE conditions must allow only those rows whose JOB_TITLE
column contains either the string “President” OR the string “Manager.”
5. The WHERE clause is
WHERE JOB_TITLE LIKE '%President%' OR JOB_TITLE LIKE '
%Manager%'
6. Sorting is accomplished with the ORDER BY clause. Composite sorting is
required using both the VARIANCE expression and the JOB_TITLE column
in descending order.
7. The ORDER BY clause is
ORDER BY VARIANCE DESC, JOB_TITLE DESC.
You may alternately specify the explicit expression in the ORDER BY clause
instead of the expression alias.

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8. Executing the statement returns a set of results matching this pattern as
shown in the following illustration.

CERTIFICATION OBJECTIVE 3.03

Ampersand Substitution
As queries are developed and perfected, they may be saved for future use. Sometimes,
queries differ very slightly, and it is desirable to have a more generic form of the
query that has a variable or placeholder defined that can be substituted at runtime.

Ampersand Substitution

143

Oracle offers this functionality in the form of ampersand substitution. Every element
of the SELECT statement may be substituted, and the reduction of queries to their
core elements to facilitate reuse can save you hours of tedious and repetitive work.
The following areas are examined in this section:
■ Substitution variables
■ The DEFINE and VERIFY commands

Substitution Variables
The key to understanding substitution variables is to regard them as placeholders.
A SQL query is composed of two or more clauses. Each clause can be divided
into subclauses, which are in turn made up of character text. Any text, subclause
or clause element, or even the entire SQL query is a candidate for substitution.
Consider the SELECT statement in its general form:
SELECT *|{[DISTINCT] column|expression [alias],…}
FROM table
[WHERE condition(s)]
[ORDER BY {col(s)|expr|numeric_pos} [ASC|DESC] [NULLS FIRST|LAST]];
Using substitution, you insert values into the italicized elements, choosing which
optional keywords to use in your queries. When the LAST_NAME column in the
EMPLOYEES table is required, the query is constructed using the general form of
the SELECT statement and substituting the column name: LAST_NAME in place
of the word column in the SELECT clause and the table name; EMPLOYEES in
place of the word table in the FROM clause.

Single Ampersand Substitution
The most basic and popular form of substitution of elements in a SQL statement
is single ampersand substitution. The ampersand character (&) is the symbol chosen
to designate a substitution variable in a statement and precedes the variable name
with no spaces between them. When the statement is executed, the Oracle server
processes the statement, notices a substitution variable, and attempts to resolve this
variable’s value in one of two ways. First, it checks whether the variable is defined in
the user session. (The DEFINE command is discussed later in this chapter.) If the
variable is not defined, the user process prompts for a value that will be substituted
in place of the variable. Once a value is submitted, the statement is complete and
is executed by the Oracle server. The ampersand substitution variable is resolved at
execution time and is sometimes known as runtime binding or runtime substitution.

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A common requirement in the sample HR department may be to retrieve the
same information for different employees at different times. Perhaps you are required
to look up contact information like PHONE_NUMBER data given either LAST_
NAME or EMPLOYEE_ID values. This generic request can be written as follows:
select employee_id, last_name, phone_number
from employees
where last_name = &LASTNAME
or employee_id = &EMPNO;

As Figure 3-20 shows, when running this query, Oracle server prompts you to
input a value for the variable called LASTNAME. You enter an employee’s last
name, if you know it, for example, 'King'. If you don’t know the last name but know
the employee ID number, you can type in any value and press the ENTER key to
submit the value. Oracle then prompts you to enter a value for the EMPNO variable.
After typing in a value, for example, 0, and hitting ENTER, there are no remaining
substitution variables for Oracle to resolve and the following statement is executed:
select employee_id, last_name, phone_number
from employees
where last_name = 'King'
or employee_id = 0;

Variables can be assigned any alphanumeric name that is a valid identifier name.
The literal you substitute when prompted for a variable must be an appropriate data
type for that context; otherwise, an ORA-00904: invalid identifier error is returned.
If the variable is meant to substitute a character or date value, the literal needs to be
enclosed in single quotes. A useful technique is to enclose the ampersand substitution
variable in single quotes when dealing with character and date values. In this way,

FIGURE 3-20

Single ampersand
substitution

Ampersand Substitution

145

the user is required to submit a literal value without worrying about enclosing it
in quotes. The following statement rewrites the previous one but encloses the
LASTNAME variable in quotes:
select employee_id, last_name, phone_number, email
from employees
where last_name = '&LASTNAME'
or employee_id = &EMPNO;

When prompted for a value to substitute for the LASTNAME variable, you may
for example, submit the value King without any single quotes, as these are already
present and when the runtime substitution is performed, the first WHERE clause
condition will resolve to: WHERE LAST_NAME = 'King'.

Double Ampersand Substitution
There are occasions when a substitution variable is referenced multiple times in the
same query. In such situations, the Oracle server will prompt you to enter a value
for every occurrence of the single ampersand substitution variable. For complex
scripts this can be very inefficient and tedious. The following statement retrieves the
FIRST_NAME and LAST_NAME columns from the EMPLOYEES table for those
rows that contain the same set of characters in both these fields:
select first_name, last_name
from employees
where last_name like '%&SEARCH%'
and first_name like '%&SEARCH%';

The two conditions are identical but apply to different columns. When this
statement is executed, you are first prompted to enter a substitution value for
the SEARCH variable used in the comparison with the LAST_NAME column.
Thereafter, you are prompted to enter a substitution value for the SEARCH variable
used in the comparison with the FIRST_NAME column. This poses two problems.
First, it is inefficient to enter the same value twice, but second and more importantly,
typographical errors may confound the query since Oracle does not verify that the
same literal value is entered each time substitution variables with the same name
are used. In this example, the logical assumption is that the contents of the variables
substituted should be the same, but the fact that the variables have the same name has
no meaning to the Oracle server and it makes no such assumption. The first example
in Figure 3-21 shows the results of running the preceding query and submitting two
distinct values for the SEARCH substitution variable. In this particular example, the
results are incorrect since the requirement was to retrieve FIRST_NAME and LAST_
NAME pairs which contained the identical string of characters.

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FIGURE 3-21

Double
ampersand
substitution

In situations when a substitution variable is referenced multiple times in the
same query and your intention is that the variable must have the same value at
each occurrence in the statement, it is preferable to make use of double ampersand
substitution. This involves prefixing the first occurrence of the substitution variable
that occurs multiple times in a query, with two ampersand symbols instead of one.
When the Oracle server encounters a double ampersand substitution variable, a session
value is defined for that variable and you are not prompted to enter a value to be
substituted for this variable in subsequent references.
The second example in Figure 3-21 demonstrates how the SEARCH variable is
preceded by two ampersands in the condition with the FIRST_NAME column and
thereafter is prefixed by one ampersand in the condition with the LAST_NAME
column. When executed, you are prompted to enter a value to be substituted for the
SEARCH variable only once for the condition with the FIRST_NAME column.
This value is then automatically resolved from the session value of the variable in
subsequent references to it, as in the condition with the LAST_NAME column.
To undefine the SEARCH variable, you need to use the UNDEFINE command
described later in this chapter.

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147

Whether you work as a developer, database administrator, or business end
user, all SQL queries you encounter may be broadly classified as either ad
hoc or repeated queries. Ad hoc queries are usually one-off statements
written during some data investigation exercise that are unlikely to be reused.
The repeated queries are those that are run frequently or periodically,
which are usually saved as script files and run with little to no modification
whenever required. Reuse prevents costly redevelopment time and allows
these consistent queries to potentially benefit from Oracle’s native automatic
tuning features geared toward improving query performance.

Substituting Column Names
Literal elements of the WHERE clause have been the focus of the discussion on
substitution thus far, but virtually any element of a SQL statement is a candidate for
substitution. In the following statement, the FIRST_NAME and JOB_ID columns
are static and will always be retrieved, but the third column selected is variable and
specified as a substitution variable named: COL. The result set is further sorted by
this variable column in the ORDER BY clause:
select first_name, job_id, &&col
from employees
where job_id in ('MK_MAN','SA_MAN')
order by &col;

As Figure 3-22 demonstrates, at runtime, you are prompted to provide a value
for the double ampersand variable called COL. You could, for example, enter the
column named SALARY and submit your input. The statement that executes
performs the substitution and retrieves the FIRST_NAME, JOB_ID, and SALARY
columns from the EMPLOYEES table sorted by SALARY.
Unlike character and date literals, column name references do not require
single quotes both when explicitly specified and when substituted via ampersand
substitution.

Substituting Expressions and Text
Almost any element of a SQL statement may be substituted at runtime. The constraint
is that Oracle requires at least the first word to be static. In the case of the SELECT
statement, at the very minimum, the SELECT keyword is required and the remainder
of the statement may be substituted as follows:
select &rest_of_statement;

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FIGURE 3-22

Substituting
column names

When executed, you are prompted to submit a value for the variable called:
REST_OF_STATEMENT, which could be any legitimate query, such as
DEPARTMENT_NAME from DEPARMENTS. If you submit this text as input for
the variable, the query that is run will be resolved to the following statement:
select department_name from departments;

Consider the general form of the SQL statement rewritten using ampersand
substitution, as shown next:
select &SELECT_CLAUSE
from &FROM_CLAUSE
where &WHERE_CLAUSE
order by &ORDER_BY_CLAUSE;

The usefulness of this statement is arguable, but it does illustrate the concept
of substitution effectively. As Figure 3-23 shows, the preceding statement allows
any query discussed so far to be submitted at runtime. The first execution queries
the REGIONS table, while the second execution queries the COUNTRIES table.
Useful candidates for ampersand substitution are statements that are run multiple
times and differ slightly from each other.

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149

FIGURE 3-23

Substituting
expressions and
text

Define and Verify
Double ampersand substitution is used to avoid repetitive input when the
same variable occurs multiple times in a statement. When a double ampersand
substitution occurs, the variable is stored as a session variable. As the statement
executes, all further occurrences of the variable are automatically resolved using the
stored session variable. Any subsequent executions of the statement within the same
session automatically resolve the substitution variables from stored session values.
This is not always desirable and indeed limits the usefulness of substitution variables.
Oracle does, however, provide a mechanism to UNDEFINE these session variables.

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INSIDE THE EXAM
There are three certification objectives in
this chapter. Limiting the rows retrieved
by a query introduced the WHERE clause,
which practically demonstrates the concept of
selection. The conditions that limit the rows
returned are based on comparisons using the
BETWEEN, IN, LIKE, Equality, and Inequality operators. Your thorough understanding of
these comparison operators and their behavior
with character, numeric, and date data types
will be examined, along with how they are
different from the Boolean NOT, AND, and
OR operators.
Using the ORDER BY clause to sort results
retrieved is optional and extremely useful.
Exam questions that test the concepts of
traditional, positional, and composite sorting,
along with how NULL values can be handled,
are common. When multiple sort terms are

present in the ORDER BY clause, it is acceptable to specify ascending sort orders for some
and descending sort orders for others. It is a
common mistake to forget that Oracle provides this mixed sorting feature as well for the
specification of the NULLS FIRST|LAST
modifier.
Your understanding of substitution using
single and double ampersands as well as the
DEFINE and UNDEFINE commands will be
tested. You may be given a statement that
includes a double ampersand substitution variable that is subsequently referenced multiple
times in the statement, along with a single
ampersand substitution variable; you will be
expected to understand the differences in
their behavior. Your understanding of column
name, expression, and text substitution will
also be measured.

The VERIFY command is specific to SQL*Plus and controls whether or not
substituted elements are echoed on the user’s screen prior to executing a SQL
statement that uses substitution variables. These commands are discussed in the next
sections.

The DEFINE and UNDEFINE Commands
Session level variables are implicitly created when they are initially referenced
in SQL statements using double ampersand substitution. They persist or remain
available for the duration of the session or until they are explicitly undefined. A
session ends when the user exits their client tool like SQL*Plus or when the user
process is terminated abnormally.

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151

The problem with persistent session variables is they tend to detract from the
generic nature of statements that use ampersand substitution variables. Fortunately,
these session variables can be removed with the UNDEFINE command. Within
a script or at the command line of SQL*Plus or SQL Developer, the syntax for
undefining session variables is as follows:
UNDEFINE variable;
Consider a simple generic example that selects a static and variable column from
the EMPLOYEES table and sorts the output based on the variable column. The
static column could be the LAST_NAME column.
select last_name, &&COLNAME
from employees
where department_id=30
order by &COLNAME;

The first time this statement executes, you are prompted to input a value for
the variable called COLNAME. Let’s assume you enter SALARY. This value is
substituted and the statement executes. A subsequent execution of this statement
within the same session does not prompt for any COLNAME values since it is
already defined as SALARY in the context of this session and can only be undefined
with the UNDEFINE COLNAME command, as shown in Figure 3-24. Once the
variable has been undefined, the next execution of the statement prompts the user
for a value for the COLNAME variable.
The DEFINE command serves two purposes. It can be used to retrieve a list of all
the variables currently defined in your SQL session; it can also be used to explicitly
define a value for a variable referenced as a substitution variable by one or more
statements during the lifetime of that session. The syntax for the two variants of the
DEFINE command are as follows:
DEFINE;
DEFINE variable=value;
As Figure 3-25 demonstrates, a variable called EMPNAME is defined explicitly
to have the value 'King'. The stand-alone DEFINE command in SQL*Plus then
returns a number of session variables prefixed with an underscore character as well as
other familiar variables, including EMPNAME and double ampersand substitution
variables implicitly defined earlier. Two different but simplistic query examples are
executed, and the explicitly defined substitution variable EMPNAME is referenced
by both queries. Finally, the variable is UNDEFINED.

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FIGURE 3-24

The UNDEFINE
command

The capacity of the SQL client tool to support session-persistent variables may
be switched off and on as required using the SET command. The SET command is
not a SQL language command, but rather a SQL environment control command. By
specifying SET DEFINE OFF, the client tool (for example, SQL*Plus) does not save
session variables or attach special meaning to the ampersand symbol. This allows
the ampersand symbol to be used as an ordinary literal character if necessary. The
SET DEFINE ON|OFF command therefore determines whether or not ampersand
substitution is available in your session.

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153

FIGURE 3-25

The DEFINE
command

The following example uses the ampersand symbol as a literal value. When
executed, you are prompted to submit a value for bind variable SID.
select 'Coda & Sid' from dual;

By turning off the ampersand substitution functionality as follows, this query may
be executed without prompts:
SET DEFINE OFF
select 'Coda & Sid' from dual;
SET DEFINE ON

Once the statement executes, the SET DEFINE ON command may be used
to switch the substitution functionality back on. If DEFINE is SET OFF and the
context that an ampersand is used in a statement cannot be resolved literally, Oracle
returns an error.

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SCENARIO & SOLUTION
The SELECT list of a query contains a single
column. Is it possible to sort the results retrieved by
this query by another column?

Yes. Unless positional sorting is used, the ORDER
BY clause is independent of the SELECT clause in a
statement.

Ampersand substitution variables support reusability
of repetitively executed SQL statements. If a
substituted value is to be used multiple times at
different parts of the same statement, is it possible to
be prompted to submit a substitution value just once
and for that value to automatically be substituted
during subsequent references to the same variable?

Yes. The two methods that may be used are double
ampersand substitution or the DEFINE command.
Both methods result in the user providing input
for a specific substitution variable once. This value
remains bound to the variable for the duration of the
session unless it is explicitly UNDEFINED.

You have been tasked to retrieve the LAST_NAME
and DEPARTMENT_ID values for all rows in the
EMPLOYEES table. The output must be sorted by
the nullable DEPARTMENT_ID column, and all
rows with NULL DEPARTMENT_ID values must
be listed last. Is it possible to provide the results as
requested?

Yes. The ORDER BY clause provides for sorting
by columns that potentially contain NULL values
by permitting the modifiers NULLS FIRST or
NULLS LAST to be specified. The following query
locates the required rows: SELECT LAST_NAME,
DEPARTMENT_ID FROM EMPLOYEES
ORDER BY DEPARTMENT_ID NULLS LAST;

The VERIFY Command
As discussed earlier, two categories of commands are available when dealing with the
Oracle server: SQL language commands and the SQL client control commands. The
SELECT statement is an example of a language command, whilst the SET command
controls the SQL client environment. There are many different language and
control commands available, but the control commands pertinent to substitution are
DEFINE and VERIFY.
The VERIFY command controls whether the substitution variable submitted is
displayed onscreen so you can verify that the correct substitution has occurred. A
message is displayed showing the old clause followed by the new clause containing the
substituted value. The VERIFY command is switched ON and OFF with the command
SET VERIFY ON|OFF. As Figure 3-26 shows, VERIFY is first switched OFF, a query
that uses ampersand substitution is executed, and you are prompted to input a value.
The value is then substituted, the statement runs, and its results are displayed.
VERIFY is then switched ON, the same query is executed, and you are prompted
to input a value. Once the value is input and before the statement commences
execution, Oracle displays the clause containing the reference to the substitution
variable as the old clause with its line number and, immediately below this, the new
clause displays the statement containing the substituted value.

Ampersand Substitution

155

FIGURE 3-26

The VERIFY
command

EXERCISE 3-3
Using Ampersand Substitution
A common calculation performed by the Human Resources department relates to
the calculation of taxes levied upon an employee. Although, this is done for all
employees, there are always a few staff members who dispute the tax deducted from
their income. The tax deducted per employee is calculated by obtaining the annual
salary for the employee and multiplying this by the current tax rate, which may
vary from year to year. You are required to write a reusable query using the current
tax rate and the EMPLOYEE_ID number as inputs and return the EMPLOYEE_ID,
FIRST_NAME, SALARY, ANNUAL SALARY (SALARY * 12), TAX_RATE, and
TAX (TAX_RATE * ANNUAL SALARY) information.
1. Start SQL*Plus and connect to the HR schema.
2. The select list must include the four specified columns as well as two expressions. The first expression aliased as ANNUAL SALARY is a simple
calculation, while the second expression aliased as TAX depends on the
TAX_RATE. Since the TAX RATE may vary, this value must be substituted
at runtime.

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3. The SELECT clause is
SELECT &&EMPLOYEE_ID, FIRST_NAME, SALARY, SALARY * 12 AS
“ANNUAL SALARY”, &&TAX_RATE, (&TAX_RATE * (SALARY * 12))
AS “TAX”
4. The double ampersand preceding EMPLOYEE_ID and TAX_RATE in the
SELECT clause stipulates to Oracle that when the statement is executed the
user must be prompted to submit a value for each substitution variable that
will be used wherever they are subsequently referenced as &EMPLOYEE_ID
and &TAX_RATE, respectively.
5. The FROM clause is
FROM EMPLOYEES
6. The WHERE clause must allow only the row whose EMPLOYEE_ID value is
specified at runtime.
7. The WHERE clause is
WHERE EMPLOYEE_ID = &EMPLOYEE_ID
8. Executing this statement returns the set of results shown in the following
illustration.

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157

CERTIFICATION SUMMARY
The WHERE clause provides the language that enables selection in SELECT
statements. The criteria for including or excluding rows take the form of conditions.
Using comparison operators, two terms are compared with each other and the
condition is evaluated as being true or false for each row. These terms may be
column values, literals, or expressions. If the Boolean sum of the results of
each condition evaluates to true for a particular row, then that row is retrieved.
Conditional operators allow terms to be compared to each other in a variety of ways
including equality, inequality, range based, set membership, and character pattern
matching comparison.
Once a set of data is isolated by your query, the ORDER BY clause facilitates
sorting the retrieved rows based on numeric, date or character columns or
expressions. Results may be sorted using combinations of columns or expressions or
both. Data is sorted in ascending order by default.
Generic, reusable statements may be constructed using ampersand substitution
variables which prompt for runtime values during execution. Session-persistent
substitution variables may be defined and are extremely convenient in situations
where many substitutions of the same variable occur in a statement or script.
The simple SELECT statement has been expanded to include a WHERE, and
ORDER BY clause. These basic building blocks offer a practical and useful language
that can be applied while you build your knowledge of SQL.

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TWO-MINUTE DRILL
Limit the Rows Retrieved by a Query
❑ The WHERE clause extends the SELECT statement by providing the language

that enables selection.
❑ One or more conditions constitute a WHERE clause. These conditions specify

rules to which the data in a row must conform to be eligible for selection.
❑ For each row tested in a condition, there are terms on the left and right of a

comparison operator. Terms in a condition can be column values, literals, or
expressions.
❑ Comparison operators may test two terms in many ways. Equality or inequality

tests are very common, but range, set, and pattern comparisons are also available.
❑ Range comparison is performed using the BETWEEN operator, which tests

whether a term falls between given start and end boundary values.
❑ Set membership is tested using the IN operator. A condition based on a set

comparison evaluates to true if the left-side term is listed in the single-quoted,
comma-delimited set on the right-side.
❑ The LIKE operator enables literal character patterns to be matched with

other literals, column values, or evaluated expressions. The percentage
symbol (%) behaves as a wildcard that matches zero or more characters. The
underscore symbol (_) behaves as a single character wildcard that matches
exactly one other character.
❑ Boolean operators include the AND, OR, and NOT operators. The AND

and OR operators enable multiple conditional clauses to be specified. These
are sometimes referred to as multiple WHERE clauses.
❑ The NOT operator negates the comparison operator involved in a condition.

Sort the Rows Retrieved by a Query
❑ Results are sorted using the ORDER BY clause. Rows retrieved may be

ordered according to one or more columns by specifying either the column
names or their numeric position in the SELECT clause.
❑ The sorted output may be arranged in descending or ascending order using

the DESC or ASC modifiers after the sort terms in the ORDER BY clause.

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159

Ampersand Substitution
❑ Ampersand substitution facilitates SQL statement reuse by providing a means

to substitute elements of a statement at runtime. The same SQL statement
may therefore be run multiple times with different input parameters.
❑ Single ampersand substitution requires user input for every occurrence of

the substitution variable in the statement. Double ampersand substitution
requires user input only once per occurrence of a substitution variable, since
it defines a session-persistent variable with the given input value.
❑ Session-persistent variables may be set explicitly using the DEFINE command.

The UNDEFINE command may be used to unset both implicitly (double
ampersand substitution) and explicitly defined session variables.
❑ The VERIFY environmental setting controls whether SQL*Plus displays the

old and new versions of statement lines which contain substitution variables.

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SELF TEST
The following questions will help you measure your understanding of the material presented in this
chapter. Read all the choices carefully because there may be more than one correct answer. Choose
all the correct answers for each question.

Limit the Rows Retrieved by a Query
1. Which two clauses of the SELECT statement facilitate selection and projection?
A. SELECT, FROM
B. ORDER BY, WHERE
C. SELECT, WHERE
D. SELECT, ORDER BY
2. Choose the query that extracts the LAST_NAME, JOB_ID, and SALARY values from the
EMPLOYEES table for records having JOB_ID values of either SA_REP or MK_MAN and
having SALARY values in the range of $1000 to $4000. The SELECT and FROM clauses are
SELECT LAST_NAME, JOB_ID, SALARY FROM EMPLOYEES:
A. WHERE JOB_ID IN ('SA_REP','MK_MAN')
AND SALARY > 1000 AND SALARY < 4000;
B. WHERE JOB_ID IN ('SA_REP','MK_MAN')
AND SALARY BETWEEN 1000 AND 4000;
C. WHERE JOB_ID LIKE 'SA_REP%' AND 'MK_MAN%'
AND SALARY > 1000 AND SALARY < 4000;
D. WHERE JOB_ID = 'SA_REP'
AND SALARY BETWEEN 1000 AND 4000
OR JOB_ID='MK_MAN';
3. Which of the following WHERE clauses contains an error? The SELECT and FROM clauses are
SELECT * FROM EMPLOYEES:
A. WHERE HIRE_DATE IN ('02-JUN-2004');
B. WHERE SALARY IN ('1000','4000','2000');
C. WHERE JOB_ID IN (SA_REP,MK_MAN);
D. WHERE COMMISSION_PCT BETWEEN 0.1 AND 0.5;

Self Test

161

4. Choose the WHERE clause that extracts the DEPARTMENT_NAME values containing the
character literal "er" from the DEPARTMENTS table. The SELECT and FROM clauses are
SELECT DEPARTMENT_NAME FROM DEPARTMENTS:
A. WHERE DEPARTMENT_NAME IN ('%e%r');
B. WHERE DEPARTMENT_NAME LIKE '%er%';
C. WHERE DEPARTMENT_NAME BETWEEN 'e' AND 'r';
D. WHERE DEPARTMENT_NAME CONTAINS 'e%r';
5. Which two of the following conditions are equivalent to each other?
A. WHERE COMMISSION_PCT IS NULL
B. WHERE COMMISSION_PCT = NULL
C. WHERE COMMISSION_PCT IN (NULL)
D. WHERE NOT(COMMISSION_PCT IS NOT NULL)
6. Which three of the following conditions are equivalent to each other?
A. WHERE SALARY <=5000 AND SALARY >=2000
B. WHERE SALARY IN (2000,3000,4000,5000)
C. WHERE SALARY BETWEEN 2000 AND 5000
D. WHERE SALARY > 1999 AND SALARY < 5001
E. WHERE SALARY >=2000 AND <=5000

Sort the Rows Retrieved by a Query
7. Choose one false statement about the ORDER BY clause.
A. When using the ORDER BY clause, it always appears as the last clause in a SELECT statement.
B. The ORDER BY clause may appear in a SELECT statement that does not contain a
WHERE clause.
C. The ORDER BY clause specifies one or more terms by which the retrieved rows are sorted.
These terms can only be column names.
D. Positional sorting is accomplished by specifying the numeric position of a column as it appears in the SELECT list, in the ORDER BY clause.
8. The following query retrieves the LAST_NAME, SALARY, and COMMISSION_PCT values
for employees whose LAST_NAME begins with the letter R. Based on the following query,
choose the ORDER BY clause that first sorts the results by the COMMISSION_PCT column,

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listing highest commission earners first, and then sorts the results in ascending order by the
SALARY column. Any records with NULL COMMISSION_PCT must appear last:
SELECT LAST_NAME, SALARY, COMMISSION_PCT
FROM EMPLOYEES
WHERE LAST_NAME LIKE 'R%'
A. ORDER BY COMMISSION_PCT DESC, 2;
B. ORDER BY 3 DESC, 2 ASC NULLS LAST;
C. ORDER BY 3 DESC NULLS LAST, 2 ASC;
D. ORDER BY COMMISSION_PCT DESC, SALARY ASC;

Ampersand Substitution
9. The DEFINE command explicitly declares a session-persistent substitution variable with a
specific value. How is this variable referenced in an SQL statement? Consider an expression
that calculates tax on an employee’s SALARY based on the current tax rate. For the following
session-persistent substitution variable, which statement correctly references the TAX_RATE
variable?
DEFINE TAX_RATE=0.14
A. SELECT SALARY * :TAX_RATE TAX FROM EMPLOYEES;
B. SELECT SALARY * &TAX_RATE TAX FROM EMPLOYEES;
C. SELECT SALARY * :&&TAX TAX FROM EMPLOYEES;
D. SELECT SALARY * TAX_RATE TAX FROM EMPLOYEES;
10. When using ampersand substitution variables in the following query, how many times will you
be prompted to input a value for the variable called JOB the first time this query is executed?
SELECT FIRST_NAME, '&JOB'
FROM EMPLOYEES
WHERE JOB_ID LIKE '%'||&JOB||'%'
AND '&&JOB' BETWEEN 'A' AND 'Z';
A. 0
B. 1
C. 2
D. 3

LAB QUESTION
Using SQL Developer or SQL*Plus, connect to the OE schema and complete the following tasks.

Lab Question

163

A customer requires a hard disk drive and a graphics card for her personal computer. She is willing
to spend between $500 and $800 on the disk drive but is unsure about the cost of a graphics card. Her
only requirement is that the resolution supported by the graphics card should be either 1024×768 or
1280×1024. As the sales representative, you have been tasked to write one query that searches the
PRODUCT_INFORMATION table where the PRODUCT_NAME value begins with HD (hard
disk) or GP (graphics processor) and their list prices. Remember the hard disk list prices must be
between $500 and $800 and the graphics processors need to support either 1024×768 or 1280×1024.
Sort the results in descending LIST_PRICE order.

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SELF TEST ANSWERS
Limit the Rows Retrieved by a Query
✓ C. The SELECT clause facilitates projection by specifying the list of columns to be
1. ®
projected from a table, whilst the WHERE clause facilitates selection by limiting the rows
retrieved based on its conditions.
®
˚ A, B, and D are incorrect because the FROM clause specifies the source of the rows being
projected and the ORDER BY clause is used for sorting the selected rows.
✓ B. The IN operator efficiently tests whether the JOB_ID for a particular row is either SA_
2. ®
REP or MK_MAN, whilst the BETWEEN operator efficiently measures whether an employee’s
SALARY value falls within the required range.
®
˚ A and C exclude employees who earn a salary of $1000 or $4000, since these SALARY
values are excluded by the inequality operators. C also selects JOB_ID values like SA_REP%
and MK_MAN%, potentially selecting incorrect JOB_ID values. D is half right. The first half
returns the rows with JOB_ID equal to SA_REP having SALARY values between $1000 and
$4000. However, the second part (the OR clause), correctly tests for JOB_ID equal to MK_
MAN but ignores the SALARY condition.
✓ C. The character literals being compared to the JOB_ID column by the IN operator must
3. ®
be enclosed by single quotation marks.
®
˚ A, B, and D are syntactically correct. Notice that B does not require quotes around the
numeric literals. Having them, however, does not cause an error.
✓ B. The LIKE operator tests the DEPARTMENT_NAME column of each row for values
4. ®
that contain the characters "er". The percentage symbols before and after the character literal
indicate that any characters enclosing the "er" literal are permissible.
®
˚ A and C are syntactically correct. A uses the IN operator, which is used to test set
membership. C tests whether the alphabetic value of the DEPARTMENT_NAME column is
between the letter "e" and the letter "r." Finally, D uses the word "contains," which cannot be
used in this context.
✓ A and D. The IS NULL operator correctly evaluates the COMMISSION_PCT column for
5. ®
NULL values. D uses the NOT operator to negate the already negative version of the IS NULL
operator, IS NOT NULL. Two negatives return a positive, and therefore A and D are equivalent.
®
˚ B and C are incorrect since NULL values cannot be tested by the equality operator or the
IN operator.

Self Test Answers

165

✓ A, C, and D. Each of these conditions tests for SALARY values in the range of $2000 to
6. ®
$5000.
®
˚ B and E are incorrect. B excludes values like $2500 from its set, and E is illegal since it is
missing the SALARY column name reference after the AND operator.

Sort the Rows Retrieved by a Query
✓ C. The terms specified in an ORDER BY clause can include column names, positional
7. ®
sorting, numeric values, and expressions.
®
˚ A, B, and D are true.
✓ C. Positional sorting is performed, and the third term in the SELECT list, COMMISSION_
8. ®
PCT, is sorted first in descending order, and any NULL COMMISSION_PCT values are listed
last. The second term in the SELECT list, SALARY, is sorted next in ascending order.
®
˚ A, B, and D are incorrect. A does not specify what to do with NULL COMMISSION_
PCT values, and the default behavior during a descending sort is to list NULLS FIRST. B
applies the NULLS LAST modifier to the SALARY column instead of the COMMISSION_
PCT column, and D ignores NULLS completely.

Ampersand Substitution
✓ B. A session-persistent substitution variable may be referenced using an ampersand symbol
9. ®
from within any SQL statement executed in that session.
®
˚ A, C, and D are incorrect. A and D attempt to reference the substitution variable using
a colon prefix to its name and the variable name on its own. These are invalid references to
substitution variables in SQL. C references a variable called TAX and not the variable
TAX_RATE.
✓ D. The first time this statement is executed, two single ampersand substitution variables are
10. ®
encountered before the third double ampersand substitution variable. If the first reference on
line one of the query contained a double ampersand substitution, you would only be prompted
to input a value once.
®
˚ A, B, and C are incorrect since you are prompted thrice to input a value for the JOB
substitution variable. In subsequent executions of this statement in the same session you will
not be prompted to input a value for this variable.

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LAB ANSWER
Using SQL Developer or SQL*Plus, connect to the OE schema and complete the following tasks.
You are required to query the PRODUCT_INFORMATION table in the OE schema for the
PRODUCT_NAME and LIST_PRICE columns. The rows selected must conform to either of two
conditions. The first condition is that the PRODUCT_NAME must begin with the characters 'HD'
and their LIST_PRICE must fall in the range between $500 and $800. Alternately, the row can conform to the second condition that the PRODUCT_NAME must begin with the characters 'GP' and
contain the characters '1024'. Finally, the results must be sorted in descending LIST_PRICE order.
1. Start SQL Developer and connect to the OE schema.
2. The SELECT clause is
SELECT PRODUCT_NAME, LIST_PRICE
3. The FROM clause is
FROM PRODUCT_INFORMATION
4. The first condition is
PRODUCT_NAME LIKE 'HD%' AND LIST_PRICE BETWEEN 500 AND 800
5. The second condition is
PRODUCT_NAME LIKE 'GP%1024%'
6. Since either the first or second condition must be fulfilled by a row in order to be retrieved,
these two conditions must be separated with the Boolean OR operator.
7. The WHERE clause is
WHERE (PRODUCT_NAME LIKE 'HD%' AND LIST_PRICE BETWEEN 500 AND 800)
OR (PRODUCT_NAME LIKE 'GP%1024%')
8. The ORDER BY clause is
ORDER BY LIST_PRICE DESC
9. Executing this statement returns the set of results matching this pattern as shown in
the illustration:

Lab Answer

167

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4
Single-Row
Functions

CERTIFICATION OBJECTIVES
4.01

Describe Various Types of Functions
Available in SQL

4.02

Use Character, Number, and Date
Functions in SELECT Statements

✓
Q&A

Two-Minute Drill
Self Test

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

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F

unctions are a wonderful extension to SQL and provide a first glimpse of the procedural
capabilities Oracle supports. Procedural languages allow a rich degree of programming
that yields an almost unlimited range of data manipulation possibilities. Oracle server
implements a proprietary procedural language called PL/SQL, or procedural SQL. A range of
named programmatic objects may be constructed using PL/SQL. These include procedures,
functions, and packages. Although writing PL/SQL is relatively straightforward, a thorough
understanding of SQL is a prerequisite and the focus of this guide. The functions discussed in this
chapter are confined to PL/SQL programs packaged and supplied by Oracle as built-in features.

CERTIFICATION OBJECTIVE 4.01

Describe Various Types of Functions
Available in SQL
SQL functions are broadly divided into those that calculate and return a value for
every row in a data set and those that return a single aggregated value for all rows.
The following two areas are explored:
■ Defining a function
■ Types of functions

Defining a Function
A function is a program written to optionally accept input parameters, perform an
operation, or return a single value. A function returns only one value per execution.
Three important components form the basis of defining a function. The first
is the input parameter list. It specifies zero or more arguments that may be passed
to a function as input for processing. These arguments or parameters may be of
differing data types, and some are mandatory while others may be optional. The
second component is the data type of its resultant value. Upon execution, only
one value is returned by the function. The third encapsulates the details of the
processing performed by the function and contains the program code that optionally
manipulates the input parameters, performs calculations and operations, and
generates a return value.
A function is often described as a black box that takes an input, performs a
calculation, and returns a value as illustrated by the following equation. Instead of

Describe Various Types of Functions Available in SQL

171

focusing on their implementation details, you are encouraged to concentrate on the
features that built-in functions provide.
F(x, y, z, …) = result;
Functions may be nested within other functions, such as F1(x, y, F2(a, b), z),
where F2, which takes two input parameters, a and b, and forms the third of four
parameters submitted to F1. Functions can operate on any available data types,
the most popular being character, date, and numeric data. These operands may be
columns or expressions.
As an example, consider a function that calculates a person’s age. The AGE
function takes one date input parameter, which is the person’s birthday. The result
returned by the AGE function is a number representing a person’s age. The black box
calculation involves obtaining the difference in years between the current date and
the birthday input parameter.

Operating on Character Data
Character data or strings are versatile since they facilitate the storage of almost any
type of data. Functions that operate on character data are broadly classified as case
conversion and character manipulation functions. The following string built-in functions
are examined in detail later in this chapter, but a brief description is provided here.
LOWER, UPPER, and INITCAP are the case conversion functions that convert
a given character column, literal, or expression into lowercase, uppercase, or initial
case, respectively:
lower('SQL') = sql
upper('sql') = SQL
initcap('sql') = Sql

The character manipulation functions are exceptionally powerful and include the
LENGTH, CONCAT, SUBSTR, INSTR, LPAD, RPAD, TRIM, and REPLACE
functions.
The LENGTH(string) function uses a character string as an input parameter and
returns a numeric value representing the number of characters present in that string:
length('A short string') = 14

The CONCAT(string 1, string 2) function takes two strings and concatenates or
joins them in the same way that the concatenation operator || does:
concat('SQL is',' easy to learn.') = SQL is easy to learn.

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The SUBSTR(string, start position, number of characters) function accepts three
parameters and returns a string consisting of the number of characters extracted from
the source string, beginning at the specified start position:
substr('http://www.domain.com',12,6) = domain

The INSTR(source string, search item, [start position],[nth occurrence of search
item]) function returns a number that represents the position in the source string,
beginning from the given start position, where the nth occurrence of the search item
begins:
instr('http://www.domain.com','.',1,2) = 18

The LPAD(string, length after padding, padding string) and RPAD(string, length after
padding, padding string) functions add a padding string of characters to the left or
right of a string until it reaches the specified length after padding.
The TRIM function literally trims off leading or trailing (or both) character
strings from a given source string:
rpad('#PASSWORD#',11,'#') = #PASSWORD##
lpad('#PASSWORD#',11,'#') = ##PASSWORD#
trim('#' from '#PASSWORD#') = PASSWORD

The REPLACE(string, search item, replacement item) function locates the search
item in a given string and replaces it with the replacement item, returning a string
with replaced values:
replace('#PASSWORD#','WORD','PORT') = #PASSPORT#

Operating on Numeric Data
Many numeric built-in functions are available. Some calculate square roots, perform
exponentiation, and convert numbers into hexadecimal format. There are too many
to mention, and many popular mathematical, scientific, and financial calculations
have been exposed as built-in functions by Oracle.
Three common numeric functions, examined later in this chapter, are ROUND,
TRUNC, and MOD. ROUND(number, decimal precision) facilitates rounding off a
number to the lowest or highest value given a decimal precision format:
round(42.39,1) = 42.4

The TRUNC(number, decimal precision) function drops off or truncates the
number given a decimal precision value:
trunc(42.39,1) = 42.3

Describe Various Types of Functions Available in SQL

173

The MOD(dividend, divisor) returns the remainder of a division operation:
mod(42,10) = 2

Operating on Date Information
Working with date values may be challenging. Performing date arithmetic that
accommodates leap years and variable month lengths can be frustrating and
error prone. Oracle addresses this challenge by providing native support for date
arithmetic and several built-in date functions such as MONTHS_BETWEEN,
ADD_MONTHS, LAST_DAY, NEXT_DAY, SYSDATE, ROUND, and TRUNC.
The MONTHS_BETWEEN(date 1, date 2) function returns the number of
months between two dates, while the ADD_MONTHS(date 1, number of months)
returns the date resulting from adding a specified number of months to a date:
months_between('01-FEB-2008','01-JAN-2008') = 1
add_months('01-JAN-2008',1) = 01-FEB-2008

The LAST_DAY(date 1) function returns the last day of the month that the
specified date falls into, while the NEXT_DAY(date 1, day of the week) returns the
date on which the next specified day of the week falls after the given date:
last_day('01-FEB-2008') = 29-FEB-2008
next_day('01-FEB-2008','Friday') = 08-FEB-2008

The SYSDATE function takes no parameters and returns a date value that
represents the current server date and time. ROUND(date, date precision format) and
TRUNC(date, date precision format) round and truncate a given date value to the
nearest date precision format like day, month, or year:
sysdate = 17-DEC-2007
round(sysdate,'month') = 01-JAN-2008
trunc(sysdate,'month') = 01-DEC-2007

Single-row functions are used in almost every query issued by analysts,
developers, and administrators. When searching for character data, the TRIM
function is frequently used to eliminate extra spaces that occur in character
fields.The case conversion functions are used to standardize the column data.
This facilitates more accurate and efficient searching since the case in which
character data is captured is often inconsistent.

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Types of Functions
Two broad types of functions operating on single and multiple rows, respectively,
are discussed next. This distinction is vital to understanding the larger context in
which functions are used. Oracle continuously strives to ensure that its commercial
interpretation of SQL conforms to international standards. This facilitates ease of
systems and skills migration across vendors and suppliers of RDBMS software. Oracle’s
implementation of SQL is compliant with the ANSI:1999 (American National
Standards Institute) standard for SQL. More recently, it claimed partial compliance
to the SQL:2003 standard endorsed by both ISO (International Organization for
Standardization) and ANSI. SQL functions have been standardized, and Oracle has
documented those that are fully or partially compliant to the SQL:2003 standard.

Single-Row Functions
There are several categories of single-row functions including character, numeric,
date, conversion, and general. The focus of this chapter is on the character, numeric,
and date single-row functions. These are functions that operate on one row of a
dataset at a time. If a query selects 10 rows, the function is executed 10 times, once
per row with the values from that row as input to the function.
As Figure 4-1 shows, two columns of the REGIONS table have been selected along
with an expression using the LENGTH function with the REGION_NAME column.
The length of the REGION_NAME column is calculated for each row, proving
that the function has executed four separate times, returning one result per row.
Single-row functions manipulate the data items in a row to extract and format
them for display purposes. The input values to a single-row function can be userspecified constants or literals, column data, variables, or expressions optionally
supplied by other nested single-row functions. The nesting of single-row functions
is a commonly used technique. Functions can return a value with a different data
type from its input parameters. As Figure 4-1 demonstrates, the LENGTH function
accepts one character input parameter and returns a numeric output.
Conversion functions like TO_CHAR, TO_NUMBER, and TO_DATE are
discussed in Chapter 5. They change the data type of column data or expressions
allowing other functions to operate on them. The general functions are also discussed
in Chapter 5. They simplify working with NULL values and facilitate conditional
logic within a SELECT statement. These include the NVL, NVL2, NULLIF,
COALESCE, CASE, and DECODE functions.
Apart from their inclusion in the SELECT list of a SQL query, single-row
functions may be used in the WHERE and ORDER BY clauses. Assume there
is a requirement to list rows from the REGIONS table where the length of the
REGION_NAME column data is at least five characters long. There is a further

Describe Various Types of Functions Available in SQL

FIGURE 4-1

175

A single-row function

need for this list to be sorted in alphabetic order based on the value of the last
character in the REGION_NAME column. The WHERE clause is shown here:
where length(region_name) > 4

To obtain the last character in a string, the SUBSTR function is used with the
REGION_NAME column as the source string. The length of the REGION_NAME
is used as the start position, producing the following ORDER BY clause:
order by substr(region_name, length(region_name),1)

As Figure 4-2 shows, only three of the four regions are returned, and the list
is sorted in alphabetic order based on the last character in the REGION_NAME
column for each row.

Multiple-Row Functions
As the name suggests, this category of functions operates on more than one row at a
time. Typical uses of multiple-row functions include calculating the sum or average of
the numeric column values or counting the total number of records in sets. These are
sometimes known as aggregation or group functions and are explored in Chapter 7.

176

Chapter 4:

FIGURE 4-2

Single-Row Functions

Functions in SELECT, WHERE, and ORDER BY clauses

Single-row functions are
executed for each row in the selected
data set.This concept is implicitly tested
via practical examples in the exam.
Functions always return only one value of a
predetermined data type.They may accept
zero or more parameters of differing data
types.The single-row character functions

like LENGTH, SUBSTR, and INSTR are
frequently used together, and a thorough
understanding of these is required.
Remember that input parameters that can
be implicitly converted to the data types
required by functions are acceptable to
Oracle.

Use Character, Number, and Date Functions in SELECT Statements

177

CERTIFICATION OBJECTIVE 4.02

Use Character, Number, and Date Functions
in SELECT Statements
This section conducts a detailed investigation of the single-row functions introduced
earlier. A structured approach will be taken that includes function descriptions,
syntax rules, parameter descriptions, and usage examples. The character case
conversion functions are examined, followed by the character manipulation
functions. Next, the numeric functions are examined, and the section concludes
with a discussion of the date functions.

Using Character Case Conversion Functions
Character data may be saved in tables from numerous sources, including application
interfaces and batch programs. It is not safe to assume that character data has been
committed in a consistent manner. The character case conversion functions serve two
important purposes. They may be used first, to modify the appearance of a character
data item for display purposes and second, to render them consistent for comparison
operations. It is simpler to search for a string using a consistent case format instead
of testing every permutation of uppercase and lowercase characters that could match
the string. It is important to remember that these functions do not alter the data
stored in tables. They still form part of the read-only SQL query.
The character functions discussed next expect string parameters. These may be
any string literal, character column value, or expression resulting in a character
value. If it is a numeric or a date value, it is implicitly converted into a string.

The LOWER Function
The LOWER function converts a string of characters into their lowercase
equivalents. It does not add extra characters or shorten the length of the initial
string. Uppercase characters are converted into their lowercase equivalents.
Numeric, punctuation, or special characters are ignored.
The LOWER function can take only one parameter. Its syntax is LOWER(s). The
following queries illustrate the usage of this function:
Query 1: select lower(100) from dual
Query 2: select lower(100+100) from dual
Query 3: select lower('The SUM '||'100+100'||' =

200') from dual

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Queries 1 and 2 return the strings 100 and 200, respectively. The parameter to the
LOWER function in query 3 is a character expression and the string returned by the
function is “The sum 100 + 100 = 200.”
Query 4: select lower(SYSDATE) from dual
Query 5: select lower(SYSDATE+2) from dual

Assume that the current system date is: 17-DEC-2007. Queries 4 and 5 return
the strings 17-dec-2007 and 19-dec-2007, respectively. The date expressions are
evaluated and implicitly converted into character data before the LOWER function
is executed.
As Figure 4-3 shows, the LOWER function is used in the WHERE clause to
locate the records with the lowercase letters “u” and “r” adjacent to each other in
the LAST_NAME field.
Consider writing an alternative query to return the same results without using the
LOWER function. It could be done as follows:
select first_name, last_name
from employees
where last_name like '%ur%'

FIGURE 4-3

The LOWER function

Use Character, Number, and Date Functions in SELECT Statements

179

or last_name like '%UR%'
or last_name like '%uR%'
or last_name like '%Ur%'

This query works but is cumbersome, and the number of OR clauses required
increases exponentially as the length of the search string increases.

The UPPER Function
The UPPER function is the logical opposite of the LOWER function and converts a
string of characters into their uppercase equivalents. It does not add extra characters
or shorten the length of the initial string. All lowercase characters are converted
into their uppercase equivalents. Numeric, punctuation, or special characters are
ignored.
The UPPER function takes only one parameter. Its syntax is UPPER(s). The
following queries illustrate the usage of this function:
Query 1: select upper(1+2.14) from dual
Query 2: select upper(SYSDATE) from dual

Query 1 returns the string 3.14. The parameter to the UPPER function in query 2
is SYSDATE, which returns the current system date. Since this value is returned in
uppercase by default, no case conversion is performed.
The UPPER function is used in Figure 4-4 to extract the rows from the COUNTRIES
table where the COUNTRY_NAME values contain the letters “U,” “S,” and “A” in
that order. The letters “U,” “S,” and “A” do not have to be adjacent to each other.
Writing an alternative query to return the same results without using the UPPER
or LOWER functions could be done using a query with eight conditions:
SELECT * FROM COUNTRIES
WHERE country_name like '%u%s%a%'
or country_name like '%u%S%a%'
or country_name like '%U%s%a%'
or country_name like '%U%S%a%'

or
or
or
or

country_name
country_name
country_name
country_name

like
like
like
like

'%u%s%A%'
'%u%S%A%'
'%U%s%A%'
'%U%S%A%'

This query works but is cumbersome. The number of OR clauses required
increases exponentially as the length of the search string increases.

The INITCAP Function
The INITCAP function converts a string of characters into capitalized case. It is
often used for data presentation purposes. The first letters of each word in the string
are converted to their uppercase equivalents, while the remaining letters of each
word are converted to their lowercase equivalents. A word is usually a string of
adjacent characters separated by a space or underscore, but other characters such as

180

Chapter 4:

FIGURE 4-4

Single-Row Functions

The UPPER function

the percentage symbol, exclamation mark, or dollar sign are valid word separators.
Punctuation or special characters are regarded as valid word separators.
The INITCAP function can take only one parameter. Its syntax is INITCAP(s).
The following queries illustrate the usage of this function:
Query 1: select initcap(21/7) from dual
Query 2: select initcap(SYSDATE) from dual
Query 3: select initcap('init cap or init_cap or init%cap') from dual

Query 1 returns the quotient 3 as a string. Query 2 returns the character string
value of the current system date, with the month portion changed from uppercase
to initial case. Assuming that the current system date is 17-DEC-2007, query 2
therefore returns 17-Dec-2007. Query 3 returns Init Cap Or Init_Cap Or Init%Cap.
The queries in Figure 4-5 select the LAST_NAME and JOB_ID values from the
EMPLOYEES table for those employees with LAST_NAME values starting with
the letter “H.” The first query applies the INITCAP function to the entire SELECT
clause. The second query shows how the INITCAP function is applied separately to
each character component. Both queries yield identical results.

Use Character, Number, and Date Functions in SELECT Statements

FIGURE 4-5

The INITCAP function

EXERCISE 4-1
Using the Case Conversion Functions
Retrieve a list of all FIRST_NAME and LAST_NAME values from the
EMPLOYEES table where FIRST_NAME contains the character string “li.”
1. Start SQL Developer and connect to the HR schema.
2. The SELECT clause is
SELECT FIRST_NAME, LAST_NAME
3. The FROM clause is
FROM EMPLOYEES

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4. The WHERE clause must compare the FIRST_NAME column values with a
pattern of characters containing all possible case combinations of the string
“li.” Therefore, if the FIRST_NAME contains the character strings “LI,” “Li,”
“lI,” or “li,” that row must be retrieved.
5. The LIKE operator is used for character matching, and four combinations
can be extracted with four WHERE clauses separated by the OR keyword.
However, the case conversion functions can simplify the condition. If the
LOWER function is used on the FIRST_NAME column, the comparison can
be done with one WHERE clause condition. The UPPER or INITCAP functions could also be used.
6. The WHERE clause is
WHERE LOWER(FIRST_NAME) LIKE '%li%'
7. Executing this statement returns employees’ names containing the characters
“li” as shown in this illustration:

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Using Character Manipulations Functions
Some of the most powerful features to emerge from Oracle are the character
manipulation functions. Their usefulness in data manipulation is almost without peer,
and many seasoned technical professionals whip together a quick script to massage
data items with SQL character manipulation functions. Nesting these functions
is common. The concatenation operator (||) is generally used instead of the
CONCAT function. The LENGTH, INSTR, SUBSTR, and REPLACE functions
often find themselves in each other’s company as do RPAD, LPAD, and TRIM.

The CONCAT Function
The CONCAT function joins two character literals, columns, or expressions to
yield one larger character expression. Numeric and date literals are implicitly cast as
characters when they occur as parameters to the CONCAT function. Numeric or date
expressions are evaluated before being converted to strings ready to be concatenated.
The CONCAT function takes two parameters. Its syntax is CONCAT(s1, s2),
where s1 and s2 represent string literals, character column values, or expressions
resulting in character values. The following queries illustrate the usage of this function:
Query 1: select concat(1+2.14,' approximates pi') from dual
Query 2: select concat('Today is:',SYSDATE) from dual

Query 1 returns the string “3.14 approximates pi.” The numeric expression is
evaluated to return the number 3.14. This number is automatically changed into
the character string “3.14,” which is concatenated to the character literal in the
second parameter. The second parameter to the CONCAT function in query 2 is
SYSDATE, which returns the current system date. This value is implicitly converted
to a string to which the literal in the first parameter is concatenated. If the system
date is 17-DEC-2007, query 2 returns the string “Today is 17-DEC-2007.”
Consider using the CONCAT function to join three terms to return one
character string. Since CONCAT takes only two parameters, it is only possible
to join two terms with one instance of this function. The solution is to nest the
CONCAT function within another CONCAT function, as shown here:
select concat('Outer1 ', concat('Inner1',' Inner2')) from dual;

The first CONCAT function has two parameters: the first is the literal “Outer1,”
while the second is a nested CONCAT function. The second CONCAT function
takes two parameters: the first is the literal “Inner1,” while the second is the literal
“Inner2.” This query results in the following string: Outer1 Inner1 Inner2. Nested
functions are described in detail in Chapter 5.

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The CONCAT function was used in Figure 4-6 to extract the rows from the
EMPLOYEES table where the DEPARTMENT_ID=100. The objective was to
produce a single string literal output from the CONCAT function of the format
FIRST_NAME LAST_NAME earns SALARY.
This simple task was transformed into a complex four-level-deep nested set of
function calls. As the second example in Figure 4-6 demonstrates, the concatenation
operator performs the equivalent task in a simpler manner.

The LENGTH Function
The LENGTH function returns the number of characters that constitute a character
string. This includes character literals, columns, or expressions. Numeric and
date literals are automatically cast as characters when they occur as parameters to
FIGURE 4-6

The CONCAT function

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185

the LENGTH function. Numeric or date expressions are evaluated before being
converted to strings ready to be measured. Blank spaces, tabs, and special characters
are all counted by the LENGTH function.
The LENGTH function takes only one parameter. Its syntax is LENGTH(s),
where s represents any string literal, character column value, or expression resulting
in a character value. The following queries illustrate the usage of this function:
Query 1: select length(1+2.14||' approximates pi') from dual
Query 2: select length(SYSDATE) from dual

Query 1 returns the number 20. The numeric expression is evaluated to return
the number 3.14. This number is cast as the character string “3.14,” which is then
concatenated to the character literal “approximates pi.” The resultant character
string contains 20 characters. Query 2 first evaluates the SYSDATE function, which
returns the current system date. This value is automatically converted to a character
string whose length is then determined. Assuming that the system date returned is
17-DEC-07, query 2 returns value 9.
The LENGTH function is used in Figure 4-7 to extract the COUNTRY_NAME
value with length greater than ten characters from the COUNTRIES table.
FIGURE 4-7

The LENGTH function

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The LPAD and RPAD Functions
The LPAD and RPAD functions, also known as left pad and right pad functions,
return a string padded with a specified number of characters to the left or right of the
source string respectively. The character strings used for padding include character
literals, column values, or expressions. Numeric and date literals are implicitly cast as
characters when they occur as parameters to the LPAD or RPAD functions. Numeric
or date expressions are evaluated before being converted to strings destined for
padding. Blank spaces, tabs, and special characters may be used as padding characters.
The LPAD and RPAD functions take three parameters. Their syntaxes are
LPAD(s, n, p) and RPAD(s, n, p), where s represents the source string, n represents
the final length of the string returned, and p specifies the character string to be used
as padding. If LPAD is used, the padding characters p are added to the left of the
source string s until it reaches length n. If RPAD is used, the padding characters p
are added to the right of the source string s until it reaches length n. Note that if
the parameter n is smaller than or equal to the length of the source string s, then no
padding occurs and only the first n characters of s are returned.
The following queries illustrate the usage of this function:
Query
Query
Query
Query

1:
2:
3:
4:

select
select
select
select

lpad(1000+200.55,14,'*') from dual
rpad(1000+200.55,14,'*') from dual
lpad(SYSDATE,14,'$#') from dual
rpad(SYSDATE,4,'$#') from dual

Query 1 returns a 14-character string: *******1200.55. The numeric expression
is evaluated to return the number 1200.55. This number is cast as the string
“1200.55” of length seven (including the decimal point). To achieve the final length
of 14 characters, 7 asterisks are left padded to the string. Query 2 returns the string
“1200.55*******.”
The LPAD function in query 3 has a target string length of 14 characters. Assume
that SYSDATE returns a 9-character date value: 17-DEC-07. This date is converted
into a string, and the padding string is systematically applied to reach the target
length. It returns: $#$#$17-DEC-07. Note that although the padding string consists
of two characters ($#), the string was not applied evenly since there are three dollar
symbols and two hash symbols. This is because LPAD and RPAD will pad the
source string as much as possible with the padding string until the target length is
reached. The RPAD function in query 4 has a target length of 4 characters, but the
SYSDATE function alone returns a 9-character value. Therefore no padding occurs
and, assuming the current system date is 17-DEC-07, the first four characters of the
converted date are returned: 17-D.
The LPAD and RPAD functions are used in Figure 4-8 to format the results in a
neater and more presentable manner. The results returned by this query are identical

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FIGURE 4-8

187

The LPAD and RPAD functions

to those in Figure 4-6 but have been rendered more user friendly using the LPAD
and RPAD functions.

The TRIM Function
The TRIM function removes characters from the beginning or end of character
literals, columns or expressions to yield one potentially shorter character item.
Numeric and date literals are automatically cast as characters when they occur as
parameters to the TRIM function. Numeric or date expressions are evaluated first
before being converted to strings ready to be trimmed.
The TRIM function takes a parameter made up of an optional and a mandatory
component. Its syntax is TRIM([trailing|leading|both] trimstring from s). The string
to be trimmed (s) is mandatory. The following points list the rules governing the use
of this function:
■ TRIM(s) removes spaces from both sides of the input string.
■ TRIM(trailing trimstring from s) removes all occurrences of trimstring from the

end of the string s if it is present.

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■ TRIM(leading trimstring from s) removes all occurrences of trimstring from the

beginning of the string s if it is present.
■ TRIM(both trimstring from s) removes all occurrences of trimstring from the

beginning and end of the string s if it is present.
The following queries illustrate the usage of this function:
Query 1: select trim(trailing 'e' from 1+2.14||' is pie') from dual
Query 2: select trim(both '*' from '*******Hidden*******') from dual
Query 3: select trim(1 from sysdate) from dual

Query 1 evaluates the numeric expression to return the number 3.14. This
number is then cast as the character string “3.14,” which is then concatenated to
the character literal to construct the string “3.14 is pie.” The TRIM function then
removes any occurrences of the character “e” from the end of the string to return
“3.14 approximates pi.” Query 2 peels away all occurrences of the asterisk trim
character from the beginning and end of the character literal and returns the string
“Hidden.” Note that although one trim character is specified, multiple occurrences
will be trimmed if they are consecutively present. Query 3 has two interesting
aspects. The trim character is not enclosed in quotes and is implicitly converted
to a character. The SYSDATE function returns the current system date, which
is assumed to be 17-DEC-07. Since no keyword is specified for trailing, leading,
or both trim directions, the default of both applies. Therefore all occurrences of
character 1 at the beginning or ending of the date string are trimmed resulting in
7-DEC-07 being returned.
The TRIM function used in Figure 4-9 does not appear to do anything but
closer examination reveals a common practical use for it. As discussed earlier, data
is frequently entered into application database tables through a variety of sources.
It may happen that spaces are accidentally entered and saved in character fields
unintentionally. The contrived trim string in the WHERE clause simulates the
LAST_NAME field padded with spaces. This could hinder searching for employees
with LAST_NAME values of Smith. Trimming the space padded LAST_NAME
field enables accurate searching and eliminates the risk of unintended spaces that
may be present in character data. Remember that when no parameters other than
the string s are specified to the TRIM function then its default behavior is to
trim(both ' ' from s).

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FIGURE 4-9

189

The TRIM function

The INSTR Function (In-string)
The INSTR function locates the position of a search string within a given string. It
returns the numeric position at which the nth occurrence of the search string begins,
relative to a specified start position. If the search string is not present the INSTR
function returns zero.
Numeric and date literals are implicitly cast as characters when they occur as
parameters to the INSTR function. Numeric or date expressions are first evaluated
before being converted to strings ready to be searched.
The INSTR function takes four parameters made up of two optional and two
mandatory arguments. The syntax is INSTR(source string, search string, [search start
position], [nth occurrence]). The default value for the search start position is 1 or the
beginning of the source string. The default value for the nth occurrence is 1 or the
first occurrence. The following queries illustrate the INSTR function with numeric
and date expressions:
Query 1: select instr(3+0.14,'.') from dual
Query 2: select instr(sysdate, 'DEC') from dual

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Query 1 evaluates the numeric expression to return the number 3.14. This
number is implicitly cast as the string 3.14. The period character is searched for
and the first occurrence of it occurs at position 2. Query 2 evaluates the SYSDATE
function and converts the returned date into a string. Assume that the current
system date is 17-DEC-07. The first occurrence of the characters DEC occurs at
position 4. Consider the following queries with character data illustrating the default
values and the third and fourth parameters of the INSTR function:
Query 3: select instr('1#3#5#7#9#','#') from dual
Query 4: select instr('1#3#5#7#9#','#',5) from dual
Query 5: select instr('1#3#5#7#9#','#',3,4) from dual

Query 3 searches for the first occurrence of the hash symbol in the source string
beginning at position 1 and returns position 2. Query 4 has the number 5 as its third
parameter indicating that the search for the hash symbol must begin at position 5 in
the source string. The subsequent occurrence of the hash symbol is at position 6,
which is returned by the query. Query 5 has the numbers 3 and 4 as its third and
fourth parameters. This indicates that the search for the hash symbol must begin
at position 3 in the source string. Query 5 then returns the number 10, which is
the position of the fourth occurrence of the hash symbol when the search begins at
position 3.
The INSTR function used in Figure 4-10 returns records from the DEPARTMENTS
table where the DEPARTMENT_NAME values have the character n as their third
character.
The INSTR function is often used in combination with the SUBSTR function in
utility programs designed to extract encoded data from electronic data streams.

The SUBSTR Function (Substring)
The SUBSTR function extracts and returns a segment from a given source string. It
extracts a substring of a specified length from the source string beginning at a given
position. If the start position is larger than the length of the source string, null is
returned. If the number of characters to extract from a given start position is greater
than the length of the source string, the segment returned is the substring from the
start position to the end of the string.
Numeric and date literals are automatically cast as characters when they occur as
parameters to the SUBSTR function. Numeric and date expressions are evaluated
before being converted to strings ready to be searched.

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FIGURE 4-10

191

The INSTR function

The SUBSTR function takes three parameters, with the first two being
mandatory. Its syntax is SUBSTR(source string, start position, [number of characters
to extract]). The default number of characters to extract is equal to the number of
characters from the start position to the end of the source string. The following queries
illustrate the SUBSTR function with numeric and date expressions:
Query 1: select substr(10000-3,3,2) from dual
Query 2: select substr(sysdate,4,3) from dual

Query 1 evaluates the numeric expression to return the number 9997. This
number is automatically changed into the character string 9997. The search for the
substring begins at position 3 and the two characters from that position onward are
extracted, yielding the substring 97. Query 2 evaluates the SYSDATE function and
converts the date returned into a character string. Assume that the current system
date is 17-DEC-07. The search for the substring begins at position 4 and the three
characters from that position onward are extracted, yielding the substring DEC.

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Consider the following queries with character data illustrating the default behavior
of the optional parameter of the SUBSTR function:
Query 3: select substr('1#3#5#7#9#',5) from dual
Query 4: select substr('1#3#5#7#9#',5,6) from dual
Query 5: select substr('1#3#5#7#9#',-3,2) from dual

Query 3 extracts the substring beginning at position 5. Since the third parameter is
not specified, the default extraction length is equal to number of characters from and
including the start position to the end of the source string, which is 6. Therefore query 3
is equivalent to query 4 and the substring returned by both queries is 5#7#9#. Query five
has the number –3 as its start position. The negative start position parameter instructs
Oracle to commence searching 3 characters from the end of the string. Therefore start
position is three characters from the end of the string, which is position 8. The third
parameter is 2, which results in the substring #9 being returned.
The SUBSTR function used in Figure 4-11 returns records from the EMPLOYEES
table, where the first two characters in the JOB_ID values are AD. This function
has been further used in the SELECT list to extract the initial character from the
FIRST_NAME field of each employee in the result set.
FIGURE 4-11

The SUBSTR function

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193

SCENARIO & SOLUTION
You would like to search for a character string stored
in the database. The case in which it is stored is
unknown and there are potentially leading and
trailing spaces surrounding the string. Can such a
search be performed?

Yes. The simplest solution is to first TRIM the
leading and trailing spaces from the column and then
convert the column data using a case conversion
function like LOWER, UPPER, or INITCAP to
simplify the number of comparisons required in the
WHERE clause condition.

You have been asked to extract the last three
characters from the LAST_NAME column in the
EMPLOYEES table. Can such a query be performed
without using the LENGTH function?

Yes. The SUBSTR(source string, start position, number
of characters) function takes three parameters. If
the start position is set to –3, and the number of
characters parameter is set to three or is omitted, the
last three characters of the LAST_NAME column
data is retrieved. The following query may be used:
SELECT SUBSTR(LAST_NAME,–3) FROM
EMPLOYEES;

You would like to extract a consistent 10-character
string based on the SALARY column in the
EMPLOYEES table. If the SALARY value is less
than 10 characters long, zeros must be added to the
left of the value to yield a 10-character string. Is this
possible?

Yes. The LPAD function may be used as follows:
SELECT LPAD(SALARY,10,0)
FROM EMPLOYEES;

The REPLACE Function
The REPLACE function replaces all occurrences of a search item in a source string
with a replacement term and returns the modified source string. If the length of the
replacement term is different from that of the search item, then the lengths of the
returned and source strings will be different. If the search string is not found, the
source string is returned unchanged. Numeric and date literals and expressions are
evaluated before being implicitly cast as characters when they occur as parameters to
the REPLACE function.
The REPLACE function takes three parameters, with the first two being
mandatory. Its syntax is REPLACE(source string, search item, [replacement term]).
If the replacement term parameter is omitted, each occurrence of the search item is
removed from the source string. In other words, the search item is replaced by an
empty string. The following queries illustrate the REPLACE function with numeric
and date expressions:
Query 1: select replace(10000-3,'9','85') from dual
Query 2: select replace(sysdate, 'DEC','NOV') from dual

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Query 1 evaluates the numeric expression to return the number 9997, which is
cast as the character string “9997.” The search string is the character “9,” which
occurs three times in the source. Each search character is substituted with the
replacement string “85,” yielding the string “8585857.” Query 2 evaluates the
SYSDATE function and converts the date returned into a character string. Assume
that the current system date is 17-DEC-07. The search string “DEC” occurs once
in the source string and is replaced with the characters “NOV,” yielding the result
17-NOV-07. Note that this is a character string and not a date value. Consider the
following queries with character data, which illustrate the default behavior of the
optional parameter of the REPLACE function:
Query 3: select replace('1#3#5#7#9#','#','->') from dual
Query 4: select replace('1#3#5#7#9#','#') from dual

The hash symbol in query 3 is specified as the search character and the
replacement string is specified as ->. The hash symbol occurs five times in the
source, and the resultant string is: 1->3->5->7->9->. Query 4 does not specify a
replacement string. The default behavior is therefore to replace the search string
with an empty string which, in effect, removes the search character completely from
the source, resulting in the string “13579” being returned.
The REPLACE function used in Figure 4-12 returns records from the EMPLOYEES
table where the JOB_ID values are SA_MAN, but it modifies the SALARY column
by replacing each 0 with 000 and aliasing the new expression as Dream Salary.

EXERCISE 4-2
Using the Case Manipulation Functions
Envelope printing restricts the addressee field to 16 characters. Ideally, the addressee
field contains employees’ FIRST_NAME and LAST_NAME values separated by
a single space. When the combined length of an employee’s FIRST_NAME and
LAST_NAME exceeds 15 characters, the addressee field should contain their formal
name. An employee’s formal name is made up of the first letter of their FIRST_
NAME and the first 14 characters of their LAST_NAME.
You are required to retrieve a list of FIRST_NAME and LAST_NAME values
and formal names for employees where the combined length of FIRST_NAME and
LAST_NAME exceeds 15 characters.
1. Start SQL*Plus and connect to the HR schema.

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FIGURE 4-12

195

The REPLACE function

2. The formal name is constructed by concatenating the first character in the
FIRST_NAME field with a space and the first 14 characters of the LAST_
NAME field to return a string that is 16 characters long. The SUBSTR
function is used to extract the initial and surname portions.
3. The SELECT clause is
SELECT FIRST_NAME, LAST_NAME, SUBSTR(FIRST_NAME,1,1)||'
'||SUBSTR(LAST_NAME,1,14) FORMAL_NAME
4. The FROM clause is
FROM EMPLOYEES
5. The WHERE clause must limit the records returned to only those where
the combined lengths of their FIRST_NAME and LAST_NAME exceeds
15 characters.
6. The WHERE clause is
WHERE LENGTH(FIRST_NAME) + LENGTH(LAST_NAME) > 15

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7. Executing this statement returns the following set of results:

Using Numeric Functions
There is a range of built-in numeric functions provided by Oracle that rivals the
mathematical toolboxes of popular spreadsheet software packages. A significant
differentiator between numeric and other functions is that they accept and return
only numeric data. Oracle provides numeric functions for solving trigonometric,
exponentiation, and logarithmic problems, amongst others. This guide focuses on
three numeric single-row functions: ROUND, TRUNC, and MOD, discussed next.

The Numeric ROUND Function
The ROUND function performs a rounding operation on a numeric value based on
the decimal precision specified. The value returned is either rounded up or down
based on the numeric value of the significant digit at the specified decimal precision
position. If the specified decimal precision is n, the digit significant to the rounding

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197

is found (n + 1) places to the RIGHT of the decimal point. If it is negative, the digit
significant to the rounding is found n places to the LEFT of the decimal point. If
the numeric value of the significant digit is greater than or equal to 5, a “round up”
occurs, else a “round down” occurs.
The ROUND function takes two parameters. Its syntax is ROUND(source number,
decimal precision). The source number parameter represents any numeric literal, column,
or expression. The decimal precision parameter specifies the degree of rounding and is
optional. If the decimal precision parameter is absent, the default degree of rounding is
zero, which means the source is rounded to the nearest whole number.
Consider the decimal degrees listed in Table 4-1 for the number 1234.5678. The
negative decimal precision values are located to the left of the decimal point while
the positive values are found to the right.
If the decimal precision parameter is one, then the source number is rounded to
the nearest tenth. If it is two, then the source is rounded to the nearest hundredth,
and so on. The following queries illustrate the usage of this function:
Query
Query
Query
Query

1:
2:
3:
4:

select
select
select
select

round(1601.916718,1) from dual
round(1601.916718,2) from dual
round(1601.916718,-3) from dual
round(1601.916718) from dual

Query 1 has a decimal precision parameter (n) of 1, which implies that the source
number is rounded to the nearest tenth. Since the hundredths (n + 1) digit is 1
(less than 5), no rounding occurs and the number returned is 1601.9. The decimal
precision parameter in query 2 is 2, so the source number is rounded to the nearest
hundredth. Since the thousandths unit is 6 (greater than 5), rounding up occurs and
the number returned is 1601.92. The decimal precision parameter of the query 3 is –3.
Since it is negative, the digit significant for rounding is found 3 places to the left of
TABLE 4-1

Decimal Precision
Descriptions

Decimal Precision

Significant Rounding Digit

Decimal Position

-4

1

Thousands (n ×1000)

-3

2

Hundreds (n ×100)

-2

3

Tens (n ×10)

-1

4

Units (n ×1)

1

5

Tenths (n ÷10)

2

6

Hundredths (n ÷100)

3

7

Thousandths (n ÷1000)

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the decimal point, at the hundreds digit, which is 6. Since the hundreds unit is 6,
rounding up occurs and the number returned is 2000. Query 4 has dispensed with the
decimal precision parameter. This implies that rounding is done to the nearest whole
number. Since the tenth unit is 9, the number is rounded up and 1602 is returned.
The example shown in Figure 4-13 selects employees working as sales managers
and computes a loyalty bonus based on the number of days employed rounded to the
nearest whole number. The ROUND function is used to round the fractional part of
the difference between the current system date and the HIRE_DATE for each sales
manager.

The Numeric TRUNC Function (Truncate)
The TRUNC function performs a truncation operation on a numeric value based
on the decimal precision specified. A numeric truncation is different from rounding
because the resulting value drops the numbers at the decimal precision specified and
does not attempt to round up or down if the decimal precision is positive. However,
if the decimal precision specified (n) is negative, the input value is zeroed down from
the nth decimal position.
FIGURE 4-13

The numeric ROUND function

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199

The TRUNC function takes two parameters. Its syntax is TRUNC (source
number, decimal precision). Source number represents any numeric literal, column, or
expression. Decimal precision specifies the degree of truncation and is optional. If the
decimal precision parameter is absent, the default degree of rounding is zero, which
means the source number is truncated to the nearest whole number.
If the decimal precision parameter is 1, then the source number is truncated at its
tenths unit. If it is 2, it is truncated at its hundredths unit, and so on. The following
queries illustrate the usage of this function:
Query
Query
Query
Query

1:
2:
3:
4:

select
select
select
select

trunc(1601.916718,1) from dual
trunc(1601.916718,2) from dual
trunc(1601.916718,-3) from dual
trunc(1601.916718) from dual

Query 1 has a decimal precision parameter of 1, which implies that the source
number is truncated at its tenths unit and the number returned is 1601.9. The
decimal precision parameter (n) in query 2 is 2, so the source number is truncated at its
hundredths unit and the number returned is 1601.91. Note that this result would be
different if a rounding operation was performed since the digit in position (n + 1) is
6 (greater than 5). Query 3 specifies a negative number (−3) as its decimal precision.
Three places to the left of the decimal point implies that the truncation happens
at the hundreds digit as shown earlier in Table 4-1. Therefore, the source number is
zeroed down from its hundreds digit (6) and the number returned is 1000. Finally,
query 4 does not have a decimal precision parameter implying that truncation is done
at the whole number degree of precision. The number returned is 1601.
The finance department has qualified for a top departmental award for which
the company decided to reward its finance staff by adjusting their salaries. Since the
fractional salary adjustment results in numbers with three decimal places, the TRUNC
function is used to truncate the proposed salary increase to a whole number, as shown
in Figure 4-14.

The MOD Function (Modulus)
The MOD function returns the numeric remainder of a division operation. Two
numbers, the dividend (number being divided) and the divisor (number to divide by)
are provided, and a division operation is performed. If the divisor is a factor of the
dividend, MOD returns zero since there is no remainder. If the divisor is zero, no division
by zero error is returned and the MOD function returns a zero instead. If the divisor is
larger than the dividend, then the MOD function returns the dividend as its result.
This is because it divides zero times into the divisor, leaving the remainder equal to
the dividend.

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FIGURE 4-14

Single-Row Functions

The numeric TRUNC function

The MOD function takes two parameters. Its syntax is MOD(dividend, divisor).
The dividend and divisor parameters represent a numeric literal, column, or expression,
which may be negative or positive. The following queries illustrate the usage of this
function:
Query
Query
Query
Query

1:
2:
3:
4:

select
select
select
select

mod(6,2) from dual
mod(5,3) from dual
mod(7,35) from dual
mod(5.2,3) from dual

Query 1 divides 6 by 2 perfectly, yielding 0 as the remainder. Query 2 divides 5 by 3,
yielding 1 with remainder 2. Query 3 attempts to divide 7 by 35. Since the divisor is
larger than the dividend, the number 7 is returned as the modulus value. Query 4 has
an improper fraction as the dividend. Dividing 5.2 by 3 yields 1 with remainder 2.2.
Any even number divided by 2 naturally has no remainder, but odd numbers
divided by 2 always have a remainder of 1.Therefore, the MOD function is
often used distinguish between even and odd numbers.

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201

The EMPLOYEE_ID column in the EMPLOYEES table stores a unique
sequential number for each record beginning with employee number 100. The
first 12 employees must be allocated to one of four teams in a round-robin manner
for a particular task. Figure 4-15 shows how this is accomplished using the MOD
function.
The 12 employees’ records are isolated with a BETWEEN operator in the WHERE
clause. The MOD function is applied to the division of the EMPLOYEE_ID column
values by the numeric literal 4. As Figure 4-15 shows, the MOD function allocates
the numbers 0 to 3 to each row in a round-robin manner.

FIGURE 4-15

The MOD function

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The default values assumed
by the optional parameters of functions are
not always intuitive but are often tested.
For example, calling the SUBSTR function
with just the first two parameters results
in the function extracting a substring from
a start position to the end of the given
source string.The optional parameter
for both the numeric and date TRUNC

and ROUND functions is the degree of
precision. For example, calling the numeric
TRUNC function without specifying degree
of truncation results in the number being
truncated to the nearest whole number.
It is useful to be familiar with the default
values assumed by optional parameters for
these functions.

Working with Dates
The date built-in functions provide a convenient way to solve date-related problems
without needing to keep track of leap years or the number of days in particular
months. We’ll discuss storage of dates by Oracle and the default date format masks
before we conduct a detailed examination of the SYSDATE function. We’ll follow
by discussing date arithmetic and the date manipulation functions: ADD_MONTHS,
MONTHS_BETWEEN, LAST_DAY, NEXT_DAY, ROUND, and TRUNC.

Date Storage in the Database
Dates are stored internally in a numeric format that supports the storage of century,
year, month, and day details, as well as time information such as hours, minutes,
and seconds. These date attributes are available for every literal, column value, or
expression that is of date data type.
When date information is accessed from a table, the default format of the results
comprises two digits that represent the day, a three-letter abbreviation of the month,
and two digits representing the year component. By default, these components are
separated with hyphens in SQL*Plus and forward slashes in SQL Developer. Figure 4-16
shows the contents of the START_DATE column from the JOB_HISTORY table.
Note the query is performed in SQL Developer, so the date elements are separated by
forward slashes.
Although the century component is not displayed by default, it is stored in the
database when the date value is inserted or updated and is available for retrieval.

Use Character, Number, and Date Functions in SELECT Statements

FIGURE 4-16

203

Default date storage in the database

The format in which a date is displayed is referred to as its format mask. There are
several formatting codes or date format masks available, as shown in Table 4-2.
The language to format date items using the full range of date format masks is
discussed in Chapter 5. The DD-MON-RR format mask is the default for display
and input. When inserting or updating date information, the century component
is obtained from the SYSDATE function if it is not supplied. The RR date format
mask differs from the YY format mask since it is may be used to specify different
centuries based on the current and specified years. The century component assigned

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TABLE 4-2

Date Format
Masks

Single-Row Functions

Format Mask

Format Description

DD

Day of the month

MON

Month of the year

YY

Two-digit year

YYYY

Four-digit year including century

RR

Two-digit year (Year 2000–compliant)

CC

Two-digit century

HH

Hours with AM and PM

HH24

Twenty-four-hour time

MI

Minutes

SS

Seconds

to a date with its year specified with the RR date format may be better understood by
considering the following principles:
■ If the two digits of the current year and specified year lie between 0 and 49,

the current century is returned. Suppose the present date is 02-JUN-2007.
The century returned for the date 24-JUL-04 in DD-MON-RR format is 20.
■ If the two digits of the current year lie between 0 and 49 and the specified

year falls between 50 and 99, the previous century is returned. Suppose the
current date is 02-JUN-2007. The century returned for 24-JUL-94 is 19.
■ If the two digits of the current and specified years lie between 50 and 99, the

current century is returned by default. Suppose the current date is 02-JUN1975. The century returned for 24-JUL-94 is 19.
■ If the two digits of the current year lie between 50 and 99 and the specified

year falls between 0 and 49, the next century is returned. Suppose the current
date is 02-JUN-1975. The century returned for 24-JUL-07 is 20.

The SYSDATE Function
The SYSDATE function takes no parameters and returns the current system date
and time according to the database server. By default the SYSDATE function
returns the DD-MON-RR components of the current system date. It is important
to remember that SYSDATE does not return the date and time as specified by your

Use Character, Number, and Date Functions in SELECT Statements

205

local system clock. If the database server is located in a different time zone from a
client querying the database, the date and time returned will differ from the local
operating system clock on the client machine. The query to retrieve the database
server date is as follows:
select sysdate from dual

Date Arithmetic
Arithmetic with date columns and expressions were briefly discussed in Chapter 2.
The following equations illustrate an important principle regarding date arithmetic:
Date1 – Date2 = Num1
Date1 – Num1 = Date2
Date1 = Date2 + Num1

A date can be subtracted from another date. The difference between two
date items represents the number of days between them. Any number, including
fractions, may be added to or subtracted from a date item. In this context the
number represents a number of days. The sum or difference between a number and a
date item always returns a date item. This principle implies that adding, multiplying,
or dividing two date items is not permitted.
To illustrate the time component of the SYSDATE function as it pertains to
date arithmetic, the SQL Developer environment has been temporarily modified to
display time information as well as date information.
To modify the SQL Developer environment to display time information for
date columns, by default, navigate to Tools | Preferences | Database | NLS
Parameters | Date Format. Change the default display mask (DD/MON/RR) to
(DD/MON/RR HH24:MI:SS).
A conversion function, which will be discussed in detail in Chapter 5, is
introduced here to aid this example. Figure 4-17 demonstrates how the TO_DATE
conversion function is used to convert the date literal 02-JUN-2008 with time
component 12.10pm into a date data type.
The first query in the figure is dissected as follows: two days prior to the second
of June, 12.10pm is the thirty-first of May, 12.10pm, which is the date and time
returned by expression 1. Adding 0.5 days or 12 hours to 02/JUN/08 12:10pm, as
expression 2 demonstrates, results in the date 03/JUN/08 and the time 00.10 being
returned. Expression 3 adds 6/24 or six hours, resulting in the date 02/JUN/08, 18.10
being returned.

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FIGURE 4-17

Single-Row Functions

The SYSDATE function and date arithmetic

The HIREDATE column for employees with DEPARTMENT_ID values of 30 is
subtracted from the date item 02/JUN/96 12:10pm in the figure’s second query. The
number of days between these two dates is returned for each row. Notice that when the
HIREDATE column value occurs later than 02/JUN/96, a negative number is returned.

Using Date Functions
The date manipulation functions provide a reliable and accurate means of working with
date items. These functions provide such ease and flexibility for date manipulation
that many integration specialists, database administrators, and other developers make
frequent use of them.

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207

The MONTHS_BETWEEN Function
The MONTHS_BETWEEN function returns a numeric value representing the
number of months between two date values. Date literals in the format DD-MONRR or DD-MON-YYYY are automatically cast as date items when they occur as
parameters to the MONTHS_BETWEEN function.
The MONTHS_BETWEEN function takes two mandatory parameters. Its syntax
is MONTHS_BETWEEN(start date, end date). The function computes the difference
in months between start date and end date. If the end date occurs before the start date,
a negative number is returned. The difference between the two date parameters
may consist of a whole number and a fractional component. The whole number
represents the number of months between the two dates. The fractional component
represents the days and time remaining after the integer difference between years
and months is calculated and is based on a 31-day month. A whole number with no
fractional part is returned if the day components of the dates being compared are
either the same or the last day of their respective months.
The following queries illustrate the MONTHS_BETWEEN function:
Query 1: select sysdate, sysdate+31, sysdate+62, sysdate+91,
months_between(sysdate+91, sysdate) from dual
Query 2: select months_between('29-mar-2008','28-feb-2008') from dual
Query 3: select months_between('29-mar-2008','28-feb-2008') * 31 from dual
Query 4: select months_between(to_date('29-feb-2008'),
to_date('28-feb-2008 12:00:00','dd-mon-yyyy hh24:mi:ss'))* 31 from dual;

Assume that the current date is 29-DEC-2007.
The first expression in query 1 returns the number 1, as the month between
29-DEC-2007 is 29-JAN-2008 (31 days later). The second expression similarly returns
2 months between 29-DEC-2007 and 29-FEB-2008 (62 days later). Since February
2008 has 29 days, 91 days must be added to 29-DEC-2007 to get 29-MAR-2008, and
the MONTHS_BETWEEN (29-MAR-2008, 29-DEC-2007) function returns exactly
three months in the third expression in query 1.
Query 2 implicitly converts the date literals into date items of the format
DD-MON-YYYY. Since no time information is provided, Oracle assumes the time to
be midnight on both days, or 00:00:00. The MONTHS_BETWEEN function returns
approximately 1.03225806. The whole number component indicates that there is
one month between these two dates. Closer examination of the fractional
component interestingly reveals that there is exactly one month between 28-MAR-2008
and 28-FEB-2008. The fractional component must therefore represent the

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one-day difference. It would include differences in hours, minutes, and seconds
as well, but for this example, the time components are identical. Multiplying
0.03225806 by 31 returns 1, since the fractional component returned by MONTHS_
BETWEEN is based on a 31-day month.
Similarly, query 3 returns the whole number 32. Query 4 demonstrates how the
time component is factored into the computation by the MONTHS_BETWEEN
function, which returns approximately 0.016129. There is a 12-hour difference
between the start and end date parameters, so zero months between them is correct.
Multiplying the fractional part by 31 yields 0.5 days, which corresponds to the
12-hour difference.
The MONTHS_BETWEEN function used in Figure 4-18 returns records from
the JOB_HISTORY table. The months between the dates an employee started
in a particular job and ended that job are computed, and the results are sorted in
descending order.

A common mistake is to
assume that the return data type of singlerow functions are the same as the category
the function belongs to.This is only true of
the numeric functions. Character and date
functions can return values of any data
type. For example the INSTR character

function and the MONTHS_BETWEEN
date function both return a numeric
value. It is important to be familiar with
the principles of date arithmetic, as it is
common to erroneously assume that the
difference between two dates is a date,
when in fact it is a number.

The ADD_MONTHS Function
The ADD_MONTHS function returns a date item calculated by adding a specified
number of months to a given date value. Date literals in the format DD-MON-RR or
DD-MON-YYYY are automatically cast as date items when they occur as parameters
to the ADD_MONTHS function.
The ADD_MONTHS function takes two mandatory parameters. Its syntax is
ADD_MONTHS (start date, number of months). The function computes the target
date after adding the specified number of months to the start date. The number of
months may be negative, resulting in a target date earlier than the start date being
returned. The number of months may be fractional, but the fractional component is
ignored and the integer component is used.

Use Character, Number, and Date Functions in SELECT Statements

FIGURE 4-18

209

The MONTHS_BETWEEN function

The three queries in Figure 4-19 illustrate the behavior of the ADD_MONTHS
function.
The first query in the figure returns 07-MAY-2009 since the day component
remains the same if possible and the month is incremented by one. The second
query has two interesting dimensions. The parameter specifying the number of
months to add contains a fractional component, which is ignored. Therefore, it is
equivalent to ADD_MONTHS ('31-dec-2008',2). Adding two months to the date
31-DEC-2008 should return the date 31-FEB-2009, but there is no such date, so the
last day of the month, 28-FEB-2009, is returned. Since the number of months added
in the third query is –12, the date 07-APR-2008 is returned, which is 12 months
prior to the start date.

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Chapter 4:

FIGURE 4-19

Single-Row Functions

The ADD_MONTHS function

EXERCISE 4-3
Using the Date Functions
You are required to obtain a list of EMPLOYEE_ID, LAST_NAME, and HIRE_
DATE values for the employees who have worked more than 100 months between
the date they were hired and 01-JAN-2000.
1. Start SQL Developer and connect to the HR schema.
2. The SELECT clause is
SELECT EMPLOYEE_ID, LAST_NAME, HIRE_DATE
3. The FROM clause is
FROM EMPLOYEES

Use Character, Number, and Date Functions in SELECT Statements

211

4. The WHERE clause must compare the months between the given date literal
and the HIRE_DATE value with the numeric literal 100.
5. The MONTHS_BETWEEN function may be used in the WHERE clause.
6. The WHERE clause is
WHERE MONTHS_BETWEEN('01-JAN-2000', HIRE_DATE) > 100
7. Executing this statement returns the set of results shown in the following
illustration:

The NEXT_DAY Function
The NEXT_DAY function returns the date when the next occurrence of a specified
day of the week occurs. Literals that may be implicitly cast as date items are
acceptable when they occur as parameters to the NEXT_DAY function.
The NEXT_DAY function takes two mandatory parameters. Its syntax is
NEXT_DAY (start date, day of the week). The function computes the date on which
the day of the week parameter next occurs after the start date. The day of the week

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parameter may be either a character value or an integer value. The acceptable values
are determined by the NLS_DATE_LANGUAGE database parameter but the
default values are at least the first three characters of the day name or integer values,
where 1 represents Sunday, 2 represents Monday, and so on. The character values
representing the days of the week may be specified in any case. The short name may
be longer than three characters, for example, Sunday may be referenced as sun, sund,
sunda or Sunday.
The three queries in Figure 4-20 illustrate the behavior of the NEXT_DAY
function.
01-JAN-2009 is a Thursday. Therefore, the next time a Tuesday occurs will be five
days later on 06-JAN-2009, which is what the first query in the figure retrieves. The
second query specifies the character literal WEDNE, which is interpreted as Wednesday.

FIGURE 4-20

The NEXT_DAY function

Use Character, Number, and Date Functions in SELECT Statements

213

SCENARIO & SOLUTION
You wish to retrieve the duration of employment in
days for each employee. Is it possible to perform such
a calculation?

Yes. The SYSDATE function may be used to
obtain the current system date. The following
query computes the duration by subtracting the
HIRE_DATE column from the value returned by the
SYSDATE function:
SELECT SYSDATE-HIRE_DATE
FROM EMPLOYEES;

You are tasked with identifying the date the end of
year staff bonus will be paid. Bonuses are usually paid
on the last Friday in December. Can the bonus date
be computed using the NEXT_DAY function?

Yes. If the NEXT_DAY function is called with the
start date parameter set to the last day in December
and the search day set to Friday, then the first
Friday in January is returned. Subtracting seven days
from this date yields the date of the last Friday in
December. Consider the following query for the year
2009:
SELECT NEXT_DAY('31-DEC-2009', 'Friday') -7
FROM DUAL;

Employees working in the IT department have
moved to new offices and, although the last four
digits of their phone numbers are the same, the set
of the three digits 423 is changed to 623. A typical
phone number of an IT staff member is 590-4234567. You are required to provide a list of employees’
names with their old and new phone numbers. Can
this list be provided?

Yes. The REPLACE function is used. To replace every
4 with a 6 will change digits that should not be changed
as well, so the string to be replaced must be uniquely
specified. The following query provides the list:
SELECT FIRST_NAME, LAST_NAME,
REPLACE( PHONE_NUMBER, '.423.','.623.')
FROM EMPLOYEES WHERE DEPARTMENT_
ID=60

The next Wednesday after 01-JAN-2009 is 07-JAN-2009. The third query uses the
integer form to specify the fifth day of the week. Assuming the default values where
Sunday is represented by the number 1, the fifth day is Thursday. The next time another
Thursday occurs after 01-JAN-2009 is 08-JAN-2009.

The LAST_DAY Function
The LAST_DAY function returns the date of the last day in the month a specified
day belongs to. Literals that may be implicitly cast as date items are acceptable when
they occur as parameters to the LAST_DAY function.
The LAST_DAY function takes one mandatory parameter. Its syntax is LAST_
DAY(start date). The function extracts the month that the start date parameter
belongs to and calculates the date of the last day of that month. The two queries in
Figure 4-21 illustrate the behavior of the LAST_DAY function.

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Chapter 4:

FIGURE 4-21

Single-Row Functions

The LAST_DAY function

The last day in the month of January 2009 is 31-JAN-2009, which is returned by
the LAST_DAY('01-JAN-2009') function call in the first query in the figure. The
second query extracts the employees with JOB_ID values of IT_PROG. The number
of days worked by these employees in their first month of employment is calculated
by subtracting the HIRE_DATE values from the LAST_DAY of that month.

The Date ROUND Function
The date ROUND function performs a rounding operation on a value based on a
specified date precision format. The value returned is either rounded up or down to
the nearest date precision format.

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215

INSIDE THE EXAM
There are two certification objectives in this
chapter. Various types of SQL functions are
described and the concept of a function is
defined. A distinction is made between singlerow functions, which execute once for each
row in a dataset, and multiple-row functions,
which execute once for all the rows in a dataset. Single-row functions may be used in the
SELECT, WHERE, and ORDER BY clauses of
the SELECT statement.
The second objective relates to the use
of character, numeric, and date functions
in queries. The exam tests your understanding of these functions by providing practical
examples of their usage. You may be asked to
predict the results returned or to identify errors inherent in the syntax of these examples.
Functions can take zero or more input
parameters, some of which may be mandatory

while others are optional. Mandatory parameters are listed first, and optional parameters
are always last. Common errors relate to
confusion about the meaning of positions of
parameters in functions. A character function like INSTR takes four parameters, with
the first two being mandatory. The first is
the source string; the second is the search
string, while the third and fourth are not
always intuitive and may be easily forgotten or
mixed up. Be sure to remember the meaning
of parameters in different positions. Another
mistake related to parameters relates to confusion about the default values used by Oracle
when optional parameters are not specified.
You may be expected to predict the results
returned from function calls that do not have
all their optional parameters specified.

The date ROUND function takes one mandatory and one optional parameter.
Its syntax is ROUND(source date, [date precision format]). The source date parameter
represents any value that can be implicitly converted into a date item. The date
precision format parameter specifies the degree of rounding and is optional. If it is
absent, the default degree of rounding is day. This means the source date is rounded
to the nearest day. The date precision formats include century (CC), year (YYYY),
quarter (Q), month (MM), week (W), day (DD), hour (HH), and minute (MI). Many
of these formats are discussed in Chapter 5.
Rounding up to century is equivalent to adding one to the current century. Rounding
up to the next month occurs if the day component is greater than 16, else rounding
down to the beginning of the current month occurs. If the month falls between one and
six, then rounding to year returns the date at the beginning of the current year, else it
returns the date at the beginning of the following year. Figure 4-22 shows four items in
the SELECT list, each rounding a date literal to a different degree of precision.

216

Chapter 4:

FIGURE 4-22

Single-Row Functions

The date ROUND function

The first item rounds the date to the nearest day. Since the time is 13:00, which
is after 12:00, the date is rounded to midnight on the following day, or 03-JUN2009 00:00. The second item rounds the date to the same day of the week as the
first day of the month and returns 01-JUN-2009. The third item rounds the date to
the beginning of the following month, since the day component is 16 and returns
01-JUL-2009. The fourth item is rounded up to the date at the beginning of the
following year since the month component is 7, and 01-JAN-2010 is returned.

The Date TRUNC Function
The date TRUNC function performs a truncation operation on a date value based
on a specified date precision format.
The date TRUNC function takes one mandatory and one optional parameter.
Its syntax is TRUNC(source date, [date precision format]). The source date parameter
represents any value that can be implicitly converted into a date item. The date
precision format parameter specifies the degree of truncation and is optional. If it is
absent, the default degree of truncation is day. This means that any time component

Certification Summary

FIGURE 4-23

217

The date TRUNC function

of the source date is set to midnight or 00:00:00 (00 hours, 00 minutes and 00 seconds).
Truncating at the month level sets the date of the source date to the first day of the
month. Truncating at the year level returns the date at the beginning of the current
year. Figure 4-23 shows four items in the SELECT list, each truncating a date literal to
a different degree of precision.
The first item sets the time component of 13:00 to 00:00 and returns the current
day. The second item truncates the date to the same day of the week as the first
day of the month and returns 01-JUN-2009. The third item truncates the date to
the beginning of the current month and returns 01-JUN-2009. The fourth item is
truncated to the date at the beginning of the current year and returns 01-JAN-2009.

CERTIFICATION SUMMARY
Single-row functions exponentially enhance the data manipulation possibilities
offered by SQL statements. These functions execute once for each row of data
selected. They may be used in SELECT, WHERE, and ORDER BY clauses in a
SELECT statement.

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The black box nature of the built-in PL/SQL functions was discussed and
a distinction between multiple and single-row functions was made. A highlevel overview describing how character, numeric, and date information may be
manipulated by single-row functions was provided before systematically exploring
several key functions in detail.
Character-case conversion functions were described before introducing the
character manipulation functions. The numeric functions ROUND, TRUNC, and
MOD were discussed, but these represent the tip of the iceberg since Oracle provides
a vast toolbox of mathematical and numeric functions. Date arithmetic and storage
was briefly explored before taking a detailed look at the date functions.
There are numerous single-row functions available, and you are not required
to memorize their every detail. Understanding the broad categories of single-row
functions and being introduced to the common character, numeric, and date
functions provides a starting point for your discovery of their usefulness.

Two-Minute Drill

✓

219

TWO-MINUTE DRILL
Describe Various Types of Functions Available in SQL
❑ Functions accept zero or more input parameters but always return one result

of a predetermined data type.
❑ Single-row functions execute once for each row selected, while multiple-row

functions execute once for the entire set of rows queried.
❑ Character functions are either case-conversion or character-manipulation

functions.

Use Character, Number, and Date Functions
in SELECT Statements
❑ The INITCAP function accepts a string of characters and returns each word

in title case.
❑ The function that computes the number of characters in a string including

spaces and special characters is the LENGTH function.
❑ The INSTR function returns the positional location of the nth occurrence of

a specified string of characters in a source string.
❑ The SUBSTR function extracts and returns a segment from a given source

string.
❑ The REPLACE function substitutes each occurrence of a search item in the

source string with a replacement term and returns the modified source string.
❑ A modulus operation returns the remainder of a division operation and is

available via the MOD function.
❑ The numeric ROUND function rounds numbers either up or down to the

specified degree of precision.
❑ The SYSDATE function is traditionally executed against the DUAL table

and returns current date and time of the database server.
❑ Date types store century, year, month, day, hour, minutes, and seconds

information.
❑ The difference between two date items is always a number that represents the

number of days between these two items.

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❑ Any number, including fractions, may be added to or subtracted from a date

item and in this context the number represents a specified number of days.
❑ The MONTHS_BETWEEN function computes the number of months

between two given date parameters and is based on a 31-day month.
❑ The LAST_DAY function is used to obtain the last day in a month given any

valid date item.

Self Test

221

SELF TEST
The following questions will help you measure your understanding of the material presented in this
chapter. Read all the choices carefully because there might be more than one correct answer. Choose
all the correct answers for each question.

Describe Various Types of Functions Available in SQL
1. Which statements regarding single-row functions are true? (Choose all that apply.)
A. They may return more than one result.
B. They execute once for each record processed.
C. They may have zero or more input parameters.
D. They must have at least one mandatory parameter.
2. Which of these are single-row character-case conversion functions? (Choose all that apply.)
A. LOWER
B. SMALLER
C. INITCASE
D. INITCAP

Use Character, Number, and Date Functions in SELECT Statements
3. What value is returned after executing the following statement:
SELECT LENGTH('How_long_is_a_piece_of_string?') FROM DUAL; (Choose the best answer.)
A. 29
B. 30
C. 24
D. None of the above
4. What value is returned after executing the following statement:
SELECT SUBSTR('How_long_is_a_piece_of_string?', 5,4) FROM DUAL; (Choose the best
answer.)
A. long
B. _long
C. string?
D. None of the above

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5. What value is returned after executing the following statement?
SELECT INSTR('How_long_is_a_piece_of_string?','_',5,3) FROM DUAL; (Choose the best
answer.)
A. 4
B. 14
C. 12
D. None of the above
6. What value is returned after executing the following statement?
SELECT REPLACE('How_long_is_a_piece_of_string?','_','') FROM DUAL; (Choose the best
answer.)
A. How long is a piece of string?
B. How_long_is_a_piece_of_string?
C. Howlongisapieceofstring?
D. None of the above
7. What value is returned after executing the following statement?
SELECT MOD(14,3) FROM DUAL; (Choose the best answer.)
A. 3
B. 42
C. 2
D. None of the above
8. Assuming SYSDATE=07-JUN-1996 12:05pm, what value is returned after executing the
following statement?
SELECT ADD_MONTHS(SYSDATE,-1) FROM DUAL; (Choose the best answer.)
A. 07-MAY-1996 12:05pm
B. 06-JUN-1996 12:05pm
C. 07-JUL-1996 12:05pm
D. None of the above
9. What value is returned after executing the following statement? Take note that 01-JAN-2009
occurs on a Thursday. (Choose the best answer.)
SELECT NEXT_DAY('01-JAN-2009','wed') FROM DUAL;
A. 07-JAN-2009
B. 31-JAN-2009
C. Wednesday
D. None of the above

Lab Question

223

10. Assuming SYSDATE=30-DEC-2007, what value is returned after executing the following
statement?
SELECT TRUNC(SYSDATE,'YEAR') FROM DUAL; (Choose the best answer.)
A. 31-DEC-2007
B. 01-JAN-2008
C. 01-JAN-2007
D. None of the above

LAB QUESTION
Using SQL Developer or SQL*Plus, connect to the OE schema and complete the following tasks.
Several quotations were requested for prices on color printers. The supplier information is not available from the usual source, but you know that the supplier identification number is embedded in the
CATALOG_URL column from the PRODUCT_INFORMATION table. You are required to retrieve
the PRODUCT_NAME and CATALOG_URL values and to extract the supplier number from the
CATALOG_URL column for all products which have both the words COLOR and PRINTER in
the PRODUCT_DESCRIPTION column stored in any case.

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SELF TEST ANSWERS
Describe Various Types of Functions Available in SQL
✓ B and C. Single-row functions execute once for every record selected in a dataset and may
1. ®
either take no input parameters, like SYSDATE, or many input parameters.
®
˚ A and D are incorrect because a function by definition returns only one result and there are
many functions with no parameters.
✓ A and D. The LOWER function converts the case of the input string parameter to its
2. ®
lowercase equivalent, while INITCAP converts the given input parameter to title case.
®
˚ B and C are not valid function names.

Use Character, Number, and Date Functions in SELECT Statements
✓ B. The LENGTH function computes the number of characters in a given input string
3. ®
including spaces, tabs, punctuation mark, and other nonprintable special characters.
®
˚ A, C, and D are incorrect.
✓ A. The SUBSTR function extracts a four-character substring from the given input string
4. ®
starting with and including the fifth character. The characters at positions 1 to 4 are How_.
Starting with the character at position 5, the next four characters form the word “long.”
®
˚ B, C, and D are incorrect because B is a five-character substring beginning at position 4,
while ring?, which is also five characters long, starts five characters from the end of the given
string.
✓ B. The INSTR function returns the position that the nth occurrence of the search string
5. ®
may be found after starting the search from a given start position. The search string is the
underscore character, and the third occurrence of this character starting from position 5 in the
source string occurs at position 14.
®
˚ A, C, and D are incorrect since position 4 is the first occurrence of the search string and
position 12 is the third occurrence if the search began at position 1.
✓ C. All occurrences of the underscore character are replaced by an empty string, which
6. ®
removes them from the string.
®
˚ A, B, and D are incorrect. A is incorrect because the underscore characters are not
replaced by spaces, and B does not change the source string.
✓ C. When 14 is divided by 3, the answer is 4 with remainder 2.
7. ®
®
˚ A, B, and D are incorrect.

Lab Answer

225

✓ A. The minus one parameter indicates to the ADD_MONTHS function that the date to be
8. ®
returned must be one month prior to the given date.
®
˚ B, C, and D are incorrect. B is one day and not one month prior to the given date. C is one
month after the given date.
✓ A. Since the first of January 2009 falls on a Thursday, the date of the following Wednesday
9. ®
is six days later.
®
˚ B, C, and D are incorrect. B returns the last day of the month in which the given date falls,
and C returns a character string instead of a date.
✓ C. The date TRUNC function does not perform rounding and since the degree of
10. ®
truncation is YEAR, the day and month components of the given date are ignored and the first
day of the year it belongs to is returned.
®
˚ A, B, and D are incorrect. A returns the last day in the month in which the given date
occurs, and B returns a result achieved by rounding instead of truncation.

LAB ANSWER
Using SQL Developer or SQL*Plus, connect to the OE schema, complete the following tasks.
1. Start SQL Developer and connect to the OE schema.
2. A typical CATALOG_URL entry looks as follows: www.supp-102094.com/cat/hw/p1797.html.
The supplier identification number is consistently six characters long and starts from the
seventeenth character of the CATALOG_URL value. The SUBSTR function is used to extract
this value.
3. The SELECT clause is therefore
SELECT PRODUCT_NAME, CATALOG_URL, SUBSTR(CATALOG_URL, 17, 6)
SUPPLIER
4. The FROM clause is
FROM PRODUCT_INFORMATION
5. The records retrieved must be limited to those containing both the words COLOR and
PRINTER. These words may occur in any order and may be present in uppercase or lowercase
or mixed case. Any of the case conversion functions may be used to deal with case issues,
but because the two words can occur in any order, two conditions are necessary. The UPPER
function will be used for case conversion for comparison.
6. The first condition is
UPPER(PRODUCT_DESCRIPTION) LIKE '%COLOR%'

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Chapter 4:

Single-Row Functions

7. The second condition is
UPPER(PRODUCT_DESCRIPTION) LIKE '%PRINTER%'
8. The WHERE clause is
WHERE UPPER(PRODUCT_DESCRIPTION) LIKE '%COLOR%' AND
UPPER(PRODUCT_DESCRIPTION) LIKE '%PRINTER%'
9. Executing the statement returns the set of results matching this pattern as shown in the
following illustration:

5
Using Conversion
Functions and
Conditional
Expressions

CERTIFICATION OBJECTIVES
5.01

Describe Various Types of Conversion
Functions Available in SQL

5.03

5.02

Use the TO_CHAR, TO_NUMBER, and
TO_DATE Conversion Functions

✓
Q&A

Apply Conditional Expressions in
a SELECT Statement
Two-Minute Drill
Self Test

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

228

Chapter 5:

Using Conversion Functions and Conditional Expressions

F

unctions that operate on numeric, date, and character information were discussed in
Chapter 4, and familiarity with that content is assumed in this chapter. Sometimes data is
not available in the exact format a function is defined to accept, resulting in a data type
mismatch. To avoid mismatch errors, Oracle implicitly converts compatible data types. Implicit
conversion is discussed before introducing explicit conversion functions, which are used for
reliable data type conversions.
The concept of nesting functions is defined and a category of general functions
aimed at simplifying interactions with NULL values is introduced. These include the
NVL, NVL2, NULLIF, and COALESCE functions.
Conditional logic, or the ability to display different results depending on data
values, is exposed by the conditional functions CASE and DECODE. These
functions provide if-then-else logic in the context of a SQL query.

CERTIFICATION OBJECTIVE 5.01

Describe Various Types of Conversion
Functions Available in SQL
SQL conversion functions are single row functions designed to alter the nature of the
data type of a column value, expression or literal. TO_CHAR, TO_NUMBER and
TO_DATE are the three most widely used conversion functions and are discussed
in detail. The TO_CHAR function converts numeric and date information into
characters, while TO_NUMBER and TO_DATE convert character data into numbers
and dates, respectively. The concepts of implicit and explicit data type conversion are
discussed in the next section.

Conversion Functions
Oracle allows columns to be defined with ANSI, DB2, and SQL/DS data types. These
are converted internally to Oracle data types. This approach allows applications written
for other database systems to be migrated to Oracle with ease.
Table definitions are obtained using the DESCRIBE command discussed in
Chapter 2. Each column has an associated data type that constrains the nature of

Describe Various Types of Conversion Functions Available in SQL

229

the data it can store. A NUMBER column cannot store character information. A
DATE column cannot store random characters or numbers. However, the character
equivalents of both number and date information can be stored in a VARCHAR2 field.
If a function that accepts a character input parameter finds a number instead,
Oracle automatically converts it into its character equivalent. If a function that
accepts a number or a date parameter encounters a character value, there are
specific conditions under which automatic data type conversion occurs. DATE and
NUMBER data types are very strict compared to VARCHAR2 and CHAR.
Although implicit data type conversions are available, it is more reliable to
explicitly convert values from one data type to another using single-row conversion
functions. Converting character information to NUMBER and DATE relies on
format masks, which are discussed later in this section.
When numeric values are supplied as input to functions expecting character
parameters, implicit data type conversion ensures that they are treated as
character values. Similarly, character strings consisting of numeric digits are
implicitly converted into numeric values if possible when a data type mismatch
occurs. But be wary of implicit conversions.There are some cases when it does
not work as expected, as in the following WHERE clause. Consider limiting
data from a table T based on a character column C, which contains the string
'100'.The condition clause WHERE C='100' works as you might expect, but the
condition WHERE C=100 returns an invalid number error.

Implicit Data Type Conversion
Values that do not share identical data types with function parameters are implicitly
converted to the required format if possible. VARCHAR2 and CHAR data types are
collectively referred to as character types. Character fields are flexible and allow the
storage of almost any type of information. Therefore, DATE and NUMBER values
can easily be converted to their character equivalents. These conversions are known
as number to character and date to character conversions. Consider the following
queries:
Query 1: select length(1234567890) from dual
Query 2: select length(SYSDATE) from dual

Both queries use the LENGTH function, which takes a character string
parameter. The number 1234567890 in query 1 is implicitly converted into a
character string, '1234567890', before being evaluated by the LENGTH function,

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Chapter 5:

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which returns the number 10. Query 2 first evaluates the SYSDATE function, which
is assumed to be 07-APR-38. This date is implicitly converted into the character
string '07-APR-38', and the LENGTH function returns the number 9.
It is uncommon for character data to be implicitly converted into numeric data
types since the only condition under which this occurs is if the character data
represents a valid number. The character string '11' will be implicitly converted to a
number, but '11.123.456' will not be, as the following queries demonstrate:
Query
Query
Query
Query

3:
4:
5:
6:

select
select
select
select

mod('11',2) from dual
mod('11.123',2) from dual
mod('11.123.456',2) from dual
mod('$11',2) from dual

Queries 3 and 4 implicitly convert the character strings '11' and '11.123' into the
numbers 11 and 11.123, respectively, before the MOD function evaluates them and
returns the results 1 and 1.123. Query 5 returns the error “ORA-1722: invalid number,”
when Oracle tries to perform an implicit character to number conversion. It fails because
the string '11.123.456' is not a valid number. Query 6 also fails with the invalid number
error, since the dollar symbol cannot be implicitly converted into a number.
Implicit character to date conversions are possible when the character string conforms
to the following date patterns: [D|DD] separator1 [MON|MONTH] separator2
[R|RR|YY|YYYY]. D and DD represent a single and 2-digit day of the month. MON
is a 3-character abbreviation, while MONTH is the full name for a month. R and
RR represent a single and 2-digit year. YY and YYYY represent a 2- and 4-digit year,
respectively. The separator1 and separator2 elements may be most punctuation marks,
spaces, and tabs. Table 5-1 demonstrates implicit character to date conversion, listing
several function calls and the results SQL Developer returns.
TABLE 5-1

Examples
of Implicit
Character to
Date Conversion

Function Call

Format

Results

add_months('24-JAN-09',1)

DD-MON-RR

24/FEB/09

add_months('1\january/8',1)

D\MONTH/R

01/FEB/08

months_between('13*jan*8',
'13/feb/2008')

DD*MON*R,
DD/MON/YYYY

–1

add_months('01$jan/08',1)

DD$MON/RR

01/FEB/08

add_months('13!jana08',1)

JANA is an invalid
month

ORA-1841: (full) year must
be between –4713 and +9999
and not be 0

add_months('24-JAN-09 18:45',1)

DD-MON-RR
HH24:MI

ORA-1830: date format
picture ends before
converting entire input string

Use the TO_CHAR, TO_NUMBER, and TO_DATE Conversion Functions

231

Explicit Data Type Conversion
Oracle offers many functions to convert items from one data type to another, known
as explicit data type conversion functions. These return a value guaranteed to be the
type required and offer a safe and reliable method of converting data items.
NUMBER and DATE items can be converted explicitly into character items
using the TO_CHAR function. A character string can be explicitly changed into
a NUMBER using the TO_NUMBER function. The TO_DATE function is used
to convert character strings into DATE items. Oracle’s format masks enable a wide
range of control over character to number and character to date conversions.

The explicit conversion
functions are critical to manipulating
date, character, and numeric information.
Questions on this topic test your
understanding of commonly used format
models or masks. Practical usage questions
typically take the form, “What is returned
when the TO_DATE,TO_CHAR, and

TO_NUMBER functions are applied to the
following data values and format masks?”
These are often nested within broader
functions, and it is common to be asked
to predict the result of a function call such
as TO_CHAR(TO_DATE('01-JAN-00',
'DD-MON-RR),'Day').

CERTIFICATION OBJECTIVE 5.02

Use the TO_CHAR,TO_NUMBER,
and TO_DATE Conversion Functions
This certification objective contains a systematic description of the TO_NUMBER,
TO_DATE, and TO_CHAR functions, with examples. The discussion of TO_CHAR
is divided into the conversion of two types of items to characters: DATE and
NUMBER. This separation is warranted by the availability of different format masks
for controlling conversion to character values. These conversion functions exist
alongside many others but tend to be the most widely used. This section focuses on
the practicalities of using the conversion functions.

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Using the Conversion Functions
Many situations demand the use of conversion functions. They may range from
formatting DATE fields in a report to ensuring that numeric digits extracted from
character fields are correctly converted into numbers before applying them in an
arithmetic expression.
Table 5-2 illustrates the syntax of the single-row explicit data type conversion
functions.
Optional national language support parameters (nls_parameters) are useful for
specifying the language and format in which the names of date and numeric elements
are returned. These parameters are usually absent, and the default values for elements
such as day or month names and abbreviations are used. As Figure 5-1 shows, there
is a publicly available view called NLS_SESSION_PARAMETERS that contains
the NLS parameters for your current session. The default NLS_CURRENCY value is
the dollar symbol, but this can be changed at the user session level. For example, to
change the currency to the 3-character-long string GBP, the following command may
be issued:
ALTER SESSION set NLS_CURRENCY='GBP';

Converting Numbers to Characters
Using the TO_CHAR Function
The TO_CHAR function returns an item of data type VARCHAR2. When applied
to items of type NUMBER, several formatting options are available. The syntax is as
follows:
TO_CHAR(number1, [format], [nls_parameter]),
The number1 parameter is mandatory and must be a value that either is or can be
implicitly converted into a number. The optional format parameter may be used to
specify numeric formatting information like width, currency symbol, the position of
a decimal point, and group (or thousands) separators and must be enclosed in single

TABLE 5-2

Syntax of
Explicit Data
Type Conversion
Functions

TO_NUMBER(char1, [format mask],
[nls_parameters]) = num1

TO_CHAR(num1, [format mask],
[nls_parameters]) = char1

TO_DATE(char1, [format mask],
[nls_parameters]) = date1

TO_CHAR(date1, [format mask],
[nls_parameters]) = char1

Use the TO_CHAR, TO_NUMBER, and TO_DATE Conversion Functions

FIGURE 5-1

233

National Language Support (NLS) session parameters

quotation marks. There are other formatting options for numbers being converted into
characters, some of which are listed in Table 5-3. Consider the following two queries:
Query 1: select to_char(00001)||' is a special number' from dual;
Query 2: select to_char(00001,'0999999')||' is a special number'
from dual;

Query 1 evaluates the number 00001, removes the leading zeros, converts the
number 1 into the character '1' and returns the character string ‘1 is a special
number’. Query 2 applies the numeric format mask '0999999' to the number 00001,
converting it into the character string '0000001'. After concatenation to the
character literals, the string returned is ‘0000001 is a special number’. The zero and 6
nines in the format mask indicate to the TO_CHAR function that leading zeros must
be displayed and that the display width must be set to seven characters. Therefore,
the string returned by the TO_CHAR function contains seven characters.

234

Chapter 5:

TABLE 5-3

Numeric
Format Masks

Using Conversion Functions and Conditional Expressions

Format Description
Element of Element

Format

Number Character Result

9

Numeric width

9999

12

12

0

Displays leading zeros

09999

0012

00012

.

Position of decimal
point

09999.999

030.40

00030.400

D

Decimal separator
position (period is
default)

09999D999

030.40

00030.400

,

Position of comma
symbol

09999,999

03040

00003,040

G

Group separator position 09999G999
(comma is default)

03040

00003,040

$

Dollar sign

$099999

03040

$003040

L

Local currency

L099999

03040

GBP003040 if
nls_currency is set to GBP

MI

Position of minus sign
for negatives

99999MI

−3040

3040−

PR

Wrap negatives in
parentheses

99999PR

−3040

<3040>

EEEE

Scientific notation

99.99999EEEE

121.976

1.21976E+02

U

nls_dual_currency

U099999

03040

CAD003040 if nls_dual_
currency is set to CAD

V

Multiplies by 10n times
(n is the number of
nines after V)

9999V99

3040

304000

S

+ or – sign is prefixed

S999999

3040

+3040

The query in Figure 5-2 retrieves the JOB_TITLE and MAX_SALARY columns
from the JOBS table for the rows with the word “president” in the JOB_TITLE
column. MAX_SALARY has further been formatted to have a dollar currency
symbol, a comma thousands separator, and a decimal point. When a format mask
is smaller than the number being converted, as illustrated in the fourth item in the
SELECT list, a string of hash symbols is returned instead. When a format mask
contains fewer fractional components than the number, it is first rounded to the
number of decimal places in the format mask before being converted.

Use the TO_CHAR, TO_NUMBER, and TO_DATE Conversion Functions

FIGURE 5-2

235

TO_CHAR function with numbers

Converting numbers into characters is a reliable way to ensure that functions
and general SQL syntax, which expects character input, do not return errors
when numbers are encountered. Converting numbers into character strings is
common when numeric data must be formatted for reporting purposes.The
format masks that support currency, thousands separators, and decimal point
separators are frequently used when presenting financial data.

Converting Dates to Characters Using the TO_CHAR Function
You can take advantage of a variety of format models to convert DATE items into
almost any character representation of a date using TO_CHAR. Its syntax is as follows:
TO_CHAR(date1, [format], [nls_parameter]),
Only the date1 parameter is mandatory and must take the form of a value
that can be implicitly converted to a date. The optional format parameter is case
sensitive and must be enclosed in single quotes. The format mask specifies which
date elements are extracted and whether the element should be described by a long
or an abbreviated name. The names of days and months are automatically padded

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Chapter 5:

Using Conversion Functions and Conditional Expressions

with spaces. These may be removed using a modifier to the format mask called the
fill mode (fm) operator. By prefixing the format model with the letters fm, Oracle
is instructed to trim all spaces from the names of days and months. There are many
formatting options for dates being converted into characters, some of which are
listed in Table 5-4.
Consider the following three queries:
Query 1: select to_char(sysdate)||' is today''s date' from dual;
Query 2: select to_char(sysdate,'Month')||'is a special time'
from dual;
Query 3: select to_char(sysdate,'fmMonth')||'is a special time'
from dual;

If the current system date is 03/JAN/09 and the default display format is
DD/MON/RR, then query 1 returns the character string ‘03/JAN/09 is today's date’.
There are two notable components in query 2. First, only the month component
of the current system date is extracted for conversion to a character type. Second,

TABLE 5-4

Date Format
Masks for Days,
Months, and Years

Format Element

Description

Result

Y

Last digit of year

5

YY

Last two digits of year

75

YYY

Last three digits of year

975

YYYY

Four-digit year

1975

RR

Two-digit year (see Chapter 3 for details)

75

YEAR

Case-sensitive English spelling of year

NINETEEN
SEVENTY-FIVE

MM

Two-digit month

06

MON

Three-letter abbreviation of month

JUN

MONTH

Case-sensitive English spelling of month

JUNE

D

Day of the week

2

DD

Two-digit day of month

02

DDD

Day of the year

153

DY

Three-letter abbreviation of day

MON

DAY

Case-sensitive English spelling of day

MONDAY

Use the TO_CHAR, TO_NUMBER, and TO_DATE Conversion Functions

237

since the format mask is case sensitive and 'Month' appears in title case, the string
returned is ‘January is a special time’. There is no need to add a space in front of the
literal 'is a special time' since the TO_CHAR function automatically pads the name
of the month with a space. If the format mask in query 2 was 'MONTH', the string
returned would be ‘JANUARY is a special time’. The fm modifier is applied to
query 3, and the resultant string is ‘January is a special time’. Note there is no space
between January and the literal 'is a special time'. In Table 5-4, assume the elements
are operating on the date 02-JUN-1975 and the current year is 2009.
The date format elements pertaining to weeks, quarters, centuries, and other less
commonly used format masks are listed in Table 5-5. The result column is obtained
by evaluating the TO_CHAR function using the date 24-SEP-1000 BC, with the
format mask from the format element column in the table.
The time component of a date time data type is extracted using the format models
in Table 5-6. The result is obtained by evaluating the TO_CHAR function using the
date including its time component 27-JUN-2010 21:35:13, with the format mask in
the format element column in Table 5-6.

TABLE 5-5

Less Commonly
Used Date
Format Masks

Format Element

Description

Result

W

Week of month

4

WW

Week of year

39

Q

Quarter of year

3

CC

Century

10

S preceding CC,
YYYY, or YEAR

If date is BC, a minus
is prefixed to result

–10, –1000 or –ONE THOUSAND

IYYY,IYY,IY,I

ISO dates of four, three,
two, and one digit,
respectively

1000, 000, 00, 0

BC, AD, B.C. and
A.D.

BC or AD and period
spaced B.C. or A.D.

BC

J

Julian day—days since 31
December 4713 BC

1356075

IW

ISO standard week (1 to
53)

39

RM

Roman numeral month

IX

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Chapter 5:

TABLE 5-6

Date Format
Mask for Time
Components

Using Conversion Functions and Conditional Expressions

Format Element

Description

Result

AM, PM, A.M. and P.M.

Meridian indicators

PM

HH, HH12 and HH24

Hour of day, 1–12 hours, and 0–23 hours

09, 09, 21

MI

Minute (0–59)

35

SS

Second (0–59)

13

SSSSS

Seconds past midnight (0–86399)

77713

Several other elements that may be used in date time format models are summarized
in Table 5-7. Punctuation marks are used to separate format elements. Three types
of suffixes exist to format components of date time elements. Furthermore, character
literals may be included in a date format model if they are enclosed in double
quotation marks. The results in Table 5-7 are obtained by applying the TO_CHAR
function using the date 12/SEP/08 14:31 with the format masks listed in the
description and format mask column.
The JOB_HISTORY table keeps track of jobs occupied by employees in the
company. The query in Figure 5-3 retrieves a descriptive sentence about the quitting
date for each employee based on their END_DATE, EMPLOYEE_ID, and JOB_ID
fields. A character expression is concatenated to a TO_CHAR function call with a
format model of: 'fmDay "the "ddth "of" Month YYYY'. The fm modifier is used to
trim blank spaces that trail the names of the shorter days and shorter months. The
two character literals enclosed in double quotation marks are the words: "the" and
"of". The 'th' format model is applied to the 'dd' date element to create an ordinal

Format Element

Description and
Format Mask

Result

-/.,?#!

Punctuation marks: 'MM.YY'

09.08

"any character literal"

Character literals:
'"Week" W "of " Month'

Week 2 of September

TH

Positional or ordinal text:
'DDth "of " Month'

12TH of September

SP

Spelled out number:
'MmSP Month Yyyysp'

Nine September
Two Thousand Eight

THSP or SPTH

Spelled out positional or
ordinal number:
'hh24SpTh'

Fourteenth

TABLE 5-7

Miscellaneous
Date Format
Masks

Use the TO_CHAR, TO_NUMBER, and TO_DATE Conversion Functions

FIGURE 5-3

239

TO_CHAR function with dates

day such as the 17th or 31st. The 'Month' format model displays the full name of the
month element of the END_DATE column in title case. Finally, the YYYY format
mask retrieves the 4-digit year component.

EXERCISE 5-1
Converting Dates into Characters Using the TO_CHAR Function
You are required to retrieve a list of FIRST_NAME and LAST_NAME values and an
expression based on the HIRE_DATE column for employees hired on a Saturday. The
expression must be aliased as START_DATE and a HIRE_DATE value of 17-FEB-1996
must return the following string:
Saturday, the 17th of February, One Thousand Nine Hundred Ninety-Six.
1. Start SQL Developer and connect to the HR schema.
2. The WHERE clause is
WHERE TO_CHAR(HIRE_DATE,'fmDay') = 'Saturday'

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The fm modifier is necessary to remove trailing blanks since a comparison
with a character literal is performed and an exact match is required.
3. The START_DATE expression is
TO_CHAR(HIRE_DATE, 'fmDay, "the "ddth "of " Month, Yyyysp.')
The year format mask results in it being spelled out in title case.
4. The SELECT clause is therefore
SELECT FIRST_NAME, LAST_NAME, TO_CHAR(HIRE_DATE, 'fmDay,
"the "ddth "of " Month, Yyyysp.') START_DATE
5. The FROM clause is
FROM EMPLOYEES
6. Executing this statement returns employees’ names and the START_DATE
expression as shown in the following illustration below:

Use the TO_CHAR, TO_NUMBER, and TO_DATE Conversion Functions

241

Converting Characters to Dates Using the TO_DATE Function
The TO_DATE function returns an item of type DATE. Character strings converted
to dates may contain all or just a subset of the date time elements comprising a DATE.
When strings with only a subset of the date time elements are converted, Oracle
provides default values to construct a complete date. Components of character strings
are associated with different date time elements using a format model or mask. The
syntax is as follows:
TO_DATE(string1, [format], [nls_parameter]),
Only the string1 parameter is mandatory and if no format mask is supplied, string1
must take the form of a value that can be implicitly converted into a date. The optional
format parameter is almost always used and is specified in single quotation marks. The
format masks are identical to those listed in Tables 5-4, 5-5, and 5-6. The TO_DATE
function has an fx modifier which is similar to fm used with the TO_CHAR function.
fx specifies an exact match for string1 and the format mask. When the fx modifier is
specified, character items that do not exactly match the format mask yield an error.
Consider the following five queries:
Query 1: select to_date('25-DEC-2010') from dual;
Query 2: select to_date('25-DEC') from dual;
Query 3: select to_date('25-DEC', 'DD-MON') from dual;
Query 4: select to_date('25-DEC-2010 18:03:45', 'DD-MON-YYYY
HH24:MI:SS') from dual;
Query 5: select to_date('25-DEC-10', 'fxDD-MON-YYYY') from dual;

Query 1 evaluates the string 25-DEC-2010 and has sufficient information to
implicitly convert it into a DATE item with a default mask of DD-MON-YYYY. The
hyphen separator could be substituted with another punctuation character. Since no
time components are provided, the time for this converted date is set to midnight
or 00:00:00. Query 2 cannot implicitly convert the string into a date because there
is insufficient information and an “ORA-01840: input value is not long enough for
date format” error is returned. By supplying a format mask DD-MON to the string
25-DEC in query 3, Oracle can match the number 25 to DD and the abbreviated
month name DEC to the MON component. Year and time components are absent,
so the current year returned by the SYSDATE function is used and the time is set to
midnight. If the current year is 2009, query 3 returns the date 25/DEC/09 00:00:00.
Query 4 performs a complete conversion of a string with all the date time elements
present, and no default values are supplied by Oracle. Query 5 uses the fx modifier in
its format mask. Since the year component of the string is 10 and the corresponding
format mask is YYYY, the fx modifier results in an “ORA-01862: the numeric value
does not match the length of the format item” error being returned.

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FIGURE 5-4

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The TO_DATE function

The TO_DATE function is used in the WHERE clause in Figure 5-4 to limit the
rows returned for those employees hired after 12 January 2000. The format mask
matches 01 to MM, 12 to DD and 2000 to YYYY.

Converting Characters to Numbers
Using the TO_NUMBER Function
The TO_NUMBER function returns an item of type NUMBER. Character strings
converted into numbers must be suitably formatted so that any nonnumeric
components are translated or stripped away with an appropriate format mask.
The syntax is as follows:
TO_NUMBER(string1, [format], [nls_parameter]),

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243

SCENARIO & SOLUTION
Your task is to extract the day and month portion
of a date column and compare it with the
corresponding components of the current system
date. Can such a comparison be performed?

Yes. The TO_CHAR function used on a date item
with a format mask like 'DD-MON' causes the day
and month component to be isolated. This value can
be compared with the current system date using the
following expression:
TO_CHAR(SYSDATE, 'DD-MON')

A report of profit and loss is required with the
results displayed as follows: if the amount is
negative, it must be enclosed in angle brackets.
The amount must be displayed with a leading
dollar sign. Can results be retrieved in the
specified format?

Yes. The numeric amount must be converted into a
character string using the TO_CHAR function with
a format mask that encloses it in angle brackets if it
is negative and precedes it with a dollar sign. The
following function call retrieves the results in the
required format:
TO_CHAR(AMOUNT, '$999999PR')

You are asked to input past employee data into
the JOB_HISTORY table from a paper-based
source, but the start date information is only
available as the year the employee started. Can
this value be converted into the first of January
of the year?

Yes. Consider the conversion function call TO_
DATE('2000','YYYY') for an employee who started in
the year 2000. If this date is extracted as follows, the
character string 01/01/2000 is returned
TO_CHAR(TO_DATE('2000','YYYY'),
'MM/DD/YYYY')

Only the string1 parameter is mandatory and if no format mask is supplied, it
must be a value that can be implicitly converted into a number. The optional format
parameter is specified in single quotation marks. The format masks are identical to
those listed in Table 5-3. Consider the following two queries:
Query 1: select to_number('$1,000.55') from dual;
Query 2: select to_number('$1,000.55','$999,999.99') from dual;

Query 1 cannot perform an implicit conversion to a number because of the dollar
sign, comma, and period and returns the error, “ORA-1722: invalid number.” Query
2 matches the dollar symbol, comma, and period from the string to the format mask
and, although the numeric width is larger than the string width, the number 1000.55
is returned.
Figure 5-5 shows how the SUBSTR function was first used to extract the last eight
characters from the PHONE_NUMBER character column. The TO_NUMBER

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FIGURE 5-5

Using Conversion Functions and Conditional Expressions

The TO_NUMBER function

function was then used to convert these eight characters, including a decimal
point, into a number which was multiplied by 10000, for employees belonging to
DEPARTMENT_ID 30.

Read the exam questions
very carefully.The TO_NUMBER function
converts character items into numbers.
If you convert a number using a shorter
format mask, an error is returned. If you
convert a number based on a longer format

mask, the original number is returned.
Be careful not to confuse TO_NUMBER
conversions with TO_CHAR. For example,
TO_NUMBER(123.56,'999.9') returns an
error, while TO_CHAR(123.56,'999.9')
returns 123.6.

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245

CERTIFICATION OBJECTIVE 5.03

Apply Conditional Expressions
in a SELECT Statement
Nested functions were introduced in Chapter 4, but a formal discussion of this concept
is provided in this section. Two new categories of functions are also introduced.
These include the general functions, which provide the language for dealing effectively
with NULL values, and the conditional functions, which support conditional logic in
expressions. This certification objective covers the following areas:
■ Nested functions
■ General functions
■ Conditional functions

Nested Functions
Nested functions use the output from one function as the input to another. Functions
always return exactly one result. Therefore, you can reliably consider a function call
in the same way as you would a literal value, when providing input parameters to a
function. Single row functions can be nested to any level of depth. The general form
of a function is as follows:
Function1(parameter 1, parameter2,…) = result1
Substituting function calls as parameters to other functions may lead to an
expression such as the following:
F1( param1.1, F2( param2.1, param2.2, F3( param3.1)), param1.3)
Nested functions are first evaluated before their return values are used as parametric
input to other functions. They are evaluated from the innermost to outermost levels.
The preceding expression is evaluated as follows:
1. F3(param3.1) is evaluated and its return value provides the third parameter
to function F2 and may be called: param2.3.
2. F2(param2.1,param2.2,param2.3) is evaluated and its return value provides
the second parameter to function F1 and is param1.2.

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3. F1(param1.1, param1.2, param1.3) is evaluated and the result is returned to
the calling program.
Function F3 is said to be nested three levels deep in this example. Consider the
following query:
select length(to_char(to_date('28/10/09', 'DD/MM/RR'),'fmMonth'))from dual;

There are three functions in the SELECT list which, from inner to outer levels,
are TO_DATE, TO_CHAR and LENGTH. The query is evaluated as follows:
1. The innermost function is evaluated first. TO_DATE('28/10/09','DD/MM/RR')
converts the character string 28/10/09 into the DATE value 28-OCT-2009. The
RR format mask is used for the year portion. Therefore, the century component
returned is the current century (the twenty-first), since the year component is
between 0 and 49.
2. The second innermost function is evaluated next. TO_CHAR('28-OCT-2009',
'fmMonth') converts the given date based on the Month format mask and
returns the character string October. The fm modifier trims trailing blank
spaces from the name of the month.
3. Finally, the LENGTH('October') function is evaluated and the query returns
the number 7.

SCENARIO & SOLUTION
Are nested functions evaluated from the
outermost level to the innermost level?

No. Nested functions are resolved from the innermost
nested level moving outward.

Must all functions in a nested expression return
the same data type?

No. The data types of the parameters of nested
functions may be different from each other. It is
important to ensure that the correct data types are
always supplied to functions to avoid errors.

Is there a simpler way to display the SALARY
information from the EMPLOYEES table in
the form $13,000 without using the following
statement?
SELECT '$'|| SUBSTR(SALARY,1,
MOD(LENGTH(SALARY),3))||',
'|| SUBSTR(SALARY, MOD
(LENGTH(SALARY),3)+1)

Yes. A simple and elegant solution is to use the
TO_CHAR function with the '$99G999' format mask
SELECT TO_CHAR(SALARY, '$99G999') FROM
EMPLOYEES;

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247

General Functions
General functions simplify working with columns that potentially contain null
values. These functions accept input parameters of all data types. The services they
offer are primarily relevant to null values.
The functions examined in the next sections include the NVL function, which
provides an alternative value to use if a null is encountered. The NVL2 function
performs a conditional evaluation of its first parameter and returns a value if a null is
encountered and an alternative if the parameter is not null. The NULLIF function
compares two terms and returns a null result if they are equal, otherwise it returns
the first term. The COALESCE function accepts an unlimited number of parameters
and returns the first nonnull parameter else it returns null.

The NVL Function
The NVL function evaluates whether a column or expression of any data type is null
or not. If the term is null, an alternative not null value is returned; otherwise, the
initial term is returned.
The NVL function takes two mandatory parameters. Its syntax is NVL(original,
ifnull), where original represents the term being tested and ifnull is the result
returned if the original term evaluates to null. The data types of the original and ifnull
parameters must always be compatible. They must either be of the same type, or
it must be possible to implicitly convert ifnull to the type of the original parameter.
The NVL function returns a value with the same data type as the original parameter.
Consider the following three queries:
Query 1: select nvl(1234) from dual;
Query 2: select nvl(null,1234) from dual;
Query 3: select nvl(substr('abc',4),'No substring exists') from
dual;

Since the NVL function takes two mandatory parameters, query 1 returns the
error, “ORA-00909: invalid number of arguments.” Query 2 returns 1234 after the
null keyword is tested and found to be null. Query 3 involves a nested SUBSTR
function that attempts to extract the fourth character from a 3-character string. The
inner function returns null, leaving the NVL(null,'No substring exists') function to
execute, which then returns the string ‘No substring exists’.
Figure 5-6 shows two almost identical queries. Both queries select rows where
the LAST_NAME begins with the letter E. The LAST_NAME, SALARY, and
COMMISSION_PCT columns are also selected. The difference between the queries is
in the calculated expression aliased as MONTHLY_COMMISSION. Due to the NVL

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FIGURE 5-6

Using Conversion Functions and Conditional Expressions

The NVL function

function in the first query, numeric results are returned. The second query returns some
null values and one numeric item even though 1000 is added to each row.
It is tempting to dive in and construct a complex expression comprising many
nested function calls, but this approach evolves with practice and experience.
Conceptualize a solution to a query and break it down into the component
function calls.The DUAL table is useful for ad hoc logical testing and debugging
of separate function calls. Do not be afraid to execute queries as many times as
you wish to perfect the components before assembling them into progressively
larger components.Test and debug these until the final expression is formed.

The NVL2 Function
The NVL2 function provides an enhancement to NVL but serves a very similar
purpose. It evaluates whether a column or expression of any data type is null or not.

Apply Conditional Expressions in a SELECT Statement

249

If the first term is not null, the second parameter is returned, else the third parameter is
returned. Recall that the NVL function is different since it returns the original term if it
is not null.
The NVL2 function takes three mandatory parameters. Its syntax is NVL2(original,
ifnotnull, ifnull), where original represents the term being tested. Ifnotnull is returned
if original is not null, and ifnull is returned if original is null. The data types of the
ifnotnull and ifnull parameters must be compatible, and they cannot be of type LONG.
They must either be of the same type, or it must be possible to convert ifnull to the
type of the ifnotnull parameter. The data type returned by the NVL2 function is
the same as that of the ifnotnull parameter. Consider the following three queries:
Query 1: select nvl2(1234,1,'a string') from dual;
Query 2: select nvl2(null,1234,5678) from dual;
Query 3: select nvl2(substr('abc',2),'Not bc','No substring') from dual;

The ifnotnull term in query 1 is a number and the ifnull parameter is 'a string'.
Since there is a data type incompatibility between them, an “ORA-01722: invalid
number” error is returned. Query 2 returns the ifnull parameter, which is 5678. Query
3 extracts the characters “bc” using the SUBSTR function and the NVL2('bc',Not
bc','No Substring') function is evaluated. The ifnotnull parameter, the string 'Not bc',
is returned.
Figure 5-7 shows how the NVL2 function is used to provide descriptive text
classifying employees with LAST_NAME values beginning with “F” into commission
and noncommission earners based on the nullable COMMISSION_PCT column.

The NULLIF Function
The NULLIF function tests two terms for equality. If they are equal the function
returns a null, else it returns the first of the two terms tested.
The NULLIF function takes two mandatory parameters of any data type. The
syntax is NULLIF(ifunequal, comparison_term), where the parameters ifunequal and
comparison_term are compared. If they are identical, then NULL is returned. If they
differ, the ifunequal parameter is returned. Consider the following three queries:
Query 1: select nullif(1234,1234) from dual;
Query 2: select nullif(1234,123+1) from dual;
Query 3: select nullif('24-JUL-2009','24-JUL-09') from dual;

Query 1 returns a null value since the parameters are identical. The arithmetic
equation in query 2 is not implicitly evaluated, and the NULLIF function finds

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FIGURE 5-7

Using Conversion Functions and Conditional Expressions

The NVL2 function

1234 different from 123+1, so it returns the ifunequal parameter, which is 1234. The
character literals in query 3 are not implicitly converted to DATE items and are
compared as two character strings by the NULLIF function. Since the strings are of
different lengths, the ifunequal parameter 24-JUL-2009 is returned.
Figure 5-8 shows how NULLIF is nested as a parameter to the NVL2 function.
The NULLIF function itself has the SUBSTR and UPPER character functions
embedded in the expression used as its ifunequal parameter. The EMAIL column
is compared with an expression, formed by concatenating the first character of
the FIRST_NAME to the uppercase equivalent of the LAST_NAME column, for
employees with 4-character-long first names. When these terms are equal, NULLIF
returns a null, else it returns the evaluated ifunequal parameter. This is used as a
parameter to NVL2. The NVL2 function provides descriptive text classifying rows as
matching the pattern or not.

Apply Conditional Expressions in a SELECT Statement

FIGURE 5-8

251

The NULLIF function

EXERCISE 5-2
Using NULLIF and NVL2 for Simple Conditional Logic
You are required to return a set of rows from the EMPLOYEES table with
DEPARTMENT_ID values of 100. The set must also contain FIRST_NAME
and LAST_NAME values and an expression aliased as NAME_LENGTHS. This
expression must return the string 'Different Length' if the length of the FIRST_NAME
differs from that of the LAST_NAME, else the string 'Same Length' must be returned.
1. Start SQL Developer and connect to the HR schema.
2. The NAME_LENGTHS expression may be calculated in several ways. The
solution provided uses the NULLIF function to test whether the LENGTH
values returned for the FIRST_NAME and LAST_NAME columns are

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the same. If they are, NULL is returned, else the LENGTH of the LAST_NAME
is returned. If the outer function (NVL2) gets a NULL parameter, the string
'Same Length' is returned, else the string 'Different Length' is returned.
3. The SELECT clause is therefore
SELECT FIRST_NAME, LAST_NAME, NVL2(NULLIF(LENGTH(LAST_
NAME), LENGTH(FIRST_NAME)), 'Different Length', 'Same Length')
NAME_LENGTHS
4. The FROM clause is
FROM EMPLOYEES
5. The WHERE clause is
WHERE DEPARTMENT_ID=100
6. Executing this statement returns employees’ names and the NAME_LENGTHS
expression as shown in the following illustration:

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253

The COALESCE Function
The COALESCE function returns the first nonnull value from its parameter list. If
all its parameters are null, then null is returned.
The COALESCE function takes two mandatory parameters and any number of
optional parameters. The syntax is COALESCE(expr1, expr2,…,exprn), where expr1
is returned if it is not null, else expr2 if it is not null, and so on. COALESCE is a
general form of the NVL function, as the following two equations illustrate:
COALESCE(expr1,expr2) = NVL(expr1,expr2)
COALESCE(expr1,expr2,expr3) = NVL(expr1,NVL(expr2,expr3))
The data type COALESCE returns if a not null value is found is the same as that
of the first not null parameter. To avoid an “ORA-00932: inconsistent data types”
error, all not null parameters must have data types compatible with the first not null
parameter. Consider the following three queries:
Query 1: select coalesce(null, null, null, 'a string') from dual;
Query 2: select coalesce(null, null, null) from dual;
Query 3: select coalesce(substr('abc',4),'Not bc','No substring') from dual;

Query 1 returns the fourth parameter: a string, since this is the first not null
parameter encountered. Query 2 returns null because all its parameters are null.
Query 3 evaluates its first parameter, which is a nested SUBSTR function, and finds
it to be null. The second parameter is not null so the string 'Not bc' is returned.
The STATE_PROVINCE, POSTAL_CODE, and CITY information was retrieved
from the LOCATIONS table for the rows with COUNTRY_ID values of UK, IT, or
JP. As Figure 5-9 shows, the COALESCE function returns the STATE_PROVINCE
value for a row if it is not null. If it is null the POSTAL_CODE value is returned. If
that is null, the CITY field is returned, else the result is null.

The parameters of the
general function NVL2 can be confusing
if you are already familiar with NVL.
NVL(original, ifnull) returns original if
it is not null, else ifnull is returned.The
NVL2(original, ifnotnull, ifnull) function
returns ifnotnull if original is not null,

else ifnull is returned.The confusion may
arise because the second parameter
in the NVL function is ifnull, while the
second parameter in the NVL2 function is
ifnotnull. Be mindful of the meaning of
the parameter positions in functions.

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FIGURE 5-9

Using Conversion Functions and Conditional Expressions

The COALESCE function

Conditional Functions
Conditional logic, also known as if-then-else logic, refers to choosing a path of
execution based on data values meeting certain conditions. Conditional functions,
like DECODE and the CASE expression, return different values based on evaluating
comparison conditions. These conditions are specified as parameters to the DECODE
function and the CASE expression. The DECODE function is specific to Oracle,
while the CASE expression is ANSI SQL compliant. An example of if-then-else logic
is: if the country value is Brazil or Australia, then return Southern Hemisphere, else
return Northern Hemisphere.

The DECODE Function
Although its name sounds mysterious, this function is straightforward. The
DECODE function implements if-then-else conditional logic by testing its first two
terms for equality and returns the third if they are equal and optionally returns
another term if they are not.

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255

The DECODE function takes at least three mandatory parameters, but can
take many more. The syntax of the function is DECODE(expr1,comp1, iftrue1,
[comp2,iftrue2...[ compN,iftrueN]], [iffalse]). These parameters are evaluated as
shown in the following pseudocode example:
If expr1 = comp1 then return iftrue1
else if expr1 = comp2 then return iftrue2
...
...
else if expr1 = compN then return iftrueN
else return null | iffalse;

Expr1 is compared to comp1. If they are equal, then iftrue1 is returned. If expr1
is not equal to comp1, then what happens next depends on whether the optional
parameters comp2 and iftrue2 are present. If they are, then expr1 is compared to
comp2. If they are equal, then iftrue2 is returned. If not, what happens next depends
on whether further compn,iftruen pairs exist, and the cycle continues until no
comparison terms remain. If no matches have been found and if the iffalse parameter
is defined, then iffalse is returned. If the iffalse parameter does not exist and no
matches are found, a null value is returned.
All parameters to the DECODE function may be expressions. The return data
type is the same as that of the first matching comparison item. The expression expr1
is implicitly converted to the data type of the first comparison parameter comp1.
As the other comparison parameters comp2…compn are evaluated, they too are
implicitly converted to the same data type as comp1. Decode considers two nulls to
be equivalent, so if expr1 is null and comp3 is the first null comparison parameter
encountered, then the corresponding result parameter iftrue3 is returned. Consider
the following three queries:
Query 1: select decode(1234,123,'123 is a match') from dual;
Query 2: select decode(1234,123,'123 is a match','No match') from dual;
Query 3: select decode('search','comp1','true1', 'comp2','true2',
'search','true3', substr('2search',2,6)),'true4', 'false') from dual;

Query 1 compares the number 1234 with the first comparison term 123. Since they
are not equal, the first result term cannot be returned. Further, as there is no default
iffalse parameter defined, a null is returned. Query 2 is identical to the first except
that an iffalse parameter is defined. Therefore, since 1234 is not equal to 123, the
string ‘No match’ is returned. Query 3 searches through the comparison parameters
for a match. The character terms ‘comp1’ and ‘comp2’ are not equal to search, so the

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results true1 and true2 are not returned. A match is found in the third comparison
term ‘comp3’ (parameter 6), which contains the string search. Therefore, the third
result term iftrue3 (parameter 7) containing the string ‘true3’ is returned. Note that
since a match has been found, no further searching takes place. So, although the
fourth comparison term (parameter 8) is also a match to expr1, this expression is
never evaluated, because a match was found in an earlier comparison term.
Distinct occurrences of the COUNTRY_ID column values in the LOCATIONS
table have been classified into either Northern or Southern Hemisphere countries
using a DECODE function. Figure 5-10 shows how the COUNTRY_ID column
is the expression for which a match is sought. If the COUNTRY_ID value is
BR (Brazil) or AU (Australia), then the function returns the string Southern
Hemisphere, else Northern Hemisphere is returned.

FIGURE 5-10

The DECODE function

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257

INSIDE THE EXAM
The certification objectives in this chapter are
primarily measured with practical examples
that require you to predict the results returned
by an expression. Explicit conversion functions
with their many format masks are tested. The
TO_CHAR function; fm modifier; and the
sp, th, and spth format models are commonly
examined. Use of the TO_CHAR function to
convert numbers into characters is tested, and
emphasis is often placed on the format masks
that determine numeric width, dollar symbol,
group separator, and decimal separator formats.
Many questions consist of expressions with
nested functions, and it is vital for you to know
how to interpret and trace them. The innermost
to outermost sequence of evaluation of nested
functions is important and must be remembered.
The general functions NVL, NVL2,
NULLIF, and COALESCE all pertain to

working with NULL. Since null values are
prevalent in many databases, a thorough
understanding of all these functions,
particularly NVL and NVL2, is required.
The DECODE function and CASE
expression are sometimes perceived as
complex and difficult to understand. They
are, however, two of the simplest and most
useful functions guaranteed to be examined.
Pay attention to the positional meaning of
the parameters when using the DECODE
function. The simple and searched CASE
expression differs from the functions
discussed earlier. This is due to the
CASE…END block, which encloses one or
more WHEN…OTHER pairs and optionally
an ELSE statement. Practice with this
expression and you will quickly learn its
readable and standardized structure.

The CASE Expression
Virtually all third and fourth generation programming languages implement a case
statement. Like the DECODE function, the CASE expression facilitates if-then-else
conditional logic. There are two variants of the CASE expression. The simple CASE
expression lists the conditional search item once, and equality to the search item is
tested by each comparison expression. The searched CASE expression lists a separate
condition for each comparison expression.

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The CASE expression takes at least three mandatory parameters but can take
many more. Its syntax depends on whether a simple or a searched CASE expression
is used. The syntax for the simple CASE expression is as follows:
CASE search_expr
WHEN comparison_expr1 THEN iftrue1
[WHEN comparison_expr2 THEN iftrue2
…
WHEN comparison_exprN THEN iftrueN
ELSE iffalse]
END
The simple CASE expression is enclosed within a CASE…END block and
consists of at least one WHEN…THEN statement. In its simplest form, with one
WHEN…THEN statement, the search_expr is compared with the comparison_expr1.
If they are equal, then the result iftrue1 is returned. If not, a null value is returned unless
an ELSE component is defined, in which case, the default iffalse value is returned.
When more than one WHEN…THEN statement exists in the CASE expression,
searching for a matching comparison expression continues until a match is found.
The search, comparison, and result parameters can be column values, expressions,
or literals but must all be of the same data type. Consider the following query:
select
case substr(1234,1,3)
when '134' then '1234 is a match'
when '1235' then '1235 is a match'
when concat('1','23') then concat('1','23')||' is a match'
else 'no match'
end
from dual;

The search expression derived from the SUBSTR(1234,1,3) is the character
string 123. The first WHEN...THEN statement compares the string 134 with
123. Since they are not equal, the result expression is not evaluated. The second
WHEN…THEN statement compares the string 1235 with 123 and again, they are
not equal. The third WHEN…THEN statement compares the results derived from
the CONCAT('1','23') expression, which is 123, to the search expression. Since
they are identical, the third results expression ‘123 is a match’, is returned.
The LAST_NAME and HIRE_DATE columns for employees with
DEPARTMENT_ID values of 10 or 60 are retrieved along with two numeric
expressions and one CASE expression, as shown in Figure 5-11.
Assume that SYSDATE is 01-JAN-2008. The numeric expression aliased as
YEARS returns a truncated value obtained by dividing the months of service by 12.

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FIGURE 5-11

259

The CASE expression

Five categories of loyalty classification based on years of service are defined by
truncating the quotient obtained by dividing the months of service by 60. This forms
the search expression in the CASE statement. None of the rows in the dataset match
the comparison expression in the first WHEN…THEN statement, but as Figure 5-11
shows, five rows met the remaining WHEN…THEN statements and one row is caught
by the ELSE statement.
The syntax for the searched CASE expression is as follows:
CASE
WHEN condition1 THEN iftrue1
[WHEN condition2 THEN iftrue2
…
WHEN conditionN THEN iftrueN
ELSE iffalse]
END

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The searched CASE expression is enclosed within a CASE…END block and
consists of at least one WHEN…THEN statement. In its simplest form with one
WHEN…THEN statement, condition1 is evaluated; if it is true, then the result
iftrue1 is returned. If not, a null value is returned unless an ELSE component is
defined, in which case the default iffalse value is returned. When more than one
WHEN…THEN statement exists in the CASE expression, searching for a matching
comparison expression continues until one is found. The query to retrieve the
identical set of results to those obtained in Figure 5-11, using a searched CASE
expression is listed next:
select last_name, hire_date,
trunc(months_between(sysdate,hire_date)/12) years,
trunc(months_between(sysdate,hire_date)/60) "Years divided
case
when trunc(months_between(sysdate,hire_date)/60) < 1 then
when trunc(months_between(sysdate,hire_date)/60) < 2 then
when trunc(months_between(sysdate,hire_date)/60) < 3 then
when trunc(months_between(sysdate,hire_date)/60) < 4 then
else 'Furniture'
end Loyalty
from employees
where department_id in (60,10);

by 5",
'Intern'
'Junior'
'Intermediate'
'Senior'

EXERCISE 5-3
Using the DECODE Function
You are requested to query the LOCATIONS table for rows with the value US in the
COUNTRY_ID column. An expression aliased as LOCATION_INFO is required to
evaluate the STATE_PROVINCE column values and returns different information as
per the following table. Sort the output based on the LOCATION_INFO expression.
If STATE_PROVINCE is

The value returned is

Washington

The string ‘Headquarters’

Texas

The string ‘Oil Wells’

California

The CITY column value

New Jersey

The STREET_ADDRESS column value

1. Start SQL Developer and connect to the HR schema.
2. The LOCATION_ID expression may be calculated in several different ways.
This includes using a CASE expression or a DECODE function. The solution
below uses DECODE.

Apply Conditional Expressions in a SELECT Statement

261

3. The SELECT clause is
SELECT DECODE(STATE_PROVINCE, 'Washington', 'Headquarters',
'Texas', 'Oil Wells', 'California', CITY, 'New Jersey', STREET_ADDRESS)
LOCATION_INFO
Notice the mixture of character literals and columns specified as parameters
to the DECODE function.
4. The FROM clause is
FROM EMPLOYEES
5. The WHERE clause is
WHERE COUNTRY_ID='US'
6. The ORDER BY clause is
ORDER BY LOCATION_INFO
7. The result of executing this statement is shown in the following illustration:

262

Chapter 5:

Using Conversion Functions and Conditional Expressions

CERTIFICATION SUMMARY
This chapter builds on the single row functions introduced previously. The concepts
of implicit and explicit data type conversion are explained along with the reliability
risks associated with implicit conversions.
Date to character and number to character conversions are described using
the TO_CHAR function. A variety of format models or masks are available. The
TO_NUMBER function performs character to number conversions, while the
TO_DATE function performs character to date data type conversions.
Nested functions and their evaluation is one of the most valuable lessons in this
chapter. Understanding this fundamental concept is crucial. The general functions
NVL, NVL2, NULLIF, and COALESCE are designed to simplify working with null
values and provide basic conditional logic functionality.
DECODE is an Oracle-specific function that supports if-then-else logic in the
context of an SQL statement along with the ANSI-compliant CASE expression.
The two variants are the simple CASE and searched CASE expressions. These
conditional functions are straightforward and extremely useful.
The conversion, general, and conditional functions add significantly to the
foundational knowledge of SQL you acquired from previous chapters and will hold
you in good stead as you progress with your learning.

Two-Minute Drill

✓

263

TWO-MINUTE DRILL
Describe Various Types of Conversion
Functions Available in SQL
❑ When values do not match the defined parameters of functions, Oracle

attempts to convert them into the required data types. This is known as
implicit conversion.
❑ Explicit conversion occurs when a function like TO_CHAR is invoked to

change the data type of a value.
❑ The TO_CHAR function performs date to character and number to character

data type conversions.
❑ Character items are explicitly transformed into date values using the TO_DATE

conversion function.
❑ Character items are changed into number values using the TO_NUMBER

conversion function.

Use the TO_CHAR,TO_NUMBER,
and TO_DATE Conversion Functions
❑ The TO_CHAR function returns an item of type VARCHAR2.
❑ Format models or masks prescribe patterns that character strings must match

to facilitate accurate and consistent conversion into number or date items.
❑ When the TO_CHAR function performs number to character conversions,

the format mask can specify currency, numeric width, position of decimal
operator, thousands separator, and many other formatting codes.
❑ The format masks available when TO_CHAR is used to convert character

items to date include day, week, month, quarter, year, and century.
❑ Format masks must always be specified enclosed in single quotes.
❑ When performing date to character conversion, the format mask specifies which

date elements are extracted and whether the element should be described by
a long or abbreviated name.
❑ Character terms, like month and day names, extracted from dates with the

TO_CHAR function are automatically padded with spaces that may be
trimmed by prefixing the format mask with the fm modifier.

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❑ The TO_DATE function has an fx modifier that specifies an exact match for

the character string to be converted and the date format mask.

Apply Conditional Expressions in a SELECT Statement
❑ Nesting functions use the output from one function as the input to another.
❑ The NVL function either returns the original item unchanged or an alternative

item if the initial term is null.
❑ The NVL2 function returns a new if-null item if the original item is null or an

alternative if-not-null item if the original term is not null.
❑ The NULLIF function tests two terms for equality. If they are equal, the function

returns null, else it returns the first of the two terms tested.
❑ The COALESCE function returns the first nonnull value from its parameter

list. If all its parameters are null, then a null value is returned.
❑ The DECODE function implements if-then-else conditional logic by

testing two terms for equality and returning the third term if they are equal
or, optionally, some other term if they are not.
❑ There are two variants of the CASE expression used to facilitate if-then-else

conditional logic: the simple CASE and searched CASE expressions.

Self Test

265

SELF TEST
The following questions will measure your understanding of the material presented in this chapter.
Read all the choices carefully because there may be more than one correct answer. Choose all the
correct answers for each question.

Describe Various Types of Conversion Functions Available in SQL
1. What type of conversion is performed by the following statement?
SELECT LENGTH(3.14285) FROM DUAL;
(Choose the best answer.)
A. Explicit conversion
B. Implicit conversion
C. TO_NUMBER function conversion
D. None of the above
2. Choose any incorrect statements regarding conversion functions. (Choose all that apply.)
A. TO_CHAR may convert date items to character items.
B. TO_DATE may convert character items to date items.
C. TO_CHAR may convert numbers to character items.
D. TO_DATE may convert date items to character items.

Use the TO_CHAR,TO_NUMBER, and TO_DATE Conversion Functions
3. What value is returned after executing the following statement?
SELECT TO_NUMBER(1234.49, '999999.9') FROM DUAL;
(Choose the best answer.)
A. 1234.49
B. 001234.5
C. 1234.5
D. None of the above
4. What value is returned after executing the following statement?
SELECT TO_CHAR(1234.49, '999999.9') FROM DUAL;
(Choose the best answer.)
A. 1234.49
B. 001234.5
C. 1234.5
D. None of the above

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5. If SYSDATE returns 12-JUL-2009, what is returned by the following statement?
SELECT TO_CHAR(SYSDATE, 'fmMONTH, YEAR') FROM DUAL;
(Choose the best answer.)
A. JUL, 2009
B. JULY, TWO THOUSAND NINE
C. JUL-09
D. None of the above
6. If SYSDATE returns 12-JUL-2009, what is returned by the following statement?
SELECT TO_CHAR(SYSDATE, 'fmDDth MONTH') FROM DUAL;
(Choose the best answer.)
A. 12TH JULY
B. 12th July
C. TWELFTH JULY
D. None of the above

Apply Conditional Expressions in a SELECT Statement
7. If SYSDATE returns 12-JUL-2009, what is returned by the following statement?
SELECT TO_CHAR(TO_DATE(TO_CHAR(SYSDATE,'DD'),'DD'),'YEAR') FROM DUAL;
(Choose the best answer.)
A. 2009
B. TWO THOUSAND NINE
C. 12-JUL-2009
D. None of the above
8. What value is returned after executing the following statement?
SELECT NVL2(NULLIF('CODA','SID'),'SPANIEL','TERRIER') FROM DUAL;
(Choose the best answer.)
A. SPANIEL
B. TERRIER
C. NULL
D. None of the above

Lab Question

267

9. What value is returned after executing the following statement?
SELECT NVL(SUBSTR('AM I NULL',10),'YES I AM') FROM DUAL;
(Choose the best answer.)
A. NO
B. NULL
C. YES I AM
D. None of the above
10. If SYSDATE returns 12-JUL-2009, what is returned by the following statement?
SELECT DECODE(TO_CHAR(SYSDATE,'MM'),'02','TAX DUE','PARTY') FROM DUAL;
(Choose the best answer.)
A. TAX DUE
B. PARTY
C. 02
D. None of the above

LAB QUESTION
Using SQL Developer or SQL*Plus, connect to the OE schema and complete the following tasks.
As part of a new marketing initiative, you are asked to prepare a list of customer birthdays that occur
between two days ago and seven days from now. The list should retrieve rows from the CUSTOMERS
table which include the CUST_FIRST_NAME, CUST_LAST_NAME, CUST_EMAIL, and DATE_
OF_BIRTH columns in ascending order based on the day and month components of the DATE_OF_
BIRTH value. An additional expression aliased as BIRTHDAY is required to return a descriptive message based on the following table. There are several approaches to solving this question. Your approach
may differ from the solution described here.
BIRTHDAY

CHARACTER STRING

Two days ago

Day before yesterday

One day ago

Yesterday

Today

Today

Tomorrow

Tomorrow

Two days in the future

Day after tomorrow

Within seven days from today

Later this week

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SELF TEST ANSWERS
Describe Various Types of Conversion Functions Available in SQL
✓ B. The number 3.14285 is given as a parameter to the LENGTH function. There is a data
1. ®
type mismatch, but Oracle implicitly converts the parameter to the character string '3.14285',
allowing the function to operate correctly.
®
˚ A, C, and D are incorrect. Explicit conversion occurs when a function like TO_CHAR is
executed. C is the correct length of the string '3.14285', but this is not asked for in the question.
✓ D. Dates are only converted into character strings using TO_CHAR and not the TO_DATE
2. ®
function.
®
˚ A, B, and C are correct statements.

Use the TO_CHAR, TO_NUMBER, and TO_DATE Conversion Functions
✓ D. An “ORA-1722: invalid number” error is returned because the statement is
3. ®
trying to convert a number using an incompatible format mask. If the expression was
TO_NUMBER(1234.49, '999999.99'), the number 1234.49 would be returned.
®
˚ A, B, and D are incorrect.
✓ C. For the number 1234.49 to match the character format mask with one decimal place,
4. ®
the number is first rounded to 1234.5 before TO_CHAR converts it into the string '1234.5'.
®
˚ A, B, and D are incorrect. A cannot be returned because the format mask only allows one
character after the decimal point. B would be returned if the format mask was '009999.9'.
✓ B. The MONTH and YEAR components of the format mask separated by a comma and a
5. ®
space indicate that TO_CHAR must extract the spelled out month and year values in uppercase
separated by a comma and a space. The fm modifier removes extra blanks from the spelled out
components.
®
˚ A, C, and D are incorrect. If the format mask was 'MON, YYYY' or 'MON-YY', A and C,
respectively, would be returned.
✓ A. The DD component returns the day of the month in uppercase. Since it is a number, it
6. ®
does not matter, unless the 'th' mask is applied, in which case that component is specified in
uppercase. MONTH returns the month spelled out in uppercase.
®
˚ B, C, and D are incorrect. B would be returned if the format mask was 'fmddth Month',
and C would be returned if the format mask was 'fmDDspth MONTH'.

Lab Answer

269

Apply Conditional Expressions in a SELECT Statement
✓ B. The innermost nested function is TO_CHAR(SYSDATE,'DD'), which extracts the
7. ®
day component of SYSDATE and returns the character 12. The next function executed is
TO_DATE('12','DD') where the character 12 is cast as the day component. When such an
incomplete date is provided, Oracle substitutes values from the SYSDATE function; since
SYSDATE is 12-JUL-2009, this is the date used. The outermost function executed in
TO_CHAR('12-JUL-2009','YEAR') returns the year spelled out as TWO THOUSAND NINE.
®
˚ A, C, and D are incorrect.
✓ A. The NULLIF function compares its two parameters and, since they are different, the
8. ®
first parameter is returned. The NVL2('CODA', 'SPANIEL','TERRIER') function call returns
SPANIEL since its first parameter is not null.
®
˚ B, C, and D are incorrect.
✓ C. The character literal 'AM I NULL' is nine characters long. Therefore, trying to obtain
9. ®
a substring beginning at the tenth character returns a null. The outer function then becomes
NVL(NULL,'YES I AM'), resulting in the string ‘YES I AM’ being returned.
®
˚ A, B, and D are incorrect.
✓ B. The innermost function TO_CHAR(SYSDATE, 'MM') results in the character string
10. ®
'07' being returned. The outer function is DECODE('07','02','TAX DUE','PARTY'). Since '07'
is not equal to '02', the else component 'PARTY' is returned.
®
˚ A, C, and D are incorrect. A would only be returned if the month component extracted
from SYSDATE was '02'.

LAB ANSWER
Using SQL Developer or SQL*Plus, connect to the OE schema and complete the following tasks.
1. Start SQL Developer and connect to the OE schema.
2. The dataset must be restricted to rows from the CUSTOMERS table where DATE_OF_BIRTH
values first have the same month as the current month, and second, the day component of the
DATE_OF_BIRTH is between two days ago and seven days from now. There are therefore two
conditions that make up the WHERE clause.

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3. The first condition of the WHERE clause is
TO_CHAR(DATE_OF_BIRTH,'MON')=TO_CHAR(SYSDATE,'MON')

4. The second condition is
TO_NUMBER(TO_CHAR(DATE_OF_BIRTH,'DD')) - TO_NUMBER(TO_CHAR(SYSDATE,'DD')) BETWEEN -2 AND 7

5. The BIRTHDAY expression may be calculated in several ways. Based on the preceding
WHERE clause, you are assured that the dataset is limited to the correct set of rows. The
SELECT clause has to retrieve and manipulate the records for appropriate display. The CASE
expression provides functionality for conditional logic and is well suited to this situation.
6. The case tested is based on the day component of the DATE_OF_BIRTH minus the day
component returned by the SYSDATE function. For example, if the difference is –2, then
the string 'Day before yesterday' is returned. The different CASE conditions are tested is a
similar manner. The ELSE component catches any rows not matching the CASE conditions
and returns the string 'Later in the week'.
7. The SELECT clause is therefore
SELECT CUST_FIRST_NAME, CUST_LAST_NAME, CUST_EMAIL, DATE_OF_BIRTH,
CASE TO_NUMBER(TO_CHAR(DATE_OF_BIRTH,'DD')) - TO_NUMBER(TO_CHAR(SYSDATE,'DD'))
WHEN -2 THEN 'Day before yesterday'
WHEN -1 THEN 'Yesterday'
WHEN 0 THEN 'Today'
WHEN 1 THEN 'Tomorrow'
WHEN 2 THEN 'Day after tomorrow'
ELSE 'Later this week'
END BIRTHDAY

8. The FROM clause is
FROM CUSTOMERS

9. The ORDER BY clause is
ORDER BY TO_CHAR(DATE_OF_BIRTH,'MMDD')

Lab Answer

271

10. Executing this statement returns the set of results matching this pattern as shown in
the following illustration for SYSDATE=08-JAN-2008:

This page intentionally left blank

6
Reporting Aggregated
Data Using the Group
Functions

CERTIFICATION OBJECTIVES
6.01

Describe the Group Functions

6.02

Identify the Available Group Functions

6.03

Group Data Using the GROUP BY
Clause

6.04

✓
Q&A

Include or Exclude Grouped Rows Using
the HAVING Clause
Two-Minute Drill
Self Test

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

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Reporting Aggregated Data Using the Group Functions

S

ingle-row functions, explored in Chapters 4 and 5, return a single value for each row in
a set of results. Group or aggregate functions operate on multiple rows. They are used to
count the number of rows or to find the average of specific column values in a dataset.
Many statistical operations, such as calculating standard deviation, medians, and averages, depend
on executing functions against grouped data and not just single rows.
Group functions are examined in two stages. First, their purpose and syntax is
discussed. Second, a detailed analysis of the AVG, SUM, MIN, MAX, and COUNT
functions is conducted. The concept of grouping or segregating data based on one or
more column values is explored before introducing the GROUP BY clause.
The WHERE clause restricts rows in a dataset before grouping, while the
HAVING clause restricts them after grouping. This chapter concludes with a
discussion of the HAVING clause.

CERTIFICATION OBJECTIVE 6.01

Describe the Group Functions
SQL group functions are defined and the different variants are discussed. The syntax
of selected group functions is explained, their data types are discussed, and the
DISTINCT keyword’s effect on them is explored. This discussion is divided into two
main areas:
■ Definition of group functions
■ Types and syntax of group functions

Definition of Group Functions
Group functions operate on aggregated data and return a single result per group.
These groups usually consist of zero or more rows of data. Single-row functions are
defined with the formula: F(x, y, z, …) = result, where x,y,z… are input parameters.
The function F executes on one row of the data set at a time and returns a result for
each row. Group functions may be defined using the following formula:
F(g1, g2, g3,…, gn) = result1, result2, result2,…, resultn;
The group function executes once for each cluster of rows and returns a single
result per group. These groups may be entire tables or portions of tables associated
using a common value or attribute. If all the rows in tables are presented as one

Describe the Group Functions

275

group to the group function then one result is returned. One or more group functions
may appear in the SELECT list as follows:
SELECT group_function(column or expression),…
FROM table [WHERE …] [ORDER BY…]

Consider the EMPLOYEES table. There are 107 rows in this table. Groups may
be created based on the common values that rows share. For example, the rows that
share the same DEPARTMENT_ID value may be clustered together. Thereafter,
group functions are executed separately against each unique group.
As Figure 6-1 shows, there are 12 distinct DEPARTMENT_ID values in the
EMPLOYEES table including a null value. The rows are distributed into 12 groups
based on common DEPARTMENT_ID values. The COUNT function executes
12 times, once for each group. Notice that the distinct groups do not contain the
same number of rows.
Group functions aggregate a number of values from multiple rows into a
single result.They are widely used for reporting purposes and are also known
as summary or aggregate functions. Useful aggregated data such as sum totals,
averages, and counts often form the basis of more sophisticated statistical
calculations. It is useful to have a good understanding of the data stored in
your application tables to maximize the quality of your aggregate reporting.

Types and Syntax of Group Functions
A brief description of the most commonly used group functions is provided next.
Many are examined in detail later in this chapter.
The COUNT function counts the number of rows in a group. Its syntax is as follows:
COUNT({*|[DISTINCT|ALL] expr}) ;
This syntax may be deconstructed into the following forms:
1. COUNT(*)
2. COUNT(DISTINCT expr)
3. COUNT(ALL expr)
4. COUNT(expr)
When COUNT(*) is invoked, all rows in the group, including those with nulls
or duplicate values are counted. When COUNT(DISTINCT expr) is executed, only
unique occurrences of expr are counted for each group. The ALL keyword is part of the
default syntax, so COUNT(ALL expr) and COUNT(expr) are equivalent. These count
the number of nonnull occurrences of expr in each group. The data type of expr may be

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Chapter 6:

Reporting Aggregated Data Using the Group Functions

FIGURE 6-1

Group functions
operating on 12
groups

NUMBER, DATE, CHAR, or VARCHAR2. If expr is a null, it is ignored unless it is
managed using a general function like NVL, NVL2, or COALESCE.
The AVG function calculates the average value of a numeric column or
expression in a group. Its syntax is as follows:
AVG([DISTINCT|ALL] expr) ;
This syntax may be deconstructed into the following forms:
1. AVG(DISTINCT expr)
2. AVG(ALL expr)
3. AVG(expr)

Describe the Group Functions

277

When AVG(DISTINCT expr) is invoked, the distinct values of expr are summed
and divided by the number of unique occurrences of expr. AVG(ALL expr) and
AVG(expr) add the nonnull values of expr for each row and divide the sum by the
number of nonnull rows in the group. The data type of the expr parameter is NUMBER.
The SUM function returns the aggregated total of the nonnull numeric
expression values in a group. It has the following syntax:
SUM([DISTINCT|ALL] expr) ;
This syntax may be deconstructed into the following forms:
1. SUM(DISTINCT expr)
2. SUM(ALL expr)
3. SUM(expr)
SUM(DISTINCT expr) provides a total by adding all the unique values returned
after expr is evaluated for each row in the group. SUM(expr) and SUM(ALL expr)
provide a total by adding expr for each row in the group. Null values are ignored.
The data type of expr is NUMBER.
The MAX and MIN functions return the maximum (largest) and minimum
(smallest) expr value in a group. Their syntax is as follows:
MAX([DISTINCT|ALL] expr); MIN([DISTINCT|ALL] expr)
This syntax may be deconstructed into the following forms:
1. MAX(DISTINCT expr); MIN(DISTINCT expr)
2. MAX(ALL expr); MIN(ALL expr)
3. MAX(expr); MIN(expr);
MAX(expr), MAX(ALL expr) and MAX(DISTINCT expr) examine the values
for expr in a group of rows and return the largest value. Null values are ignored.
MIN(expr), MIN(ALL expr) and MIN(DISTINCT expr) examine the values for expr
in a group of rows and return the smallest value. The data type of the expr parameter
may be NUMBER, DATE, CHAR or VARCHAR2.
The STDDEV and VARIANCE functions are two of many statistical group
functions Oracle provides. VARIANCE has the following syntax:
VARIANCE([DISTINCT|ALL] expr);
This syntax may be deconstructed into the following forms:
1. VARIANCE(DISTINCT expr)
2. VARIANCE(ALL expr)
3. VARIANCE(expr)

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STDDEV has the following syntax:
STDDEV([DISTINCT|ALL] expr);
This syntax may be deconstructed into the following forms:
1. STDDEV(DISTINCT expr)
2. STDDEV(ALL expr)
3. STDDEV(expr)
Statistical variance refers to the variability of scores in a sample or set of data.
VARIANCE(DISTINCT expr) returns the variability of unique nonnull data in
a group. VARIANCE(expr) and VARIANCE(ALL expr) return the variability of
nonnull data in the group.
STDDEV calculates statistical standard deviation, which is the degree of deviation
from the mean value in a group. It is derived by finding the square root of the variance.
STDDEV(DISTINCT expr) returns the standard deviation of unique nonnull data in
a group. STDDEV(expr) and STDDEV(ALL expr) return the standard deviation of
nonnull data in the group. The data type of the expr parameter is NUMBER.

There are two fundamental
rules to remember when studying group
functions. First, they always operate on a
single group of rows at a time.The group
may be one of many groups a dataset has
been segmented into or it may be an entire
table.The group function executes once per
group. Second, rows with nulls occurring in
group columns or expressions are ignored,
unless a general function like NVL, NVL2,
or COALESCE is provided to handle them.

Consider the following example. If the
average value for COMMISSION_PCT
is retrieved from the EMPLOYEES table,
only the nonnull values are considered.
The expression AVG(COMMISSION_PCT)
adds the 35 nonnull COMMISSION_
PCT values and divides the total by
35.The average based on all 107 rows
may be computed using the expression
AVG(NVL(COMMISSION_PCT,0)).

Identify the Available Group Functions

279

CERTIFICATION OBJECTIVE 6.02

Identify the Available Group Functions
The different variants of group functions and their syntax were just discussed. This
section provides examples demonstrating the application of these functions. The
interactions of group functions with null values and the DISTINCT keyword are
discussed, and the concept of nesting group functions is also considered. The
available group functions are identified and explored under the following headings:
■ Using the group functions
■ Nesting group functions

Using the Group Functions
The practical application of group functions is demonstrated using AVG, SUM, MIN,
MAX, and COUNT. These group functions all return numeric results. Additionally,
the MIN and MAX functions may return character and date results. These five
functions operate on nonnull values but, unlike the others, the COUNT(*) function
call also counts rows with null values. The DISTINCT keyword is used to constrain
the rows submitted to the group functions.
Analysts frequently wish to know the aggregated total or average value of a
column or an expression. This is simple to achieve using most spreadsheet packages.
Using the SQL group functions for reporting provides two advantages over using
a spreadsheet for analysis. First, they offer a convenient platform for performing
calculations using real-time live data. Second, they allow for easy analysis of every
value in a dataset or of specific groups of values.

The COUNT Function
The execution of COUNT on a column or an expression returns an integer value
that represents the number of rows in the group. The COUNT function has the
following syntax:
COUNT({*|[DISTINCT|ALL ] expr});
There is one parameter that can be either *, which represents all columns
including null values, or a specific column or expression. It may be preceded by the
DISTINCT or ALL keywords. Consider the following queries:
Query 1: select count(*) from employees
Query 2: select count(commission_pct) from employees

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Chapter 6:

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Query 3: select count(distinct commission_pct) from employees
Query 4: select count(hire_date), count(manager_id) from employees

Query 1 counts the rows in the EMPLOYEES table and returns the integer 107.
Query 2 counts the rows with nonnull COMMISSION_PCT values and returns 35.
Query 3 considers the 35 nonnull rows, determines the number of unique values, and
returns 7. Query 4 demonstrates two features. First, multiple group functions may
be used in the same SELECT list and second, the COUNT function is used on both
a DATE column and a NUMBER column. The integers 107 and 106 are returned
since there are 107 nonnull HIRE_DATE values and 106 nonnull MANAGER_ID
values in the group.
Three adjacent expressions using the COUNT function are shown in Figure 6-2.
This query illustrates that there are 107 employee records in the EMPLOYEES
table. Further, these 107 employees are allocated to 12 departments, including null
departments, and work in 19 unique jobs.

The SUM Function
The aggregated total of a column or an expression is computed with the SUM
function. Its syntax is as follows:
SUM([DISTINCT|ALL ] expr);
One numeric parameter, optionally preceded by the DISTINCT or ALL
keywords, is provided to the SUM function, which returns a numeric value.
Consider the following queries:
Query
Query
Query
Query

FIGURE 6-2

The COUNT
function

1:
2:
3:
4:

select
select
select
select

sum(2) from employees
sum(salary) from employees
sum(distinct salary) from employees
sum(commission_pct) from employees

Identify the Available Group Functions

281

There are 107 rows in the EMPLOYEES table. Query 1 adds the number 2 across
107 rows and returns 214. Query 2 takes the SALARY column value for every row in
the group, which in this case is the entire table, and returns the total salary amount
of 691400. Query 3 returns a total of 397900 since many employees get paid the
same salary and the DISTINCT keyword only adds unique values in the column to
the total. Query 4 returns 7.8 after adding the nonnull values.
Figure 6-3 shows two queries. The first calculates the number of days between the
current system date and the value in the HIRE_DATE column. The date arithmetic is
performed for every row. The number returned is added using one SUM function call.
The result is divided by 365.25 to give the total number of years worked by all current
employees. The second query shows that the SUM function returns an “ORA-00932:
inconsistent datatypes” error if it is provided with a nonnumeric argument.

The AVG Function
The average value of a column or expression divides the sum by the number of
nonnull rows in the group. The AVG function has the following syntax:
AVG([DISTINCT|ALL ] expr);
One numeric parameter, preceded by the DISTINCT or ALL keywords, is
provided to the AVG function, which returns a numeric value. Consider the
following queries:
Query
Query
Query
Query

FIGURE 6-3

The SUM
function

1:
2:
3:
4:

select
select
select
select

avg(2) from employees
avg(salary) from employees
avg(distinct salary) from employees
avg(commission_pct) from employees

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There are 107 rows in the EMPLOYEES table. Query 1 adds the number 2 across
107 rows and divides the total by the number of rows to return the number 2.
Numeric literals submitted to the AVG function are returned unchanged. Query 2
adds the SALARY value for each row to obtain the total salary amount of 691400.
This is divided by the 107 rows with nonnull SALARY values to return the average
6461.68224. You may expect query 3 to return a smaller result than query 2, but
it does not. There are 57 unique salary values, which when added, yields a total of
397900. Dividing 397900 by 57 returns 6980.70175 as the average of the distinct
salary values. Query 4 may produce unanticipated results if not properly understood.
Adding the nonnull values, including duplicates, produces a total of 7.8. There are
35 employee records with nonnull COMMISSION_PCT values. Dividing 7.8 by
35 yields an average COMMISSION_PCT of 0.222857143.
Figure 6-4 shows two queries. The first lists the LAST_NAME and JOB_ID
columns with an expression that calculates the total number of years worked by
programmers in the organization. The second uses the AVG function to calculate
the average number of years for which current programmers have been employed.

The MAX and MIN Functions
The MAX and MIN functions operate on NUMBER, DATE, CHAR, and
VARCHAR2 data types. They return a value of the same data type as their input
arguments, which are either the largest or smallest items in the group. When applied
to DATE items, MAX returns the latest date and MIN returns the earliest one.

FIGURE 6-4

The AVG function

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283

Character strings are converted to numeric representations of their constituent
characters based on the NLS settings in the database. When the MIN function is
applied to a group of character strings, the word that appears first alphabetically is
returned, while MAX returns the word that would appear last. The MAX and MIN
functions have the following syntax:
MAX([DISTINCT|ALL] expr); MIN([DISTINCT|ALL] expr)
They take one parameter preceded by the DISTINCT or ALL keywords. Consider
the following queries:
Query 1: select min(commission_pct), max(commission_pct) from
employees
Query 2: select min(start_date),max(end_date) from job_history
Query 3: select min(job_id),max(job_id) from employees

Query 1 returns the numeric values 0.1 and 0.4 for the minimum and maximum
COMMISSION_PCT values in the EMPLOYEES table. Note that null values for
COMMISSION_PCT are ignored. Query 2 evaluates a DATE column and indicates
that the earliest START_DATE in the JOB_HISTORY table is 17-SEP-1987
and the latest END_DATE is 31-DEC-1999. Query 3 returns AC_ACCOUNT
and ST_MAN as the JOB_ID values appearing first and last alphabetically in the
EMPLOYEES table.
The first query shown in Figure 6-5 uses the MAX and MIN functions to obtain
information about employees with JOB_ID values of SA_REP. The results indicate
that the sales representatives working for the shortest and longest durations were
FIGURE 6-5

The MIN and
MAX functions

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hired on 21-APR-2000 and 30-JAN-1996, respectively. Furthermore, the sales
representatives earning the largest and smallest salaries earn 11500 and 6100,
respectively. The second query fetches the LAST_NAME values for the sales
representatives to whom the minimum and maximum HIRE_DATE and SALARY
values apply.

EXERCISE 6-1
Using the Group Functions
The COUNTRIES table stores a list of COUNTRY_NAME values. You are required
to calculate the average length of all the country names. Any fractional components
must be rounded to the nearest whole number.
1. Start SQL*Plus and connect to the HR schema.
2. The length of the country name value for each row is calculated using the
LENGTH function. The average length may be determined using the AVG
function. It may be rounded to the nearest whole number using the ROUND
function.
3. The SELECT clause using alias AVERAGE_COUNTRY_NAME_LENGTH is
SELECT ROUND(AVG( LENGTH(COUNTRY_NAME)))
AVERAGE_COUNTRY_NAME_LENGTH
4. The FROM clause is
FROM COUNTRIES
5. Executing this statement returns a single row representing the average
length of all the country names in the COUNTRIES table, as shown in the
following illustration:

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285

SCENARIO & SOLUTION
You would like to retrieve the earliest date from a
column that stores DATE information. Can a group
function be utilized to retrieve this value?

Yes. The MIN function operates on numeric, date,
and character data. When the MIN function is
executed against a DATE column, the earliest date
value is returned.

Summary statistics are required by senior
management. This includes details like number of
employees, total staff salary cost, lowest salary, and
highest salary values. Can such a report be drawn
using one query?

Yes. There is no restriction to the number of group
functions listed in the SELECT clause. The requested
report can be drawn using the following query:
SELECT COUNT(*) Num_Employees,
SUM(SALARY) Tot_Salary_Cost,
MIN(SALARY) Lowest Salary,
MAX(SALARY) Maximum Salary
FROM EMPLOYEES;

You are asked to list the number of unique jobs
performed by employees in the organization.
Counting the JOB_ID records will give you all the
jobs. Is it possible to count the unique jobs?

Yes. The DISTINCT keyword may be used with the
aggregate functions. To count unique JOB_ID values
in the EMPLOYEES table, you can issue the query:
SELECT COUNT(DISTINCT JOB_ID)
FROM EMPLOYEES;

Nested Group Functions
Recall that single-row functions may be nested or embedded to any level of depth.
Group functions may only be nested two levels deep. Three formats using group
functions are shown here:
G1(group_item) = result
G1(G2(group_item ) = result
G1(G2(G3(group_item))) is NOT allowed.
Group functions are represented by the letter G followed by a number. The first
simple form contains no nested functions. Examples include the SUM(group_item)
or AVG(group_item) functions that return a single result per group. The second form
supports two nested group functions, like SUM(AVG(group_item)). In this case, a
GROUP BY clause is mandatory since the average value of the group_item per group
is calculated before being aggregated by the SUM function.
The third form is disallowed by Oracle. Consider an expression that nests
three group functions. If the MAX function is applied to the previous example,
the expression MAX(SUM(AVG(group_item))) is formed. The two inner group
functions return a single value representing the sum of a set of average values. This
expression becomes MAX(single value). A group function cannot be applied to a
single value.

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Figure 6-6 demonstrates two queries. Both restrict the rows returned to those with
DEPARTMENT_ID values of null, 40, and 80. These are then partitioned by their
DEPARTMENT_ID values into three groups. The first query calculates the sum of
the COMMISSION_PCT values for each group and returns the values 0.15, null,
and 7.65. Query 2 contains the nested group functions, which may be evaluated as
follows: AVG(SUM(COMMISSION_PCT)) = (0.15 + 7.65) /2 = 3.9.
FIGURE 6-6

Nested group
functions

Group Data Using the GROUP BY Clause

Single-row functions
may be nested to any level, but group
functions may be nested to, at most,
two levels deep.The nested function
call COUNT(SUM(AVG( X))) returns
the error, “ORA-00935: group function
is nested too deeply.” It is acceptable to

287

nest single-row functions within group
functions. Consider the following query:
SELECT SUM(AVG(LENGTH(LAST_
NAME))) FROM EMPLOYEES GROUP BY
DEPARTMENT_ID. It calculates the sum of
the average length of LAST_NAME values
per department.

CERTIFICATION OBJECTIVE 6.03

Group Data Using the GROUP BY Clause
The group functions discussed earlier use groups of rows comprising the entire table.
This section explores partitioning a set of data into groups using the GROUP BY
clause. Group functions may be applied to these subsets or clusters of rows. The syntax
of group functions and the GROUP BY clause are discussed in the following areas:
■ Creating groups of data
■ The GROUP BY clause
■ Grouping by multiple columns

Creating Groups of Data
A table has at least one column and zero or more rows of data. In many tables
this data requires analysis to transform it into useful information. It is a common
reporting requirement to calculate statistics from a set of data divided into groups
using different attributes. Previous examples using group functions operated against
all the rows in a table. The entire table was treated as one large group.
Groups of data within a set are created by associating rows with common properties
or attributes with each other. Thereafter, group functions can execute against each of
these groups. Groups of data include entire rows and not specific columns.

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Consider the EMPLOYEES table. It comprises 11 columns and 107 rows. You
could create groups of rows that share a common DEPARTMENT_ID value. The
SUM function may then be used to create salary totals per department. Another
possible set of groups may share common JOB_ID column values. The AVG group
function may then be used to identify the average salary paid to employees in
different jobs.
A group is defined as a subset of the entire dataset sharing one or more common
attributes. These attributes are typically column values but may also be expressions.
The number of groups created depends on the distinct values present in the common
attribute.
As Figure 6-7 shows, there are 12 unique DEPARTMENT_ID values in the
EMPLOYEES table. If rows are grouped using common DEPARTMENT_ID values,
there will be 12 groups. If a group function is executed against these groups, there
will be 12 values returned, as it will execute once for each group.
Grouping data and using summary functions are widely utilized for reporting
purposes. It is valuable to practice the segmentation of a set of data into
different groupings. Oracle provides the analytical language to deconstruct
datasets into groups, divide these into further subgroups, and so on. Aggregate
grouping functions can then be executed against these groups and subgroups.

The GROUP BY Clause
The SELECT statement is enhanced by the addition of the GROUP BY clause.
This clause facilitates the creation of groups. It appears after the WHERE clause but
before the ORDER BY clause, as follows:
SELECT column|expression|group_function(column|expression [alias]),…}
FROM table
[WHERE condition(s)]
[GROUP BY {col(s)|expr}]
[ORDER BY {col(s)|expr|numeric_pos} [ASC|DESC] [NULLS FIRST|LAST]];
The column or expression specified in the GROUP BY clause is also known as
the grouping attribute and is the component that rows are grouped by. The dataset is
segmented based on the grouping attribute. Consider the following query:
select max(salary), count(*)
from employees
group by department_id
order by department_id

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289

FIGURE 6-7

Unique
DEPARTMENT_
ID values in the
EMPLOYEES table

The grouping attribute in this example is the DEPARTMENT_ID column. The
dataset, on which the group functions in the SELECT list must operate, is divided
into 12 groups, one for each department. For each group (department), the maximum
salary value and the number of rows are returned. Since the results are sorted by
DEPARTMENT_ID, the third row in the set of results contains the values 11000 and 6.
This indicates that 6 employees have a DEPARTMENT_ID value of 30. Of these 6,
the highest earner has a SALARY value of 11000. This query demonstrates that the
grouping attribute does not have to be included in the SELECT list.

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It is common to see the grouping attribute in the SELECT list alongside grouping
functions. If an item, which is not a group function, appears in the SELECT list and
there is no GROUP BY clause, an “ORA-00937: not a single-group group function”
error is raised. If a GROUP BY clause is present but that item is not a grouping
attribute, then an “ORA-00979: not a GROUP BY expression” error is returned.
Any item in the SELECT list that is not a group function must be a grouping attribute
of the GROUP BY clause.
If a group function is placed in a WHERE clause, an “ORA-00934: group
function is not allowed here” error is returned. Imposing group-level conditions is
achieved using the HAVING clause discussed in the next section. Group functions
may, however, be used as part of the ORDER BY clause.
The first query in Figure 6-8 raises an error since the END_DATE column is
in the SELECT list with a group function and there is no GROUP BY clause. An
“ORA-00979” error is returned from the second query since the START_DATE
item is listed in the SELECT clause, but it is not a grouping attribute.
The third query divides the JOB_HISTORY rows into groups based on the 4-digit
year component from the END_DATE column. Four groups are created using this
grouping attribute. These represent different years when employees ended their
jobs. The COUNT shows the number of employees who quit their jobs during each

FIGURE 6-8

The GROUP BY
clause

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291

of these years. The results are listed in descending order based on the “Number of
Employees” expression. Note that the COUNT group function is present in the
ORDER BY clause.

A dataset is divided into
groups using the GROUP BY clause.
The grouping attribute is the common
key shared by members of each group.
The grouping attribute is usually a single
column but may be multiple columns or

an expression that cannot be based on
group functions. Note that only grouping
attributes and group functions are
permitted in the SELECT clause when using
GROUP BY.

Grouping by Multiple Columns
A powerful extension to the GROUP BY clause uses multiple grouping attributes.
Oracle permits datasets to be partitioned into groups and allows these groups to be
further divided into subgroups using a different grouping attribute. Consider the
following two queries:
Query 1: select department_id, sum(commission_pct) from employees
where commission_pct is not null group by department_id
Query 2: select department_id, job_id, sum(commission_pct) from
employees
where commission_pct is not null group by department_id, job_id

Query 1 restricts the rows returned from the EMPLOYEES table to the 35 rows
with nonnull COMMISSION_PCT values. These rows are then divided into two
groups: 80 and NULL based on the DEPARTMENT_ID grouping attribute. The
result set contains two rows, which return the sum of the COMMISSION_PCT
values for each group.
Query 2 is similar to the first one except it has an additional item: JOB_ID in both
the SELECT and GROUP BY clauses. This second grouping attribute decomposes
the two groups based on DEPARTMENT_ID into the constituent JOB_ID
components belonging to the rows in each group. The distinct JOB_ID values for
rows with DEPARTMENT_ID=80 are SA_REP and SA_MAN. The distinct JOB_ID
value for rows with null DEPARTMENT_ID is SA_REP. Therefore, query 2 returns
two groupings, one which consists of two subgroups, and the other with only one, as
shown in Figure 6-9.

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FIGURE 6-9

The GROUP
BY clause with
multiple columns

EXERCISE 6-2
Grouping Data Based on Multiple Columns
Analysis of staff turnover is a common reporting requirement. You are required to
create a report containing the number of employees who left their jobs, grouped by
the year in which they left. The jobs they performed are also required. The results
must be sorted in descending order based on the number of employees in each group.
The report must list the year, the JOB_ID, and the number of employees who left a
particular job in that year.
1. Start SQL Developer and connect to the HR schema.
2. The JOB_HISTORY table contains the END_DATE and JOB_ID columns,
which constitute the source data for this report.
3. The year component is extracted using the TO_CHAR function. The number
of employees who quit a particular job in each year is obtained using the
COUNT(*) function.
4. The SELECT clause is
TO_CHAR(END_DATE,‘YYYY’) “Quitting Year”, JOB_ID, COUNT(*)
“Number of Employees”
5. The FROM clause is
FROM EMPLOYEES

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293

6. There is no WHERE clause.
7. Since the report requires employees to be listed by year and JOB_ID,
these two items must appear in the GROUP BY clause, which is
GROUP BY TO_CHAR(END_DATE,‘YYYY’), JOB_ID
8. The sorting is performed with
ORDER BY COUNT(*) DESC
9. Executing this statement returns the staff turnover report requested as shown
in the following illustration:

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SCENARIO & SOLUTION
You wish to print name badges for the staff who
work as sales representatives. Can the length of
the shortest and longest LAST_NAME values be
determined for these employees?

Yes. The MAX and MIN functions applied to the
LAST_NAME field will determine the shortest and
longest names as shown in the following query:
SELECT MIN(LENGTH(LAST_NAME)),
MAX(LENGTH(LAST_NAME)) FROM
EMPLOYEES WHERE JOB_ID='SA_REP';

Is it possible to count the records in each group,
first by dividing the employee records by year of
employment, then by job, and finally by salary?

Yes. Grouping by multiple columns is a powerful
option allowing fine-grained analysis as shown in the
following query: SELECT COUNT(*),
TO_CHAR( HIRE_DATE, 'YYYY'), JOB_ID,
SALARY FROM EMPLOYEES
GROUP BY TO_CHAR(HIRE_DATE,'YYYY') ,
JOB_ID, SALARY

Is there a limit to the number of groups within groups
that can be formed?

No. There is no limit to the number of groups and
subgroups that can be formed.

CERTIFICATION OBJECTIVE 6.04

Include or Exclude Grouped Rows
Using the HAVING Clause
Creating groups of data and applying aggregate functions is very useful. A refinement
to these features is the ability to include or exclude results based on group-level
conditions. This section introduces the HAVING clause. A clear distinction is made
between the WHERE clause and the HAVING clause. The HAVING clause is
explained in the following areas:
■ Restricting group results
■ The HAVING clause

Restricting Group Results
WHERE clause conditions restrict rows returned by a query. Rows are included based
on whether they fulfill the conditions listed and are sometimes known as row-level
results. Clustering rows using the GROUP BY clause and applying an aggregate

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295

function to these groups returns results often referred to as group-level results. The
HAVING clause provides the language to restrict group-level results.
The following query limits the rows retrieved from the JOB_HISTORY table by
specifying a WHERE condition based on the DEPARTMENT_ID column values.
select department_id
from job_history
where department_id in (50,60,80,110);

This query returns seven rows. If the WHERE clause was absent, all ten rows
would be retrieved. Suppose you want to know how many employees were previously
employed in each of these departments. There are seven rows that can be manually
grouped and counted. However, if there are a large number of rows, an aggregate
function like COUNT may be used, as shown in the following query:
select department_id, count(*)
from job_history
where department_id in (50,60,80,110)
group by department_id;

This query is very similar to the previous statement. The aggregate function
COUNT was added to the SELECT list, and a GROUP BY DEPARTMENT_ID
clause was also added. Four rows with their aggregate row count are returned and it
is clear that the original seven rows restricted by the WHERE clause were clustered
into four groups based on common DEPARTMENT_ID values, as shown in the
following table:
DEPARTMENT_ID

COUNT(*)

50

2

60

1

80

2

110

2

Suppose you wanted to refine this list to include only those departments with
more than one employee. The HAVING clause limits or restricts the group-level
rows as required.
This query must perform the following steps:
1. Consider the entire row-level dataset.
2. Limit the dataset based on any WHERE clause conditions.

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3. Segment the data into one or more groups using the grouping attributes
specified in the GROUP BY clause.
4. Apply any aggregate functions to create a new group-level dataset. Each row
may be regarded as an aggregation of its source row-level data based on the
groups created.
5. Limit or restrict the group-level data with a HAVING clause condition. Only
group-level results matching these conditions are returned.
Choosing the appropriate context to use a WHERE or a HAVING clause
depends on whether physical or group-level rows are to be restricted. Rows
containing data stored in columns are sometimes called actual or physical
rows. When actual (physical) rows are restricted, one or more conditions are
imposed using a WHERE clause. When these rows are grouped together, one
or more aggregate functions may be applied, yielding one or more group-level
rows.These are not physical rows, but temporary aggregations of data. Grouplevel rows are restricted using conditions imposed by a HAVING clause.

The HAVING Clause
The general form of the SELECT statement is further enhanced by the addition of
the HAVING clause and becomes:
SELECT column|expression|group_function(column|expression [alias]),…}
FROM table
[WHERE condition(s)]
[GROUP BY {col(s)|expr}]
[HAVING group_condition(s)]
[ORDER BY {col(s)|expr|numeric_pos} [ASC|DESC] [NULLS FIRST|LAST]];
An important difference between the HAVING clause and the other SELECT
statement clauses is that it may only be specified if a GROUP BY clause is present.
This dependency is sensible since group-level rows must exist before they can be
restricted. The HAVING clause can occur before the GROUP BY clause in the
SELECT statement. However, it is more common to place the HAVING clause after
the GROUP BY clause. All grouping is performed and group functions are executed
prior to evaluating the HAVING clause.
The following query shows how the HAVING clause is used to restrict an
aggregated dataset. Records from the JOB_HISTORY table are divided into four
groups. The rows that meet the HAVING clause condition (contributing more than
one row to the group row count) are returned:

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297

select department_id, count(*)
from job_history
where department_id in (50,60,80,110)
group by department_id
having count(*)>1

Three rows with DEPARTMENT_ID values of 50, 80, and 110, each with a
COUNT(*) value of 2, are returned.
Figure 6-10 shows three queries. Query 1 divides the 107 records from the
EMPLOYEES table into 19 groups based on common JOB_ID values. The average
salary for each JOB_ID group and the aggregate row count is computed. Query
2 refines the results by conditionally excluding those aggregated rows where the
FIGURE 6-10

The HAVING
clause

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average salary is less than or equal to 10000, using a HAVING clause. Query 3
demonstrates that the Boolean operators may be used to specify multiple HAVING
clause conditions.

The HAVING clause may
only be specified when a GROUP BY
clause is present. A GROUP BY clause can
be specified without a HAVING clause.
Multiple conditions may be imposed by a

HAVING clause using the Boolean AND,
OR, and NOT operators.The HAVING
clause conditions restrict group-level data
and must contain a group function or an
expression that uses one.

EXERCISE 6-3
Using the HAVING Clause
The company is planning a recruitment drive and wants to identify the days of the
week on which 20 or more staff members were hired. Your report must list the days
and the number of employees hired on each of them.
1. Start SQL*Plus and connect to the HR schema.
2. EMPLOYEES records must be divided into groups based on the day
component of the HIRE_DATE column. The number of employees per group
may be obtained using the COUNT function.
3. The SELECT clause is
SELECT TO_CHAR(HIRE_DATE,'DAY'), COUNT(*)
4. No WHERE clause is required since all the physical rows of the EMPLOYEES
table are considered.
5. The GROUP BY clause is
GROUP BY TO_CHAR(HIRE_DATE,'DAY')
This GROUP BY clause potentially creates seven group-level rows, one for
each day of the week.
6. The COUNT function in the SELECT clause then lists the number of staff
members employed on each day. The HAVING clause must be used to restrict
these seven rows to only those where the count is greater than or equal to 20.

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299

7. The HAVING clause is
HAVING COUNT(*) >= 20
8. The FROM clause is
FROM EMPLOYEES
9. Executing this statement returns the days of the week on which 20 or more
employees were hired as shown in the following illustration:

INSIDE THE EXAM
The certification objectives in this chapter
are examined using practical examples and
scenarios that require you to predict the results returned from SQL queries. These queries
focus on dividing datasets into groups using
one or more grouping attributes.
Understand the limitations of nesting group
functions. Remember that group functions
must exist in the SELECT clause if there is a
GROUP BY clause. The HAVING clause may
occur before the GROUP BY clause but is
usually followed by it. An error is returned
if the HAVING clause is used without a
GROUP BY clause. Recall that the HAVING

clause may contain multiple conditions, but
each one must contain a group function.
Knowing how to interpret and trace nested
group and single-row functions is vital, as
many questions contain expressions with
nested functions. The innermost to outermost
order of evaluation of nested group and single
functions is identical and must be remembered.
There are many built-in group functions
available, but the exams will test your understanding of the COUNT, SUM, AVG, MAX,
and MIN functions. Ensure that you have a
thorough understanding of these functions
and how they interact with the DISTINCT
keyword and NULL values.

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CERTIFICATION SUMMARY
Multiple row functions and the concept of dividing data into groups are described
in this chapter. There is a multitude of group or aggregate functions available. The
key functions for creating sum totals; calculating averages, minimums, or maximums;
and obtaining a record count are explored in detail.
The differences between nesting group functions and single-row functions are
investigated, and the limitations of the former are explained. Creating groups using
common grouping attributes is concretized with the introduction of the GROUP BY
clause to the SELECT statement. Row-level data is limited by conditions specified
in the WHERE clause. The restriction of group-level data using the HAVING clause
is also discussed.

Two-Minute Drill

✓

301

TWO-MINUTE DRILL
Describe the Group Functions
❑ Group functions are also known as multiple row, aggregate, or summary

functions. They execute once for each group of data and aggregate the data
from multiple rows into a single result for each group.
❑ Groups may be entire tables or portions of a table grouped together by a common

grouping attribute.

Identify the Available Group Functions
❑ The COUNT of a function returns an integer value representing the number

of rows in a group.
❑ The SUM function returns an aggregated total of all the nonnull numeric

expression values in a group.
❑ The AVG function divides the sum of a column or expression by the number

of nonnull rows in a group.
❑ The MAX and MIN functions operate on NUMBER, DATE, CHAR, and

VARCHAR2 data types. They return a value that is either the largest or
smallest item in the group.
❑ Group functions may only be nested two levels deep.

Group Data Using the GROUP BY Clause
❑ The GROUP BY clause specifies the grouping attribute rows must have in

common for them to be clustered together.
❑ The GROUP BY clause facilitates the creation of groups within a selected set

of data and appears after the WHERE clause but before the ORDER BY clause.
❑ Any item on the SELECT list that is not a group function must be a grouping

attribute.
❑ Group functions may not be placed in a WHERE clause.
❑ Datasets may be partitioned into groups and further divided into subgroups

based on multiple grouping attributes.

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Include or Exclude Grouped Rows Using the HAVING Clause
❑ Clustering rows using a common grouping attribute with the GROUP BY

clause and applying an aggregate function to each of these groups returns
group-level results.
❑ The HAVING clause provides the language to limit the group-level results

returned.
❑ The HAVING clause may only be specified if there is a GROUP BY clause

present.
❑ All grouping is performed and group functions are executed prior to evaluating

the HAVING clause.

Self Test

303

SELF TEST
The following questions will help you measure your understanding of the material presented in this
chapter. Read all the choices carefully because there may be more than one correct answer. Choose
all the correct answers for each question.

Describe the Group Functions
1. What result is returned by the following statement?
SELECT COUNT(*) FROM DUAL; (Choose the best answer.)
A. NULL
B. 0
C. 1
D. None of the above
2. Choose one correct statement regarding group functions.
A. Group functions may only be used when a GROUP BY clause is present.
B. Group functions can operate on multiple rows at a time.
C. Group functions only operate on a single row at a time.
D. Group functions can execute multiple times within a single group.

Identify the Available Group Functions
3. What value is returned after executing the following statement?
SELECT SUM(SALARY) FROM EMPLOYEES;
Assume there are 10 employee records and each contains a SALARY value of 100, except for 1,
which has a null value in the SALARY field. (Choose the best answer.)
A. 900
B. 1000
C. NULL
D. None of the above
4. Which values are returned after executing the following statement?
SELECT COUNT(*), COUNT(SALARY) FROM EMPLOYEES;
Assume there are 10 employee records and each contains a SALARY value of 100, except for 1,
which has a null value in their SALARY field. (Choose all that apply.)
A. 10 and 10
B. 10 and NULL
C. 10 and 9
D. None of the above

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5. What value is returned after executing the following statement?
SELECT AVG(NVL(SALARY,100)) FROM EMPLOYEES;
Assume there are ten employee records and each contains a SALARY value of 100, except for
one employee, who has a null value in the SALARY field. (Choose the best answer.)
A. NULL
B. 90
C. 100
D. None of the above

Group Data Using the GROUP BY Clause
6. What value is returned after executing the following statement?
SELECT SUM((AVG(LENGTH(NVL(SALARY,0)))))
FROM EMPLOYEES
GROUP BY SALARY;
Assume there are ten employee records and each contains a SALARY value of 100, except for
one, which has a null value in the SALARY field. (Choose the best answer.)
A. An error is returned
B. 3
C. 4
D. None of the above
7. How many records are returned by the following query?
SELECT SUM(SALARY), DEPARTMENT_ID FROM EMPLOYEES
GROUP BY DEPARTMENT_ID;
Assume there are 11 nonnull and 1 null unique DEPARTMENT_ID values. All records have a
nonnull SALARY value. (Choose the best answer.)
A. 12
B. 11
C. NULL
D. None of the above
8. What values are returned after executing the following statement?
SELECT JOB_ID, MAX_SALARY FROM JOBS GROUP BY MAX_SALARY;
Assume that the JOBS table has ten records with the same JOB_ID value of DBA and the same
MAX_SALARY value of 100. (Choose the best answer.)
A. One row of output with the values DBA, 100
B. Ten rows of output with the values DBA, 100
C. An error is returned
D. None of the above

Lab Question

305

Include or Exclude Grouped Rows Using the HAVING Clause
9. How many rows of data are returned after executing the following statement?
SELECT DEPT_ID, SUM(NVL(SALARY,100)) FROM EMP
GROUP BY DEPT_ID HAVING SUM(SALARY) > 400;
Assume the EMP table has ten rows and each contains a SALARY value of 100, except for one,
which has a null value in the SALARY field. The first and second five rows have DEPT_ID
values of 10 and 20, respectively. (Choose the best answer.)
A. Two rows
B. One row
C. Zero rows
D. None of the above
10. How many rows of data are returned after executing the following statement?
SELECT DEPT_ID, SUM(SALARY) FROM EMP GROUP BY DEPT_ID HAVING
SUM(NVL(SALARY,100)) > 400;
Assume the EMP table has ten rows and each contains a SALARY value of 100, except for one,
which has a null value in the SALARY field. The first and second five rows have DEPT_ID
values of 10 and 20, respectively. (Choose the best answer.)
A. Two rows
B. One row
C. Zero rows
D. None of the above

LAB QUESTION
Using SQL Developer or SQL*Plus, connect to the OE schema and complete the following tasks.
The PRODUCT_INFORMATION table lists items that are orderable and others that are
planned, obsolete, or under development. You are required to prepare a report that groups the nonorderable products by their PRODUCT_STATUS and shows the number of products in each group
and the sum of the LIST_PRICE of the products per group. Further, only the group-level rows, where
the sum of the LIST_PRICE is greater than 4000, must be displayed. A product is nonorderable if the
PRODUCT_STATUS value is not equal to the string ‘orderable’. There are several approaches to
solving this question. Your approach may differ from the solution proposed.

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SELF TEST ANSWERS
Describe the Group Functions
✓ C. The DUAL table has one row and one column. The COUNT(*) function returns the
1. ®
number of rows in a table or group.
®
˚ A, B, and D are incorrect.
✓ B. By definition, group functions can operate on multiple rows at a time, unlike single-row
2. ®
functions.
®
˚ A, C, and D are incorrect statements. A group function may be used without a GROUP BY
clause. In this case, the entire dataset is operated on as a group. The COUNT function is often
executed against an entire table, which behaves as one group. D is incorrect. Once a dataset has
been partitioned into different groups, any group functions execute once per group.

Identify the Available Group Functions
✓ A. The SUM aggregate function ignores null values and adds nonnull values. Since nine
3. ®
rows contain the SALARY value 100, 900 is returned.
®
˚ B, C, and D are incorrect. B would be returned if SUM(NVL(SALARY,100)) was
executed. C is a tempting choice since regular arithmetic with NULL values returns a NULL
result. However, the aggregate functions, except for COUNT(*), ignore NULL values.
✓ C. COUNT(*) considers all rows including those with NULL values. COUNT(SALARY)
4. ®
only considers the nonnull rows.
®
˚ A, B, and D are incorrect.
✓ C. The NVL function converts the one NULL value into 100. Thereafter, the average
5. ®
function adds the SALARY values and obtains 1000. Dividing this by the number of records
returns 100.
®
˚ A, B, and D are incorrect. B would be returned if AVG(NVL(SALARY,0)) was selected. It
is interesting to note that if AVG(SALARY) was selected, 100 would have also been returned,
since the AVG function would sum the nonnull values and divide the total by the number of
rows with nonnull SALARY values. So AVG(SALARY) would be calculated as: 900/9=100.

Group Data Using the GROUP BY Clause
✓ C. The dataset is segmented based on the SALARY column. This creates two groups: one
6. ®
with SALARY values of 100 and the other with a null SALARY value. The average length of
SALARY value 100 is 3 for the rows in the first group. The NULL salary value is first converted

Self Test Answers

307

into the number 0 by the NVL function, and the average length of SALARY is 1. The SUM
function operates across the two groups adding the values 3 and 1 returning 4.
®
˚ A, B, and D are incorrect. A seems plausible since group functions may not be nested more
than two levels deep. Although there are four functions, only two are group functions while the
others are single-row functions evaluated before the group functions. B would be returned if the
expression SUM(AVG(LENGTH(SALARY))) was selected.
✓ A. There are 12 distinct DEPARTMENT_ID values. Since this is the grouping attribute,
7. ®
12 groups are created, including 1 with a null DEPARTMENT_ID value. Therefore 12 rows are
returned.
®
˚ B, C, and D are incorrect.

Include or Exclude Grouped Rows Using the HAVING Clause
✓ C. For a GROUP BY clause to be used, a group function must appear in the SELECT list.
8. ®
®
˚ A, B, and D are incorrect since the statement is syntactically inaccurate and is disallowed
by Oracle. Do not mistake the column named MAX_SALARY for the MAX(SALARY)
function.
✓ B. Two groups are created based on their common DEPT_ID values. The group with
9. ®
DEPT_ID values of ten consists of five rows with SALARY values of 100 in each of them.
Therefore, the SUM(SALARY) function returns 500 for this group, and it satisfies the
HAVING SUM(SALARY) > 400 clause. The group with DEPT_ID values of 20 has four
rows with SALARY values of 100 and one row with a NULL SALARY. SUM(SALARY) only
returns 400 and this group does not satisfy the HAVING clause.
®
˚ A, C, and D are incorrect. Beware of the SUM(NVL(SALARY,100)) expression in the
SELECT clause. This expression selects the format of the output. It does not restrict or limit the
dataset in anyway.
✓ A. Two groups are created based on their common DEPT_ID values. The group with
10. ®
DEPT_ID values of 10 consists of five rows with SALARY values of 100 in each of them.
Therefore the SUM(NVL(SALARY,100)) function returns 500 for this group and it satisfies the
HAVING SUM(SALARY) > 400 clause. The group with DEPT_ID values of 20 has four rows
with SALARY values of 100 and one row with a null SALARY. SUM(NVL(SALARY,100))
returns 500 and this group satisfies the HAVING clause. Therefore two rows are returned.
®
˚ B, C, and D are incorrect. Although the SELECT clause contains SUM(SALARY),
which returns 500 and 400 for the two groups, the HAVING clause contains the
SUM(NVL(SALARY,100)) expression, which specifies the inclusion or exclusion criteria for a
group-level row.

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LAB ANSWER
Using SQL Developer or SQL*Plus, connect to the OE schema and complete the following tasks.
There are several approaches to solving this question. Your approach may differ from the solution
proposed here.
1. Start SQL Developer and connect to the OE schema.
2. The dataset must be restricted to rows from the PRODUCT_INFORMATION table where the
PRODUCT_STATUS is not equal to the string ‘orderable’. Since this character literal may
have been input in mixed case, a case conversion function like UPPER can be used.
3. The WHERE clause is
WHERE UPPER(PRODUCT_STATUS) <> 'ORDERABLE'
4. Since the dataset must be segmented into groups based on the PRODUCT_STATUS column,
the GROUP BY statement is
GROUP BY PRODUCT_STATUS
5. The dataset is now partitioned into different groups based on their PRODUCT_STATUS
values. Therefore, the COUNT(*) function may be used to obtain the number of products in
each group. The SUM(LIST_PRICE) aggregate function can be used to calculate the sum of
the LIST_PRICE values for all the rows in each group.
6. The SELECT clause is therefore
SELECT COUNT(*), SUM(LIST_PRICE), PRODUCT_STATUS
7. The HAVING clause which restricts group-level rows is therefore
HAVING SUM(LIST_PRICE) > 4000
8. The FROM clause is
FROM PRODUCT_INFORMATION
9. Executing this statement returns the report required as shown in the following illustration.

7
Displaying Data
from Multiple
Tables

CERTIFICATION OBJECTIVES
7.01

Write SELECT Statements to Access
Data from More Than One Table Using
Equijoins and Nonequijoins

7.02

Join a Table to Itself Using a Self-Join

7.03

View Data that Does Not Meet a Join
Condition Using Outer Joins

7.04

✓
Q&A

Generate a Cartesian Product of Two
or More Tables
Two-Minute Drill
Self Test

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

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T

he three pillars of relational theory are selection, projection, and joining. This chapter
focuses on the practical implementation of joining. Rows from different tables are
associated with each other using joins. Support for joining has implications for the way
data is stored in database tables. Many data models such as third normal form or star schemas
have emerged to exploit this feature.
Tables may be joined in several ways. The most common technique is called
an equijoin. A row is associated with one or more rows in another table based on
the equality of column values or expressions. Tables may also be joined using a
nonequijoin. In this case, a row is associated with one or more rows in another table if
its column values fall into a range determined by inequality operators.
A less common technique is to associate rows with other rows in the same
table. This association is based on columns with logical and usually hierarchical
relationships with each other. This is called a self-join. Rows with null or differing
entries in common join columns are excluded when equijoins and nonequijoins are
performed. An outer join is available to fetch these one-legged or orphaned rows if
necessary.
A cross join or Cartesian product is formed when every row from one table is joined
to all rows in another. This join is often the result of missing or inadequate join
conditions but is occasionally intentional.

CERTIFICATION OBJECTIVE 7.01

Write SELECT Statements to Access Data
from More Than One Table Using Equijoins
and Nonequijoins
This certification objective receives extensive coverage in this chapter. It is crucial
to learning the concepts and language for performing joins. Different types of joins
are introduced in their primitive forms outlining the broad categories that are
available. An in-depth discussion of the various join clauses is then conducted.
The modern ANSI-compliant and traditional Oracle syntaxes are discussed, but
emphasis is placed on the modern syntax. This section concludes with a discussion

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311

of nonequijoins and additional join conditions. Joining is described by focusing on
the following eight areas:
■ Types of joins
■ Joining tables using SQL:1999 syntax
■ Qualifying ambiguous column names
■ The NATURAL JOIN clause
■ The natural JOIN USING clause
■ The natural JOIN ON clause
■ N-way joins and additional join conditions
■ Nonequijoins

Types of Joins
Two basic joins are the equijoin and the nonequijoin. Equijoins are more frequently
used. Joins may be performed between multiple tables, but much of the following
discussion will use two hypothetical tables to illustrate the concepts and language of
joins. The first table is called the source and the second is called the target. Rows in
the source and target tables comprise one or more columns. As an example, assume
that the source and target are the COUNTRIES and REGIONS tables from the HR
schema, respectively.
The COUNTRIES table contains three columns named COUNTRY_ID,
COUNTRY_NAME, and REGION_ID. The REGIONS table is comprised of two
columns named REGION_ID and REGION_NAME. The data in these two tables
is related to each other based on the common REGION_ID column. Consider the
following queries:
Query 1: select * from countries where country_id='CA';
Query 2: select region_name from regions where region_id='2';

The name of the region to which a country belongs may be determined by
obtaining its REGION_ID value. This value is used to join it with the row in the
REGIONS table with the same REGION_ID. Query 1 retrieves the column values
associated with the row from the COUNTRIES table where the COUNTRY_
ID='CA'. The REGION_ID value of this row is 2. Query 2 fetches the Americas
REGION_NAME from the REGIONS table for the row with REGION_ID=2.
Equijoining facilitates the retrieval of column values from multiple tables using a
single query.

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The source and target tables can be swapped, so the REGIONS table could be the
source and the COUNTRIES table could be the target. Consider the following two
queries:
Query 1: select * from regions where region_name='Americas';
Query 2: select country_name from countries where region_id='2';

Query 1 fetches one row with a REGION_ID value of 2. Joining in this reversed
manner allows the following question to be asked: What countries belong to the
Americas region? The answers from the second query are five COUNTRY_NAME
values: Argentina, Brazil, Canada, Mexico, and the United States of America.
These results may be obtained from a single query that joins the tables together.
The language to perform equijoins, nonequijoins, outer joins, and cross joins is
introduced next, along with a discussion of the traditional Oracle join syntax.

Natural Joins
The natural join is implemented using three possible join clauses that use the
following keywords in different combinations: NATURAL JOIN, USING, and ON.
When the source and target tables share identically named columns, it is possible
to perform a natural join between them without specifying a join column. This is
sometimes referred to as a pure natural join. In this scenario, columns with the same
names in the source and target tables are automatically associated with each other.
Rows with matching column values in both tables are retrieved. The REGIONS and
COUNTRIES table both share the REGION_ID column. They may be naturally
joined without specifying join columns, as shown in the first two queries in Figure 7-1.
The NATURAL JOIN keywords instruct Oracle to identify columns with
identical names between the source and target tables. Thereafter, a join is implicitly
performed between them. In the first query, the REGION_ID column is identified as
the only commonly named column in both tables. REGIONS is the source table and
appears after the FROM clause. The target table is therefore COUNTRIES. For each
row in the REGIONS table, a match for the REGION_ID value is sought from all
the rows in the COUNTRIES table. An interim result set is constructed containing
rows matching the join condition. This set is then restricted by the WHERE clause.
In this case, because the COUNTRY_NAME value must be Canada, REGION_
NAME of Americas is returned.
The second query shows a natural join where COUNTRIES is the source table.
The REGION_ID value for each row in the COUNTRIES table is identified and a
search for a matching row in the REGIONS table is initiated. If matches are found,
the interim results are limited by any WHERE conditions. The COUNTRY_NAME
from rows with Americas as their REGION_NAME are returned.

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313

FIGURE 7-1

Natural joins

Sometimes more control must be exercised regarding which columns to use for
joins. When there are identical column names in the source and target tables you
want to exclude as join columns, the JOIN…USING format may be used. Remember
that Oracle does not impose any rules stating that columns with the same name
in two discrete tables must have a relationship with each other. The third query
explicitly specifies that the REGIONS table be joined to the COUNTRIES table
based on common values in their REGION_ID columns. This syntax allows natural
joins to be formed on specific columns instead of on all commonly named columns.
The fourth query demonstrates the JOIN…ON format of the natural join, which
allows join columns to be explicitly stated. This format does not depend on the
columns in the source and target tables having identical names. This form is more
general and is the most widely used natural join format.

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Be wary when using pure natural joins since database designers may assign
the same name to key or unique columns.These columns may have names
like ID or SEQ_NO. If a pure natural join is attempted between such tables,
ambiguous and unexpected results may be returned.

Outer Joins
Not all tables share a perfect relationship, where every record in the source table can
be matched to at least one row in the target table. It is occasionally required that
rows with nonmatching join column values also be retrieved by a query. This may
seem to defeat the purpose of joins but has some practical benefits.
Suppose the EMPLOYEES and DEPARTMENTS tables are joined with common
DEPARTMENT_ID values. EMPLOYEES records with null DEPARTMENT_ID
values are excluded along with values absent from the DEPARTMENTS table. An
outer join fetches these rows.

Cross Joins
A cross join or Cartesian product derives its names from mathematics, where it is also
referred to as a cross product between two sets or matrices. This join creates one row
of output for every combination of source and target table rows.
If the source and target tables have three and four rows, respectively, a cross join
between them results in (3 × 4 = 12) rows being returned. Consider the row counts
retrieved from the queries in Figure 7-2.
The first two row counts are performed on the COUNTRIES and REGIONS
tables yielding 25 and 4 rows respectively. The third query counts the number
of rows returned from a cross join of these tables and yields 100. Query 4 would
return 100 records if the WHERE clause was absent. Each of the four rows in the
REGIONS table is joined to the one row from the COUNTRIES table. Each row
returned contains every column from both tables.

Oracle Join Syntax
A proprietary Oracle join syntax has evolved that is stable and understood by
millions of users. This traditional syntax is supported by Oracle and is present in
software systems across the world. You will no doubt encounter the traditional
Oracle join syntax that is now making way for the standardized ANSI-compliant
syntax discussed in this chapter.

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315

FIGURE 7-2

Cross join

The traditional Oracle join syntax supports natural joining, outer joins, and
Cartesian joins, as shown in the following queries:
Query 1: select regions.region_name, countries.country_name
from regions, countries
where regions.region_id=countries.region_id;
Query 2: select last_name, department_name
from employees, departments
where employees.department_id (+) = departments.department_id;
Query 3: select * from regions,countries;

Query 1 performs a natural join by specifying the join as a condition in the
WHERE clause. This is the most significant difference between the traditional and
ANSI SQL join syntaxes. Take note of the column aliasing using the TABLE
.COLUMN_NAME notation to disambiguate the identical column names. This
notation is discussed in detail later in this chapter. Query 2 specifies the join
between the source and target tables as a WHERE condition. There is a plus symbol
enclosed in brackets (+) to the left of the equal sign that indicates to Oracle that a
right outer join must be performed. This query returns employees’ LAST_NAME and

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their matching DEPARTMENT_NAME values. In addition, the outer join retrieves
DEPARTMENT_NAME from the rows with DEPARTMENT_ID values not
currently assigned to any employee records. Query 3 performs a Cartesian or cross
join by excluding the join condition.

The traditional Oracle
join syntax is widely used. However, the
exam assesses your understanding of joins
and the ANSI SQL forms of its syntax. Be
prepared, though: some questions may tap

your knowledge of the traditional syntax.
This knowledge is useful since traditional
Oracle syntax is deeply embedded across
software systems worldwide.

Joining Tables Using SQL:1999 Syntax
Prior to Oracle 9i, the traditional join syntax was the only language available to join
tables. Since then, Oracle has introduced a new language that is compliant to the
ANSI SQL:1999 standards. It offers no performance benefits over the traditional
syntax. Natural, outer, and cross joins may be written using both SQL:1999 and
traditional Oracle SQL.
The general form of the SELECT statement using ANSI SQL:1999 syntax is as
follows:
SELECT table1.column, table2.column
FROM table1
[NATURAL JOIN table2] |
[JOIN table2 USING (column_name)] |
[JOIN table2 ON (table1.column_name = table2.column_name)] |
[LEFT | RIGHT | FULL OUTER JOIN table2
ON (table1.column_name = table2.column_name)] |
[CROSS JOIN table2];
This is dissected and examples are explained in the following sections. The
general form of the traditional Oracle-proprietary syntax relevant to joins is as
follows:
SELECT table1.column, table2.column
FROM table1, table2
[WHERE (table1.column_name = table2.column_name)] |
[WHERE (table1.column_name(+)= table2.column_name)] |
[WHERE (table1.column_name)= table2.column_name) (+)] ;

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317

If no joins or fewer than N-1 joins are specified in the WHERE clause conditions,
where N refers to the number of tables in the query, then a Cartesian or cross join
is performed. If an adequate number of join conditions is specified, then the first
optional conditional clause specifies a natural join, while the second two optional
clauses specify the syntax for right and left outer joins.

Qualifying Ambiguous Column Names
Columns with the same names may occur in tables involved in a join. The
columns named DEPARTMENT_ID and MANAGER_ID are found in both the
EMPLOYEES and DEPARTMENTS tables. The REGION_ID column is present
in both the REGIONS and COUNTRIES tables. Listing such columns in a query
becomes problematic when Oracle cannot resolve their origin. Columns with unique
names across the tables involved in a join cause no ambiguity, and Oracle can easily
resolve their source table.
The problem of ambiguous column names is addressed with dot notation. A
column may be prefixed by its table name and a dot or period symbol to designate
its origin. This differentiates it from a column with the same name in another table.
Dot notation may be used in queries involving any number of tables. Referencing
some columns using dot notation does not imply that all columns must be referenced
in this way.
Dot notation is enhanced with table aliases. A table alias provides an alternate,
usually shorter name for a table. A column may be referenced as TABLE_NAME
.COLUMN_NAME or TABLE_ALIAS.COLUMN_NAME. Consider the query
shown in Figure 7-3.
The EMPLOYEES table is aliased with the short name EMP while the
DEPARTMENTS table is not. The SELECT clause references the EMPLOYEE_ID
and MANAGER_ID columns as EMP.EMPLOYEE_ID and EMP.MANAGER_ID.
The MANAGER_ID column from the DEPARTMENTS table is referred to as
DEPARTMENTS.MANAGER_ID. Qualifying the EMPLOYEE_ID column using
dot notation is unnecessary because there is only one column with this name
between the two tables. Therefore, there is no ambiguity.
The MANAGER_ID column must be qualified to avoid ambiguity because
it is featured in both tables. Since the JOIN…USING format is applied, only
DEPARTMENT_ID is used as the join column. If a NATURAL JOIN was employed,
both the DEPARTMENT_ID and MANAGER_ID columns would be used. If the
MANAGER_ID column was not qualified, an “ORA-00918:column ambiguously
defined” error would be returned. If DEPARTMENT_ID was aliased, an “ORA25154:column part of USING clause cannot have qualifier” error would be raised.

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FIGURE 7-3

Dot notation

SQL Developer provides the heading MANAGER_ID to the first reference made
in the SELECT clause. The string "_1" is automatically appended to the second
reference, creating the heading MANAGER_ID_1.
Qualifying column references with dot notation to indicate a column’s table
of origin has a performance benefit.Time is saved because Oracle is directed
instantaneously to the appropriate table and does not have to resolve the
table name.

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319

The NATURAL JOIN Clause
The general syntax for the NATURAL JOIN clause is as follows:
SELECT table1.column, table2.column
FROM table1
NATURAL JOIN table2;
The pure natural join identifies the columns with common names in table1 and
table2 and implicitly joins the tables using all these columns. The columns in the
SELECT clause may be qualified using dot notation unless they are one of the join
columns. Consider the following queries:
Query
Query
where
Query
Query

1: select * from locations natural join countries;
2: select * from locations, countries
locations.country_id = countries.country_id;
3: select * from jobs natural join countries;
4: select * from jobs, countries;

The natural join identifies columns with common names between the two tables.
In query 1, COUNTRY_ID occurs in both tables and becomes the join column.
Query 2 is written using traditional Oracle syntax and retrieves the same rows as
query 1. Unless you are familiar with the columns in the source and target tables,
natural joins must be used with caution, as join conditions are automatically formed
between all columns with shared names.
Query 3 performs a natural join between the JOBS and COUNTRIES tables. There
are no columns with identical names, resulting in a Cartesian product. Query 4 is
equivalent to query 3, and a Cartesian join is performed using traditional Oracle syntax.
The natural join is simple but prone to a fundamental weakness. It suffers the
risk that two columns with the same name might have no relationship and may
not even have compatible data types. In Figure 7-4, the COUNTRIES, REGIONS,
and SALE_REGIONS tables are described. The SALES_REGIONS table was
constructed to illustrate the following important point: Although it has REGION_
ID in common with the COUNTRIES table, it cannot be naturally joined to it
because their data types are incompatible. The data types of the COUNTRIES.
REGION_ID and SALES_REGIONS.REGION_ID columns are NUMBER and
VARCHAR2, respectively. The character data cannot be implicitly converted into
numeric data and an “ORA-01722: invalid number” error is raised. The REGIONS
.REGION_ID column is of type NUMBER and its data is related to the data in
the COUNTRIES table. Therefore, the natural join between the REGIONS and
COUNTRIES table works perfectly.

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FIGURE 7-4

The natural join

EXERCISE 7-1
Using the NATURAL JOIN
The JOB_HISTORY table shares three identically named columns with the
EMPLOYEES table: EMPLOYEE_ID, JOB_ID, and DEPARTMENT_ID. You
are required to describe the tables and fetch the EMPLOYEE_ID, JOB_ID,
DEPARTMENT_ID, LAST_NAME, HIRE_DATE, and END_DATE values for all
rows retrieved using a pure natural join. Alias the EMPLOYEES table as EMP and
the JOB_HISTORY table as JH and use dot notation where possible.
1. Start SQL*Plus and connect to the HR schema.
2. The tables are described using the commands DESC EMPLOYEES and
DESC JOB_HISTORY, and the columns with identical names and their data
types may be examined.

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3. The FROM clause is
FROM JOB_HISTORY JH
4. The JOIN clause is
NATURAL JOIN EMPLOYEES EMP
5. The SELECT clause is
SELECT EMP.LAST_NAME, EMP.HIRE_DATE, JH.END_DATE
6. Executing this statement returns a single row with the same EMPLOYEE_ID,
JOB_ID, and DEPARTMENT_ID values in both tables and is shown in the
following illustration:

The Natural JOIN USING Clause
The format of the syntax for the natural JOIN USING clause is as follows:
SELECT table1.column, table2.column
FROM table1
JOIN table2 USING (join_column1, join_column2…);

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While the pure natural join contains the NATURAL keyword in its syntax, the
JOIN…USING syntax does not. An error is raised if the keywords NATURAL
and USING occur in the same join clause. The JOIN…USING clause allows one
or more equijoin columns to be explicitly specified in brackets after the USING
keyword. This avoids the shortcomings associated with the pure natural join. Many
situations demand that tables be joined only on certain columns, and this format
caters to this requirement. Consider the following queries:
Query
Query
where
Query

1: select * from locations join countries using (country_id);
2: select * from locations, countries
locations.country_id = countries.country_id;
3: select * from jobs join countries using ;

Query 1 specifies that the LOCATIONS and COUNTRIES tables must be
joined on common COUNTRY_ID column values. All columns from these
tables are retrieved for the rows with matching join column values. Query 2
shows a traditionally specified query that retrieves the same rows as query 1.
Query 3 illustrates that a Cartesian join cannot be accidentally specified with the
JOIN…USING syntax since only columns with shared names are permitted after
the USING keyword. The join columns cannot be qualified using table names or
aliases when they are referenced. Since this join syntax potentially excludes some
columns with identical names from the join clause, these must be qualified if they
are referenced to avoid ambiguity.
As Figure 7-5 shows, the JOB_HISTORY and EMPLOYEES tables were joined
based on the presence of equal values in their JOB_ID and EMPLOYEE_ID
columns. Rows conforming to this join condition are retrieved. These tables
share three identically named columns. The JOIN…USING syntax allows the
specification of only two of these as join columns. Notice that although the third
identically named column is DEPARTMENT_ID, it is qualified with a table alias
to avoid ambiguity. However, the join columns in the SELECT clause cannot be
qualified with table aliases.

The Natural JOIN ON Clause
The format of the syntax for the natural JOIN ON clause is as follows:
SELECT table1.column, table2.column
FROM table1
JOIN table2 ON (table1.column_name = table2.column_name);

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FIGURE 7-5

Natural join using
the JOIN…
USING clause

The pure natural join and the JOIN…USING clauses depend on join columns
with identical column names. The JOIN…ON clause allows the explicit
specification of join columns, regardless of their column names. This is the most
flexible and widely used form of the join clauses. The ON and NATURAL keywords
cannot appear together in a join clause. The equijoin columns are fully qualified as
table1.column1 = table2.column2 and are optionally specified in brackets after the
ON keyword. The following queries illustrate the JOIN…ON clause:
Query 1: select * from departments d
join employees e on (e.employee_id=d.department_id);
Query 2: select * from employees e, departments d
where e.employee_id=d.department_id;

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SCENARIO & SOLUTION
You are required to retrieve information from
multiple tables, group the results, and apply an
aggregate function to them. Can a group function be
used against data from multiple table sources?

Yes. Joining multiple tables ultimately yields a set
of data comprising one or more rows and columns.
Once the dataset is created, aggregate functions treat
it as if the data originated from one source.

When joining two tables, there is a risk that between
them they contain common column names. Does
Oracle know which tables to fetch data from if such
columns are present in the SELECT list?

No. Oracle does not know from which tables such
columns originate, and an error is raised. Ambiguous
column references can be avoided using qualifiers.
Qualifiers employ dot notation to clarify a column’s
table of origin.

The NATURAL JOIN clause is used to join rows
from two tables based on columns with common
names sharing identical values. Is it possible to join
two tables based on some of the shared columns and
not all of them?

Yes. The clause recommended to naturally join two
tables based on one or more of the columns with
identical names is JOIN…USING. A pair of brackets
follows the USING clause in which the unqualified
join columns are specified.

Query 1 retrieves all column values from both the DEPARTMENTS and
EMPLOYEES tables for the rows that meet an equijoin condition. This condition
is fulfilled by EMPLOYEE_ID values matching DEPARTMENT_ID values in the
DEPARTMENTS table. The traditional Oracle syntax in query 2 returns the same
results as query 1. Notice the similarities between the traditional join condition
specified in the WHERE clause and the join condition specified after the ON
keyword.
The START_DATE column in the JOB_HISTORY table is joined to the HIRE_
DATE column in the EMPLOYEES table in Figure 7-6. This equijoin retrieves the
details of employees who worked for the organization and changed jobs.

EXERCISE 7-2
Using the NATURAL JOIN…ON Clause
Each record in the DEPARTMENTS table has a MANAGER_ID column matching
an EMPLOYEE_ID value in the EMPLOYEES table. You are required to produce a
report with one column aliased as Managers. Each row must contain a sentence of the
format FIRST_NAME LAST_NAME is manager of the DEPARTMENT_NAME

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FIGURE 7-6

Natural join using
the JOIN…ON
clause

department. Alias the EMPLOYEES table as E and the DEPARTMENTS table as D
and use dot notation where possible.
1. Start SQL Developer and connect to the HR schema.
2. The Managers column may be constructed by concatenating the required
items and separating them with spaces.
3. The SELECT clause is
SELECT E.FIRST_NAME||' '||E.LAST_NAME||' is manager of the '||
D.DEPARTMENT_NAME||' department.' "Managers"

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4. The FROM clause is
FROM EMPLOYEES E
5. The JOIN…ON clause is
JOIN DEPARTMENTS D
ON (E. EMPLOYEE_ID=D.MANAGER_ID).
6. Executing this statement returns 11 rows describing the managers of each
department as shown in the following illustration:

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327

N-Way Joins and Additional Join Conditions
The joins just discussed were demonstrated using two tables. There is no restriction
on the number of tables that may be related using joins. Third normal form consists
of a set of tables connected through a series of primary and foreign key relationships.
Traversing these relationships using joins enables consistent and reliable retrieval of
data. When multiple joins exist in a statement, they are evaluated from left to right.
Consider the following query using pure natural joins:
select r.region_name, c.country_name, l.city, d.department_name
from departments d natural join locations l
natural join countries c natural join regions r

The join between DEPARTMENTS and LOCATIONS creates an interim
result set consisting of 27 rows. These tables provide the DEPARTMENT_NAME
and CITY columns. This set is naturally joined to the COUNTRIES table. Since
the interim set does not contain the COUNTRY_ID column, a Cartesian join is
performed. The 27 interim rows are joined to the 25 rows in the COUNTRIES
table, yielding a new interim results set with 675 (27 × 25) rows and three columns:
DEPARTMENT_NAME, CITY, and COUNTRY_NAME. This set is naturally
joined to the REGIONS table. Once again, a Cartesian join occurs because the
REGION_ID column is absent from the interim set. The final result set contains
2700 (675 × 4) rows and four columns. Using pure natural joins with multiple tables
is error prone and not recommended.
The JOIN…USING and JOIN…ON syntaxes are better suited for joining multiple
tables. The following query joins four tables using the pure natural join syntax:
select region_id, country_id, c.country_name, l.city, d.department_name
from departments d natural join locations l
natural join countries c natural join regions r

This query correctly yields 27 rows in the final results set since the required join
columns are listed in the SELECT clause. The following query demonstrates how the
JOIN…ON clause is used to fetch the same 27 rows. A join condition can reference
only columns in its scope. In the following example, the join from DEPARTMENTS
to LOCATIONS may not reference columns in the COUNTRIES or REGIONS
tables, but the join between COUNTRIES and REGIONS may reference any
column from the four tables involved in the query.
select r.region_name, c.country_name, l.city, d.department_name
from departments d
join locations l on (l.location_id=d.location_id)
join countries c on (c.country_id=l.country_id)
join regions r on (r.region_id=c.region_id)

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The JOIN…USING clause can also be used to join these four tables as follows:
select r.region_name, c.country_name, l.city, d.department_name
from departments d
join locations l using (location_id)
join countries c using (country_id)
join regions r using (region_id)

The WHERE clause is used to specify conditions that restrict the results set of a
query whether it contains joins or not. The JOIN…ON clause is also used to specify
conditions that limit the results set created by the join. Consider the following two
queries:
Query 1: select d.department_name from departments d
join locations l on (l.LOCATION_ID=d.LOCATION_ID)
where d.department_name like 'P%'
Query 2: select d.department_name from departments d
join locations l on
(l.LOCATION_ID=d.LOCATION_ID and d.department_name like 'P%')

Query 1 uses a WHERE clause to restrict the 27 rows created by equijoining the
DEPARTMENTS and LOCATIONS tables based on their LOCATION_ID values
to the three that contain DEPARTMENT_ID values beginning with the letter P.
Query 2 implements the condition within the brackets of the ON subclause and
returns the same three rows.
Five tables are joined in Figure 7-7, resulting in a list describing the top earning
employees and geographical information about their departments.

There are three natural
join formats.The pure natural join uses
the NATURAL JOIN clause and joins two
tables based on all columns with shared
names.The other two formats use the
JOIN…USING and JOIN…ON clauses
and are also referred to as natural joins.
They do not use the NATURAL keyword.

Pay attention to the syntax, since a
join clause such as SELECT * FROM
TABLE1 NATURAL JOIN TABLE2 USING
(COLUMN) may appear correct, but is, in
fact, syntactically incorrect. Remember the
USING, ON, and NATURAL keywords are
mutually exclusive in the context of the
same join clause.

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FIGURE 7-7

N-way joins and
additional join
conditions

Nonequijoins
Nonequijoins match column values from different tables based on an inequality
expression. The value of the join column in each row in the source table is
compared to the corresponding values in the target table. A match is found if the
expression used in the join, based on an inequality operator, evaluates to true. When
such a join is constructed, a nonequijoin is performed.
A nonequijoin is specified using the JOIN…ON syntax, but the join condition
contains an inequality operator instead of an equal sign.

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The format of the syntax for a nonequijoin clause is as follows:
SELECT table1.column, table2.column
FROM table1
[JOIN table2 ON (table1.column_name < table2.column_name)]|
[JOIN table2 ON (table1.column_name > table2.column_name)]|
[JOIN table2 ON (table1.column_name <= table2.column_name)]|
[JOIN table2 ON (table1.column_name >= table2.column_name)]|
[JOIN table2 ON (table1.column BETWEEN table2.col1 AND table2.col2)]|
Consider the first 15 rows returned by the query in Figure 7-8. The EMPLOYEES
table is nonequijoined to the JOBS table based on the inequality join condition
(2*E.SALARY < J.MAX_SALARY). The JOBS table stores the salary range for
different jobs in the organization. The SALARY value for each employee record is

FIGURE 7-8

Nonequijoins

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331

doubled and compared with all MAX_SALARY values in the JOBS table. If the join
condition evaluates to true, the row is returned.
The first two rows display the employee with a LAST_NAME of Abel who
currently has a JOB_ID value of SA_REP and earns a SALARY of 11000. These
are the only two rows in the JOBS table that satisfy the inequality join condition
(2*E.SALARY < J.MAX_SALARY) for this employee record.
Nonequijoins are not commonly used.The BETWEEN range operator often
appears with nonequijoin conditions. It is simpler to use one BETWEEN
operator in a condition than two nonequijoin conditions based on the less
than or equal to (<=) and the greater than or equal to (>=) operators.

CERTIFICATION OBJECTIVE 7.02

Join a Table to Itself Using a Self-Join
Storing hierarchical data in a single relational table is accomplished by allocating at
least two columns per row. One column stores an identifier of the row’s parent record
and the second stores the row’s identifier. Associating rows with each other based
on a hierarchical relationship requires Oracle to join a table to itself. This self-join
technique is discussed in the next section.

Joining a Table to Itself Using the JOIN…ON Clause
Suppose there is a need to store a family tree in a relational table. There are several
approaches one could take. One option is to use a table called FAMILY with
columns named ID, NAME, MOTHER_ID, and FATHER_ID, where each row
stores a person’s name, unique ID number, and the ID values for their parents.
When two tables are joined, each row from the source table is subjected to the
join condition with rows from the target table. If the condition evaluates to true,
then the joined row, consisting of columns from both tables, is returned.
When the join columns originate from the same table, a self-join is required.
Conceptually, the source table is duplicated to create the target table. The self-join
works like a regular join between these tables. Note that, internally, Oracle does not

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duplicate the table and this description is merely provided to explain the concept of
self-joining. Consider the following four queries:
Query 1: select id, name, father_id from family;
Query 2: select name from family where id=&father_id;
Query 3: select f1.name Dad, f2.name Child
from family f1 join family f2 on (f1.id=f2.father_id)

To identify a person’s father in the FAMILY table, you could use query 1 to get
that person’s ID, NAME, and FATHER_ID value. In query 2, the FATHER_ID
value obtained from the first query can be substituted to obtain the father’s NAME
value. Notice that both queries 1 and 2 source information from the FAMILY table.
Query 3 performs a self-join with the JOIN…ON clause by aliasing the FAMILY
table as f1 and f2. Oracle treats these as different tables even though they point to
the same physical table. The first occurrence of the FAMILY table, aliased as f1, is
designated as the source table, while the second occurrence, aliased as f2, is assigned
as the target table. The join condition in the ON clause is of the format source.child_
id=target.parent_id. Figure 7-9 shows a sample of FAMILY data and demonstrates a
three-way self-join to the same table.

EXERCISE 7-3
Performing a Self-Join
There is a hierarchical relationship between employees and their managers. For
each row in the EMPLOYEES table the MANAGER_ID column stores the
EMPLOYEE_ID of every employee’s manager. Using a self-join on the EMPLOYEES
table, you are required to retrieve the employee’s LAST_NAME, EMPLOYEE_ID,
manager’s LAST_NAME, and employee’s DEPARTMENT_ID for the rows with
DEPARMENT_ID values of 10, 20, or 30. Alias the EMPLOYEES table as E and
the second instance of the EMPLOYEES table as M. Sort the results based on the
DEPARTMENT_ID column.
1. Start SQL Developer and connect to the HR schema.
2. The SELECT clause is
SELECT E.LAST_NAME EMPLOYEE, E.EMPLOYEE_ID, E.MANAGER_ID,
M.LAST_NAME MANAGER, E.DEPARTMENT_ID.
3. The FROM clause with source table and alias is
FROM EMPLOYEES E.

Join a Table to Itself Using a Self-Join

FIGURE 7-9

Self-join

4. The JOIN…ON clause with aliased target table is
JOIN EMPLOYEES M ON (E.MANAGER_ID=M.EMPLOYEE_ID).
5. The WHERE clause is
WHERE E.DEPARTMENT_ID IN (10,20,30).
6. The ORDER BY clause is
ORDER BY E.DEPARTMENT_ID.

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7. Executing this statement returns nine rows describing the managers of each
employee in these departments as shown in the following illustration:

CERTIFICATION OBJECTIVE 7.03

View Data That Does Not Meet a Join Condition by
Using Outer Joins
Equijoins match rows between two tables based on the equality of the column data
stored in each table. Nonequijoins rely on matching rows between tables based on
a join condition containing an inequality expression. Target table rows with no
matching join column in the source table are usually not required. When they are

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335

required, however, an outer join is used to fetch them. Several variations of outer joins
may be used depending on whether join column data is missing from the source or
target tables or both. These outer join techniques are described in the following topics:
■ Inner versus outer joins
■ Left outer joins
■ Right outer joins
■ Full outer joins

Inner versus Outer Joins
When equijoins and nonequijoins are performed, rows from the source and target
tables are matched using a join condition formulated with equality and inequality
operators, respectively. These are referred to as inner joins. An outer join is performed
when rows, which are not retrieved by an inner join, are returned.
Two tables sometimes share a master-detail or parent-child relationship. In the
sample HR schema there are several pairs of tables with such a relationship. One
pair is the DEPARTMENTS and EMPLOYEES tables. The DEPARTMENTS table
stores a master list of DEPARTMENT_NAME and DEPARTMENT_ID values.
Each EMPLOYEES record has a DEPARTMENT_ID column constrained to be
either a value that exists in the DEPARTMENTS table or null. This leads to one
of the following three scenarios. The fourth scenario could occur if the constraint
between the tables was removed.
1. An employee row has a DEPARTMENT_ID value that matches a row in the
DEPARTMENTS table.
2. An employee row has a null value in its DEPARTMENT_ID column.
3. There are rows in the DEPARTMENTS table with DEPARTMENT_ID
values that are not stored in any employee records.
4. An employee row has a DEPARTMENT_ID value that is not featured in the
DEPARTMENTS table.
The first scenario describes a natural inner join between the two tables. The
second and third scenarios cause many problems. Joining the EMPLOYEES and
DEPARTMENTS tables results in employee rows being excluded. An outer join can
be used to include these orphaned rows in the results set. The fourth scenario should
rarely occur in a well designed database, because foreign key constraints would
prevent the insertion of child records with no parent values. Since this row will be
excluded by an inner join, it may be retrieved using an outer join.

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A left outer join between the source and target tables returns the results of an
inner join as well as rows from the source table excluded by that inner join. A right
outer join between the source and target tables returns the results of an inner join
as well as rows from the target table excluded by that inner join. If a join returns
the results of an inner join as well as rows from both the source and target tables
excluded by that inner join, then a full outer join has been performed.

Left Outer Joins
The format of the syntax for the LEFT OUTER JOIN clause is as follow:
SELECT table1.column, table2.column
FROM table1
LEFT OUTER JOIN table2
ON (table1.column = table2.column);
A left outer join performs an inner join of table1 and table2 based on the
condition specified after the ON keyword. Any rows from the table on the left of
the JOIN keyword excluded for not fulfilling the join condition are also returned.
Consider the following two queries:
Query 1: select e.employee_id, e.department_id EMP_DEPT_ID,
d.department_id DEPT_DEPT_ID, d.department_name
from departments d left outer join employees e
on (d.DEPARTMENT_ID=e.DEPARTMENT_ID)
where d.department_name like 'P%'
Query 2: select e.employee_id, e.department_id EMP_DEPT_ID,
d.department_id DEPT_DEPT_ID, d.department_name
from departments d join employees e
on (d.DEPARTMENT_ID=e.DEPARTMENT_ID)
where d.department_name like 'P%'

Queries 1 and 2 are identical except for the join clauses, which have the keywords
LEFT OUTER JOIN and JOIN, respectively. Query 2 performs an inner join and
seven rows are returned. These rows share identical DEPARTMENT_ID values
in both tables. Query 1 returns the same seven rows and one additional row. This
extra row is obtained from the table to the left of the JOIN keyword, which is the
DEPARTMENTS table. It is the row containing details of the Payroll department.
The inner join does not include this row since no employees are currently assigned
to the department.
A left outer join is shown in Figure 7-10. The inner join produces 27 rows with
matching LOCATION_ID values in both tables. There are 43 rows in total, which
implies that 16 rows were retrieved from the LOCATIONS table, which is on the

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337

FIGURE 7-10

Left outer join

left of the JOIN keyword. None of the rows from the DEPARTMENTS table contain
any of these 16 LOCATION_ID values.

Right Outer Joins
The format of the syntax for the RIGHT OUTER JOIN clause is as follows:
SELECT table1.column, table2.column
FROM table1
RIGHT OUTER JOIN table2
ON (table1.column = table2.column);

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A right outer join performs an inner join of table1 and table2 based on the join
condition specified after the ON keyword. Rows from the table to the right of the
JOIN keyword, excluded by the join condition, are also returned. Consider the
following query:
select e.last_name, d.department_name from departments d
right outer join employees e
on (e.department_id=d.department_id)
where e.last_name like 'G%';

The inner join produces seven rows containing details for the employees with
LAST_NAME values that begin with G. The EMPLOYEES table is to the right
of the JOIN keyword. Any employee records which do not conform to the join
condition are included, provided they conform to the WHERE clause condition. In
addition, the right outer join fetches one EMPLOYEE record with a LAST_NAME
of Grant. This record currently has a null DEPARTMENT_ID value. The inner join
excludes the record since no DEPARTMENT_ID is assigned to this employee.
A right outer join between the JOB_HISTORY and EMPLOYEES tables is
shown in Figure 7-11. The EMPLOYEES table is on the right of the JOIN keyword.
The DISTINCT keyword eliminates duplicate combinations of JOB_ID values from
the tables. The results show the jobs that employees have historically left. The jobs
that no employees have left are also returned.

There are three types of
outer join formats. Each of them performs
an inner join before including rows the
join condition excluded. If a left outer join
is performed then rows excluded by the

inner join, to the left of the JOIN keyword,
are also returned. If a right outer join is
performed then rows excluded by the inner
join, to the right of the JOIN keyword, are
returned as well.

Full Outer Joins
The format of the syntax for the FULL OUTER JOIN clause is as follows:
SELECT table1.column, table2.column
FROM table1
FULL OUTER JOIN table2
ON (table1.column = table2.column);

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339

FIGURE 7-11

Right outer join

A full outer join returns the combined results of a left and right outer join.
An inner join of table1 and table2 is performed before rows excluded by the join
condition from both tables are merged into the results set.
The traditional Oracle join syntax does not support a full outer join, which is
typically performed by combining the results from a left and right outer join using
the UNION set operator described in Chapter 9. Consider the full outer join shown
in Figure 7-12. The WHERE clause restricting the results to rows with NULL
DEPARTMENT_ID values shows the orphan rows in both tables. There is one
record in the EMPLOYEES table which has no DEPARTMENT_ID values, and
there are 16 departments to which no employees belong.

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FIGURE 7-12

Full outer join

EXERCISE 7-4
Performing an Outer-Join
The DEPARTMENTS table contains details of all departments in the organization.
You are required to retrieve the DEPARTMENT_NAME and DEPARTMENT_ID
values for those departments to which no employees are currently assigned.
1. Start SQL*Plus and connect to the HR schema.
2. The SELECT clause is
SELECT D.DEPARTMENT_NAME, D.DEPARTMENT_ID.

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SCENARIO & SOLUTION
The data in two tables you wish to join is related but
does not share any identically named columns. Is
it possible to join tables using columns that do not
share the same name?

Yes. The JOIN…ON clause is provided for this
purpose. It provides a flexible and generic solution to
joining tables based on nonidentical column names.

You wish to divide staff into four groups named after
the four regions in the REGIONS table. Is it possible
to obtain a list of EMPLOYEE_ID, LAST_NAME,
and REGION_NAME values for each employee
by joining the EMPLOYEE_ID and REGION_ID
columns in a round-robin manner?

Yes. The REGION_ID value ranges from 1 to 4.
Adding 1 to the remainder of EMPLOYEE_ID
divided by 4 creates a value in the range 1 to 4. The
round-robin assignment of employees may be done as
follows:
SELECT LAST_NAME, EMPLOYEE_ID,
REGION_NAME, FROM EMPLOYEES JOIN
REGIONS ON (MOD(EMPLOYEE_ID,4)+1=
REGION_ID)

You are required to retrieve a list of DEPARTMENT_
NAME and LAST_NAME values for all
departments, including those that currently have no
employees assigned to them. In such cases the string
‘No Employees’ should be displayed as the LAST_
NAME column value. Can this be done using joins?

Yes. Depending on which side of the JOIN keyword
the DEPARTMENTS table is placed, a left or right
outer join may be used, since this is the table from
where the orphan rows originate. The following
query satisfies the request:
SELECT DEPARTMENT_NAME, NVL(LAST_
NAME,'No Employees') FROM EMPLOYEES
RIGHT OUTER JOIN DEPARTMENTS USING
(DEPARTMENT_ID)

3. The FROM clause with source table and alias is
FROM DEPARTMENTS D.
4. The LEFT OUTER JOIN clause with aliased target table is
LEFT OUTER JOIN EMPLOYEES E ON
E.DEPARTMENT_ID=D.DEPARTMENT_ID.
5. The WHERE clause is
WHERE E.DEPARTMENT_ID IS NULL.

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6. Executing this statement returns sixteen rows describing the departments to
which no employees are currently assigned as shown in the following illustration:

CERTIFICATION OBJECTIVE 7.04

Generate a Cartesian Product of Two or More Tables
A Cartesian product of two tables may be conceptualized as joining each row of the
source table with every row in the target table. The number of rows in the result set
created by a Cartesian product is equal to the number of rows in the source table
multiplied by the number of rows in the target table. Cartesian products may be
formed intentionally using the ANSI SQL:1999 cross join syntax. This technique is
described in the next section.

Creating Cartesian Products Using Cross Joins
Cartesian product is a mathematical term. It refers to the set of data created by
merging the rows from two or more tables together. Cross join is the syntax used

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343

to create a Cartesian product by joining multiple tables. Both terms are often used
synonymously. The format of the syntax for the CROSS JOIN clause is as follows:
SELECT table1.column, table2.column
FROM table1
CROSS JOIN table2;
It is important to observe that no join condition is specified using the ON or USING
keywords. A Cartesian product freely associates the rows from table1 with every row
in table2. Conditions that limit the results are permitted in the form of WHERE
clause restrictions. If table1 and table2 contain x and y number of rows, respectively,
the Cartesian product will contain x times y number of rows. The results from a
cross join may be used to identify orphan rows or generate a large data set for use in
application testing. Consider the following queries:
Query 1: select * from jobs cross join job_history;
Query 2: select * from jobs j cross join job_history jh
where j.job_id='AD_PRES';

Query 1 takes the 19 rows and 4 columns from the JOBS table and the 10 rows and
5 columns from the JOB_HISTORY table and generates one large set of 190 records
with 9 columns. SQL*Plus presents any identically named columns as headings.
SQL Developer appends an underscore and number to each shared column name
and uses it as the heading. The JOB_ID column is common to both the JOBS and
JOB_HISTORY tables. The headings in SQL Developer are labeled JOB_ID and
JOB_ID_1, respectively. Query 2 generates the
same Cartesian product as the first, but the 190
rows are constrained by the WHERE clause
condition and only 10 rows are returned.
When using the cross
Figure 7-13 shows a cross join between
join syntax, a Cartesian product is
the REGIONS and COUNTRIES tables.
intentionally generated. Inadvertent
There are 4 rows in REGIONS and 25 rows
Cartesian products are created when
in COUNTRIES. Since the WHERE clause
there are insufficient join conditions in
limits the REGIONS table to 2 of 4 rows,
a statement. Joins that specify fewer
the Cartesian product produces 50 (25 × 2)
than N-1 join conditions when joining
records. The results are sorted alphabetically,
N tables or that specify invalid join
first on the REGION_NAME and then on the
conditions may inadvertently create
COUNTRY_NAME. The first record has
Cartesian products. A pure natural join
the pair of values, Asia and Argentina. When
between two tables sharing no identically
the REGION_NAME changes, the first record
named columns results in a Cartesian join
has the pair of values, Africa and Argentina.
since two tables are joined but less than
Notice that the COUNTRY_NAME values are
one condition is available.
repeated for every REGION_NAME.

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FIGURE 7-13

The cross join

EXERCISE 7-5
Performing a Cross-Join
You are required to obtain the number of rows in the EMPLOYEES and
DEPARTMENTS table as well as the number of records that would be created

Generate a Cartesian Product of Two or More Tables

345

by a Cartesian product of these two tables. Confirm your results by explicitly
counting and multiplying the number of rows present in each of these tables.
1. Start SQL*Plus and connect to the HR schema.
2. The SELECT clause to find the number of rows in the Cartesian product is
SELECT COUNT(*).
3. The FROM clause is
FROM EMPLOYEES.
4. The Cartesian product is performed using
CROSS JOIN DEPARTMENTS.
5. Explicit counts of the rows present in the source tables are performed using
SELECT COUNT(*) FROM EMPLOYEES;
SELECT COUNT(*) FROM DEPARTMENTS;
6. Explicit multiplication of the values resulting from the previous queries may
be performed by querying the DUAL table.
7. Executing these statements reveals that there are 107 records in the
EMPLOYEES table, 27 records in the DEPARTMENTS table, and 2889 records
in the Cartesian product of these two data sets as shown in the following
illustration:

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INSIDE THE EXAM
Joining is a fundamental relational principle.
The certification objectives in this chapter
are examined using practical scenarios in
which two tables are joined. You are required
to predict the number of rows returned by a
join query or to assess whether it is syntactically correct or not. The natural join clauses
include NATURAL JOIN, JOIN…USING,
and JOIN…ON.
Remember the following simple rules. The
keywords NATURAL, USING, and ON are
mutually exclusive. They may not be used
together in the same join clause. The pure
NATURAL join takes no join conditions.
The JOIN…USING clause requires unqualified column references in join conditions,
which must appear in brackets after the
USING keyword.

Self-joins are often used for searching
through hierarchical data stored in separate
columns in the same table. It is an uncommon
join and little emphasis is placed on testing
your knowledge of self-joins in the exam.
Outer joins, however, form a significant part
of the exam content. Ensure that you have
a solid understanding of LEFT, RIGHT, and
FULL OUTER joins.
Cartesian products may be created inadvertently or intentionally using the CROSS
JOIN clause. A mistake frequently made in
the early stages of learning about joins is to
specify fewer join conditions than are necessary when joining multiple tables. This leads
to accidental Cartesian joins and is sometimes
tested in the exams. Remember that when
joining N tables, at least N-1 join conditions
are required to avoid a Cartesian join.

CERTIFICATION SUMMARY
Data stored in separate tables may be associated with each other using various types
of joins. Joins allow data to be stored in a relational manner. This prevents the need
for multiple copies of the same data across multiple tables.
Equijoins and nonequijoins are referred to as inner joins. They associate rows
from multiple tables that conform to join conditions and are specified using either
equality or inequality operators. Rows that do not conform to these join conditions,
which are ordinarily excluded by inner joins, may be retrieved with outer joins. Left,
right, and full outer joins facilitate the retrieval of orphan rows.
The ANSI SQL:1999-compliant join syntax is discussed in detail, and three forms
of the natural join are explored. Each form has a purpose, and the advantages and
risks associated with them are considered.

Certification Summary

347

Joins associate columns from multiple tables that may share the same name.
Dot notation uses a method of qualifying columns to disambiguate them. It is
accompanied by table aliasing, which is not strictly essential but helps a great deal
when formulating joins between tables with lengthy names.
The retrieval of hierarchical data stored in a single table using self-joins
is considered. N-way joins allow more than two tables to be joined, and this
generalized option is discussed. Finally, cross joins and the unique challenges
associated with them are examined.
Joining is one of the fundamental pillars of relational theory and is critical to your
successful exploitation of the full potential that SQL offers.

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Chapter 7:

✓

Displaying Data from Multiple Tables

TWO-MINUTE DRILL
Write SELECT Statements to Access Data from More Than
One Table Using Equijoins and Nonequijoins
❑ Equijoining occurs when one query fetches column values from multiple tables

in which the rows fulfill an equality-based join condition.
❑ A pure natural join is performed using the NATURAL JOIN syntax when the

source and target tables are implicitly equijoined using all identically named
columns.
❑ The JOIN…USING syntax allows a natural join to be formed on specific

columns with shared names.
❑ Dot notation refers to qualifying a column by prefixing it with its table name

and a dot or period symbol. This designates the table a column originates
from and differentiates it from identically named columns from other tables.
❑ The JOIN…ON clause allows the explicit specification of join columns

regardless of their column names. This provides a flexible joining format.
❑ The ON, USING, and NATURAL keywords are mutually exclusive and

therefore cannot appear together in a join clause.
❑ A nonequijoin is performed when the values in the join columns fulfill the join

condition based on an inequality expression.

Join a Table to Itself Using a Self-Join
❑ A self-join is required when the join columns originate from the same table.

Conceptually, the source table is duplicated and a target table is created. The
self-join then works as a regular join between two discrete tables.
❑ Storing hierarchical data in a relational table requires a minimum of two

columns per row. One column stores an identifier of the row's parent record
and the second stores the row’s identifier.

View Data That Does Not Meet a Join Condition Using Outer Joins
❑ When equijoins and nonequijoins are performed, rows from the source and

target tables are matched. These are referred to as inner joins.

Two-Minute Drill

349

❑ An outer join is performed when rows, which are not retrieved by an inner

join, are included for retrieval.
❑ A left outer join between the source and target tables returns the results of an

inner join and the missing rows it excluded from the source table.
❑ A right outer join between the source and target tables returns the results of an

inner join and the missing rows it excluded from the target table.
❑ A full outer join returns the combined results of a left outer join and right outer

join.

Generate a Cartesian Product of Two or More Tables
❑ A Cartesian product is sometimes called a cross join. It is a mathematical

term that refers to the set of data created by merging the rows from two or
more tables.
❑ The count of the rows returned from a Cartesian product is equal to the num-

ber of rows in the source table multiplied by the number of rows in the target
table.
❑ Joins that specify fewer than N-1 join conditions when joining N tables, or

that specify invalid join conditions, inadvertently create Cartesian products.

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SELF TEST
The following questions will help you measure your understanding of the material presented in this
chapter. Read all the choices carefully because there may be more than one correct answer. Choose
all the correct answers for each question.

Write SELECT Statements to Access Data from More Than One Table Using
Equijoins and Nonequijoins
1. The EMPLOYEES and DEPARTMENTS tables have two identically named columns:
DEPARTMENT_ID and MANAGER_ID. Which of these statements joins these tables based
only on common DEPARTMENT_ID values? (Choose all that apply.)
A. SELECT * FROM EMPLOYEES NATURAL JOIN DEPARTMENTS;
B. SELECT * FROM EMPLOYEES E NATURAL JOIN DEPARTMENTS D ON
E.DEPARTMENT_ID=D.DEPARTMENT_ID;
C. SELECT * FROM EMPLOYEES NATURAL JOIN DEPARTMENTS USING
(DEPARTMENT_ID);
D. None of the above
2. The EMPLOYEES and DEPARTMENTS tables have two identically named columns:
DEPARTMENT_ID and MANAGER_ID. Which statements join these tables based on both
column values? (Choose all that apply.)
A. SELECT * FROM EMPLOYEES NATURAL JOIN DEPARTMENTS;
B. SELECT * FROM EMPLOYEES JOIN DEPARTMENTS USING (DEPARTMENT_
ID,MANAGER_ID);
C. SELECT * FROM EMPLOYEES E JOIN DEPARTMENTS D ON E.DEPARTMENT_
ID=D.DEPARTMENT_ID AND E.MANAGER_ID=D.MANAGER_ID;
D. None of the above
3. Which join is performed by the following query?
SELECT E.JOB_ID,J.JOB_ID FROM EMPLOYEES E
JOIN JOBS J ON (E.SALARY < J.MAX_SALARY); (Choose the best answer.)
A. Equijoin
B. Nonequijoin
C. Cross join
D. Outer join
4. Which of the following statements are syntactically correct? (Choose all that apply.)
A. SELECT * FROM EMPLOYEES E JOIN DEPARTMENTS D USING (DEPARTMENT_ID);
B. SELECT * FROM EMPLOYEES JOIN DEPARTMENTS D USING (D.DEPARTMENT_ID);

Self Test

351

C. SELECT D.DEPARTMENT_ID FROM EMPLOYEES JOIN DEPARTMENTS D USING
(DEPARTMENT_ID);
D. None of the above
5. Which of the following statements are syntactically correct? (Choose all that apply.)
A. SELECT E.EMPLOYEE_ID, J.JOB_ID PREVIOUS_JOB, E.JOB_ID CURRENT_JOB FROM
JOB_HISTORY J CROSS JOIN EMPLOYEES E ON (J.START_DATE=E.HIRE_DATE);
B. SELECT E.EMPLOYEE_ID, J.JOB_ID PREVIOUS_JOB, E.JOB_ID CURRENT_JOB
FROM JOB_HISTORY J JOIN EMPLOYEES E ON (J.START_DATE=E.HIRE_DATE);
C. SELECT E.EMPLOYEE_ID, J.JOB_ID PREVIOUS_JOB, E.JOB_ID CURRENT_JOB
FROM JOB_HISTORY J OUTER JOIN EMPLOYEES E ON (J.START_DATE=E.
HIRE_DATE);
D. None of the above
6. Choose one correct statement regarding the following query:
SELECT * FROM EMPLOYEES E
JOIN DEPARTMENTS D ON (D.DEPARTMENT_ID=E.DEPARTMENT_ID) JOIN
LOCATIONS L ON (L.LOCATION_ID =D.LOCATION_ID);
A. Joining three tables is not permitted.
B. A Cartesian product is generated.
C. The JOIN…ON clause may be used for joins between multiple tables.
D. None of the above

Join a Table to Itself Using a Self-Join
7. How many rows are returned after executing the following statement?
SELECT * FROM REGIONS R1 JOIN REGIONS R2 ON (R1.REGION_ID=LENGTH(R2.
REGION_NAME)/2);
The REGIONS table contains the following row data. (Choose the best answer.)

A.
B.
C.
D.

REGION_ID

REGION_NAME

1

Europe

2

Americas

3

Asia

4

Middle East and Africa

2
3
4
None of the above

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View Data That Does Not Meet a Join Condition Using Outer Joins
8. Choose one correct statement regarding the following query.
SELECT C.COUNTRY_ID
FROM LOCATIONS L RIGHT OUTER JOIN COUNTRIES C
ON (L.COUNTRY_ID=C.COUNTRY_ID) WHERE L.COUNTRY_ID is NULL
A. No rows in the LOCATIONS table have the COUNTRY_ID values returned.
B. No rows in the COUNTRIES table have the COUNTRY_ID values returned.
C. The rows returned represent the COUNTRY_ID values for all the rows in the LOCATIONS
table.
D. None of the above
9. Which of the following statements are syntactically correct? (Choose all that apply.)
A. SELECT JH.JOB_ID FROM JOB_HISTORY JH RIGHT OUTER JOIN JOBS J ON
JH.JOB_ID=J.JOB_ID
B. SELECT JOB_ID FROM JOB_HISTORY JH RIGHT OUTER JOIN JOBS J ON
(JH.JOB_ID=J.JOB_ID)
C. SELECT JOB_HISTORY.JOB_ID FROM JOB_HISTORY OUTER JOIN JOBS ON
JOB_HISTORY.JOB_ID=JOBS.JOB_ID
D. None of the above

Generate a Cartesian Product of Two or More Tables
10. If the REGIONS table, which contains 4 rows, is cross joined to the COUNTRIES table, which
contains 25 rows, how many rows appear in the final results set? (Choose the best answer.)
A. 100 rows
B. 4 rows
C. 25 rows
D. None of the above

LAB QUESTION
Using SQL Developer or SQL*Plus, connect to the OE schema and complete the following tasks.
You are required to produce a report of customers who purchased products with list prices of
more than $1000. The report must contain customer first and last names and the product names
and their list prices. Customer information is stored in the CUSTOMERS table, which has the
CUSTOMER_ID column as its primary key. The product name and list price details are stored in

Lab Question

353

the PRODUCT_INFORMATION table with the PRODUCT_ID column as its primary key. Two
other related tables may assist in generating the required report: the ORDERS table, which stores
the CUSTOMER_ID and ORDER_ID information, and the ORDER_ITEMS table, which stores
the PRODUCT_ID values associated with each ORDER_ID.
There are several approaches to solving this question. Your approach may differ from the solution
listed.

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SELF TEST ANSWERS
Write SELECT Statements to Access Data from More Than One Table Using
Equijoins and Nonequijoins
✓ D. The queries in B and C incorrectly contain the NATURAL keyword. If this is removed,
1. ®
they will join the DEPARTMENTS and EMPLOYEES tables based on the DEPARTMENT_ID
column.
®
˚ A, B, and C are incorrect. A performs a pure natural join that implicitly joins the two
tables on all columns with identical names which, in this case, are DEPARTMENT_ID and
MANAGER_ID.
✓ A, B, and C. These clauses demonstrate different techniques to join the tables on both the
2. ®
DEPARTMENT_ID and MANAGER_ID columns.
®
˚ D is incorrect.
✓ B. The join condition is an expression based on the less than inequality operator. Therefore,
3. ®
this join is a nonequijoin.
®
˚ A, C, and D are incorrect. A would be correct if the operator in the join condition
expression was an equality operator. The CROSS JOIN keywords or the absence of a join
condition would result in C being true. D would be true if one of the OUTER JOIN clause was
used instead of the JOIN…ON clause.
✓ A. This statement demonstrates the correct usage of the JOIN…USING clause.
4. ®
®
˚ B, C, and D are incorrect. B is incorrect since only nonqualified column names are allowed
in the brackets after the USING keyword. C is incorrect because the column in brackets after
the USING keyword cannot be referenced with a qualifier in the SELECT clause.
✓ B demonstrates the correct usage of the JOIN…ON clause.
5. ®
®
˚ A, C, and D are incorrect. A is incorrect since the CROSS JOIN clause cannot contain
the ON keyword. C is incorrect since the OUTER JOIN keywords must be preceded by the
LEFT, RIGHT, or FULL keyword.
✓ C. The JOIN…ON clause and the other join clauses may all be used for joins between multiple
6. ®
tables. The JOIN…ON and JOIN…USING clauses are better suited for N-way table joins.
®
˚ A, B, and D are incorrect. A is false since you may join as many tables as you wish. A
Cartesian product is not created since there are two join conditions and three tables.

Join a Table to Itself Using a Self-Join
✓ B. Three rows are returned. For the row with a REGION_ID value of 2, the REGION_NAME
7. ®
is Asia and half the length of the REGION_NAME is also 2. Therefore this row is returned.

Lab Answer

355

The same logic results in the rows with REGION_ID values of three and four and
REGION_NAME values of Europe and Americas being returned.
®
˚ A, C, and D are incorrect.

View Data That Does Not Meet a Join Condition Using Outer Joins
✓ A. The right outer join fetches the COUNTRIES rows that the inner join between the
8. ®
LOCATIONS and COUNTRIES tables have excluded. The WHERE clause then restricts the
results by eliminating the inner join results. This leaves the rows from the COUNTRIES table
with which no records from the LOCATIONS table records are associated.
®
˚ B, C, and D are incorrect.
✓ A. This statement demonstrates the correct use of the RIGHT OUTER JOIN…ON clause.
9. ®
®
˚ B, C, and D are incorrect. The JOB_ID column in the SELECT clause in B is not qualified
and is therefore ambiguous since the table from which this column comes is not specified. C
uses an OUTER JOIN without the keywords LEFT, RIGHT, or FULL.

Generate a Cartesian Product of Two or More Tables
✓ A. The cross join associates every four rows from the REGIONS table 25 times with the
10. ®
rows from the COUNTRIES table yielding a result set that contains 100 rows.
®
˚ B, C, and D are incorrect.

LAB ANSWER
Using SQL Developer or SQL*Plus, connect to the OE schema, and complete the following tasks.
There are several approaches to solving this question. Your approach may differ from the following
solution listed.
1. Start SQL Developer and connect to the OE schema.
2. The SELECT list consists of four columns from two tables, which will be associated with each
other using several joins. The SELECT clause is
SELECT CUST_FIRST_NAME, CUST_LAST_NAME, PRODUCT_NAME, LIST_PRICE.
3. The FROM clause is
FROM CUSTOMERS.
4. The WHERE clause is
WHERE LIST_PRICE > 1000.
5. The JOIN clauses are interesting since the PRODUCT_INFORMATION and CUSTOMERS
tables not directly related. They are related through two other tables.

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6. The ORDERS table must first be joined to the CUSTOMERS table based on common
CUSTOMER_ID values. The first join clause following the FROM CUSTOMERS clause is
JOIN ORDERS USING (CUSTOMER_ID).
7. This set must then be joined to the ORDER_ITEMS table based on common ORDER_ID values
since the ORDER_ITEMS table can ultimately link to the PRODUCT_INFORMATION table.
The second join clause is
JOIN ORDER_ITEMS USING (ORDER_ID).
8. The missing link to join to the PRODUCT_INFORMATION table based on common
PRODUCT_ID column values is now available. The third join clause is
JOIN PRODUCT_INFORMATION USING (PRODUCT_ID).
9. Executing this statement returns the report required as shown in the following illustration:

8
Using Subqueries
to Solve Problems

CERTIFICATION OBJECTIVES
8.01

Define Subqueries

8.02

Describe the Types of Problems
That the Subqueries Can Solve

8.03

List the Types of Subqueries

8.04

✓
Q&A

Write Single-Row and Multiple-Row
Subqueries
Two-Minute Drill
Self Test

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

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Using Subqueries to Solve Problems

T

he previous six chapters have dealt with the SELECT statement in considerable detail,
but in every case the SELECT statement has been a single, self-contained command.
This chapter is the first of two that show how two or more SELECT commands can be
combined into one statement. The first technique (covered in this chapter) is the use of subqueries.
A subquery is a SELECT statement whose output is used as input to another SELECT statement
(or indeed to a DML statement, as done in Chapter 10). The second technique is the use of set
operators, where the results of several SELECT commands are combined into a single result set.

CERTIFICATION OBJECTIVE 8.01

Define Subqueries
A subquery is a query that is nested inside a SELECT, INSERT, UPDATE, or DELETE
statement or inside another subquery. A subquery can return a set of rows or just one
row to its parent query. A scalar subquery is a query that returns exactly one value:
a single row, with a single column. Scalar subqueries can be used in most places in
a SQL statement where you could use an expression or a literal value.
The places in a query where a subquery may be used are as follows:
■ In the SELECT list used for column projection
■ In the FROM clause
■ In the WHERE clause
■ In the HAVING clause

Subqueries can be nested
to an unlimited depth in a FROM clause
but to “only” 255 levels in a WHERE
clause.They can be used in the SELECT
list and in the FROM, WHERE, and
HAVING clauses of a query.

A subquery is often referred to as an inner
query, and the statement within which it occurs
is then called the outer query. There is nothing
wrong with this terminology, except that it may
imply that you can only have two levels, inner
and outer. In fact, the Oracle implementation of
subqueries does not impose any practical limits
on the level of nesting: the depth of nesting
permitted in the FROM clause of a statement
is unlimited, and that in the WHERE clause is
up to 255.

Define Subqueries

359

A subquery can have any of the usual clauses for selection and projection. The
following are required clauses:
■ A SELECT list
■ A FROM clause

The following are optional clauses:
■ WHERE
■ GROUP BY
■ HAVING

The subquery (or subqueries) within a statement must be executed before the
parent query that calls it, in order that the results of the subquery can be passed to
the parent.

EXERCISE 8-1
Types of Subqueries
In this exercise, you will write code that demonstrates the places where subqueries
can be used. Use either SQL*Plus or SQL Developer. All the queries should be run
when connected to the HR schema.
1. Log on to your database as user HR.
2. Write a query that uses subqueries in the column projection list. The query
will report on the current numbers of departments and staff:
select sysdate Today,
(select count(*) from departments) Dept_count,
(select count(*) from employees) Emp_count
from dual;

3. Write a query to identify all the employees who are managers. This will
require using a subquery in the WHERE clause to select all the employees
whose EMPLOYEE_ID appears as a MANAGER_ID:
select last_name from employees where
(employee_id in (select manager_id from employees));

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4. Write a query to identify the highest salary paid in each country. This will
require using a subquery in the FROM clause:
select max(salary),country_id from
(select salary,department_id,location_id,country_id from
employees natural join departments natural join locations)
group by country_id;

CERTIFICATION OBJECTIVE 8.02

Describe the Types of Problems
That the Subqueries Can Solve
There are many situations where you will need the result of one query as the input
for another.

Use of a Subquery Result Set for Comparison Purposes
Which employees have a salary that is less than the average salary? This could be
answered by two statements, or by a single statement with a subquery. The following
example uses two statements:
select avg(salary) from employees;
select last_name from employees where salary < result_of_previous_query ;

Alternatively, this example uses one statement with a subquery:
select last_name from employees
where salary < (select avg(salary)from employees);

In this example, the subquery is used to substitute a value into the WHERE clause
of the parent query: it is returning a single value, used for comparison with the rows
retrieved by the parent query.
The subquery could return a set of rows. For example, you could use the following to
find all departments that do actually have one or more employees assigned to them:
select department_name from departments where department_id in
(select distinct(department_id) from employees);

Describe the Types of Problems That the Subqueries Can Solve

361

In the preceding example, the subquery is used as an alternative to a join. The
same result could have been achieved with the following:
select department_name from departments inner join employees
on employees.department_id = departments.department_id
group by department_name;

If the subquery is going to return more than one row, then the comparison
operator must be able to accept multiple values. These operators are IN, NOT IN,
ANY, and ALL. If the comparison operator is EQUAL, GREATER THAN, or LESS
THAN (which each can only accept one value), the parent query will fail.
Using NOT IN is fraught with problems because of the way SQL handles
NULLs. As a general rule, do not use NOT IN unless you are certain that the
result set will not include a NULL.

Star Transformation
An extension of the use of subqueries as an alternative to a join is to enable the star
transformation often needed in data warehouse applications. Consider a large table
recording sales. Each sale is marked as being of a particular product to a particular
buyer through a particular channel. These attributes are identified by codes, used as
foreign keys to dimension tables with rows that describe each product, buyer, and
channel. To identify all sales of books to buyers in Germany through Internet orders,
one could run a query like this:
select … from sales s, products p, buyers b, channels c
where s.prod_code=p.prod_code
and s.buy_code=b.buy_code
and s.chan_code=c.chan_code
and p.product=’Books’
and b.country=’Germany’
and c.channel=’Internet’;

This query uses the WHERE clause to join the tables and then to filter the results.
The following is an alternative query that will yield the same result:
select … from sales
where prod_code in (select prod_code from products where product=’Books’)
and buy_code in (select buy_code from buyers where country=’Germany’)
and chan_code in (select chan_code from channels where channel=’Internet);

The rewrite of the first statement to the second is the star transformation. Apart
from being an inherently more elegant structure (most SQL developers with any

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Using Subqueries to Solve Problems

sense of aesthetics will agree with that), there are technical reasons why the database
may be able to execute it more efficiently than the original query. Also, star queries
are easier to maintain; it is very simple to add more dimensions to the query or to
replace the single literals (‘Books,’ ‘Germany,’ and ‘Internet’) with lists of values.
There is an instance initialization parameter, STAR_TRANSFORMATION_
ENABLED, which (if set to true) will permit the Oracle query optimizer to
re-write code into star queries.

Generate a Table from Which to SELECT
Subqueries can also be used in the FROM clause, where they are sometimes referred
to as inline views. Consider another problem based on the HR schema: employees
are assigned to a department, and departments have a location. Each location is in a
country. How can you find the average salary of staff in a country, even though they
work for different departments? Like this:
select avg(salary),country_id from
(select salary,department_id,location_id,country_id from
employees natural join departments natural join locations)
group by country_id;

The subquery constructs a table with every employee’s salary and the country in
which his department is based. The parent query then addresses this table, averaging
the SALARY and grouping by COUNTRY_ID.

Generate Values for Projection
The third place a subquery can go is in the SELECT list of a query. How can you
identify the highest salary and the highest commission rate and thus what the
maximum commission paid would be if the highest salaried employee also had the
highest commission rate? Like this, with two subqueries:
select
(select max(salary) from employees) *
(select max(commission_pct) from employees)
/ 100
from dual;

In this usage, the SELECT list used to project columns is being populated with
the results of the subqueries. A subquery used in this manner must be scalar, or the
parent query will fail with an error.

Describe the Types of Problems That the Subqueries Can Solve

363

Generate Rows to be Passed to a DML Statement
DML statements are covered in detail in Chapter 10. For now, consider these examples:
insert into sales_hist select * from sales where date > sysdate-1;
update employees set salary = (select avg(salary) from employees);
delete from departments
where department_id not in (select department_id from employees);

The first example uses a subquery to identify a
set of rows in one table that will be inserted into
another. The second example uses a subquery to
A subquery can be used
calculate the average salary of all employees and
to select rows for insertion but not in a
passes this value (a scalar quantity) to an update
VALUES clause of an INSERT statement.
statement. The third example uses a subquery to
retrieve all DEPARTMENT_IDs that are in use
and passes the list to a DELETE command, which will remove all departments that
are not in use.
Note that it is not legal to use a subquery in the VALUES clause of an insert
statement; this is fine:
insert into dates select sysdate from dual;

But this is not:
insert into dates (date_col) values (select sysdate fom dual);

EXERCISE 8-2
More Complex Subqueries
In this exercise, you will write some more complicated subqueries. Use either
SQL*Plus or SQL Developer. All the queries should be run when connected to the
HR schema.
1. Log on to your database as user HR.
2. Write a query that will identify all employees who work in departments
located in the United Kingdom. This will require three levels of nested
subqueries:
select last_name from employees where department_id in
(select department_id from departments
where location_id in
(select location_id from locations

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where country_id =
(select country_id from countries
where country_name=’United Kingdom’)
)
);

3. Check that the result from step 2 is correct by running the subqueries independently. First, find the COUNTRY_ID for the United Kingdom:
select country_id from countries where country_name=’United Kingdom’;

The result will be UK. Then find the corresponding locations:
select location_id from locations where country_id = ’UK’;

The LOCATION_IDs returned will be 2400, 2500, and 2600. Then find the
DEPARTMENT_IDs of department in these locations:
select department_id from departments where location_id in (2400,2500,2600);

The result will be two departments, 40 and 80. Finally, find the relevant employees:
select last_name from employees where department_id in (40,80);

4. Write a query to identify all the employees who earn more than the average
and who work in any of the IT departments. This will require two subqueries,
not nested:
select last_name from employees
where department_id in
(select department_id from departments where department_name like ’IT%’)
and salary > (select avg(salary) from employees);

CERTIFICATION OBJECTIVE 8.03

List the Types of Subqueries
There are three broad divisions of subquery:
■ Single-row subqueries
■ Multiple-row subqueries
■ Correlated subqueries

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365

Single- and Multiple-Row Subqueries
The single-row subquery returns one row. A special case is the scalar subquery, which
returns a single row with one column. Scalar subqueries are acceptable (and often
very useful) in virtually any situation where you could use a literal value, a constant,
or an expression. Multiple-row subqueries return sets of rows. These queries are
commonly used to generate result sets that will be passed to a DML or SELECT
statement for further processing. Both single-row and multiple-row subqueries will be
evaluated once, before the parent query is run.
Single- and multiple-row subqueries can be used in the WHERE and HAVING
clauses of the parent query, but there are restrictions on the legal comparison
operators. If the comparison operator is any of the ones in the following table, the
subquery must be a single-row subquery:
Symbol

Meaning

=

equal

>

greater than

>=

greater than or equal

<

less than

<=

less than or equal

<>

not equal

!=

not equal

If any of the operators in the preceding table are used with a subquery that returns
more than one row, the query will fail. The operators in the following table can use
multiple-row subqueries:
Symbol

Meaning

IN

equal to any member in a list

NOT IN

not equal to any member in a list

ANY

returns rows that match any value on a list

ALL

returns rows that match all the values in a list

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The comparison operators
valid for single-row subqueries are =, >, >=,
<, <=, and <>. The comparison operators

valid for multiple-row subqueries are IN,
NOT IN, ANY, and ALL.

Correlated Subqueries
A correlated subquery has a more complex method of execution than single- and
multiple-row subqueries and is potentially much more powerful. If a subquery
references columns in the parent query, then its result will be dependent on the
parent query. This makes it impossible to evaluate the subquery before evaluating
the parent query. Consider this statement, which lists all employees who earn less
than the average salary:
select last_name from employees
where salary < (select avg(salary) from employees);

The single-row subquery need only be executed once, and its result substituted
into the parent query. But now consider a query that will list all employees whose
salary is less than the average salary of their department. In this case, the subquery
must be run for each employee to determine the average salary for her department;
it is necessary to pass the employee’s department code to the subquery. This can be
done as follows:
select p.last_name, p.department_id from employees p
where p.salary < (select avg(s.salary) from employees s
where s.department_id=p.department_id);

In this example, the subquery references a column, p.department_id, from
the select list of the parent query. This is the signal that, rather than evaluating the
subquery once, it must be evaluated for every row in the parent query. To execute
the query, Oracle will look at every row in EMPLOYEES and, as it does so, run the
subquery using the DEPARTMENT_ID of the current employee row.
The flow of execution is as follows:
1. Start at the first row of the EMPLOYEES table.
2. Read the DEPARTMENT_ID and SALARY of the current row.
3. Run the subquery using the DEPARTMENT_ID from step 2.

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367

4. Compare the result of step 3 with the SALARY from step 2, and return the
row if the SALARY is less than the result.
5. Advance to the next row in the EMPLOYEES table.
6. Repeat from step 2.
A single-row or multiple-row subquery is evaluated once, before evaluating the
outer query; a correlated subquery must be evaluated once for every row in the
outer query. A correlated subquery can be single- or multiple-row, if the comparison
operator is appropriate.
Correlated subqueries can be a very inefficient construct, due to the need
for repeated execution of the subquery. Always try to find an alternative
approach.

EXERCISE 8-3
Investigate the Different Types of Subqueries
In this exercise, you will demonstrate problems that can occur with different types
of subqueries. Use either SQL*Plus or SQL Developer. All the queries should be run
when connected to the HR schema: it is assumed that the EMPLOYEES table has
the standard sets of rows.
1. Log on to your database as user HR.
2. Write a query to determine who earns more than Mr. Tobias:
select last_name from employees where
salary > (select salary from employees where last_name=’Tobias’)
order by last_name;

This will return 86 names, in alphabetical order.
3. Write a query to determine who earns more than Mr. Taylor:
select last_name from employees where
salary > (select salary from employees where last_name=’Taylor’)
order by last_name;

This will fail with the error “ORA-01427: single-row subquery returns more than
one row.” The following illustration shows the last few lines of the output from
step 2 followed by step 3 and the error, executed with SQL*Plus.

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4. Determine why the query in step 2 succeeded but failed in step 3. The answer
lies in the state of the data:
select count(last_name) from employees where last_name=’Tobias’;
select count(last_name) from employees where last_name=’Taylor’;

The use of the “greater than” operator in the queries for steps 2 and 3 requires
a single-row subquery, but the subquery used may return any number of rows,
depending on the search predicate used.
5. Fix the code in steps 2 and 3 so that the statements will succeed no matter
what LAST_NAME is used. There are two possible solutions: one uses a
different comparison operator that can handle a multiple-row subquery; the
other uses a subquery that will always be single-row.
The first solution:
select last_name from employees where
salary > all (select salary from employees where last_name=’Taylor’)
order by last_name;

The second solution:
select last_name from employees where
salary > (select max(salary) from employees where last_name=’Taylor’)
order by last_name;

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369

SCENARIO & SOLUTION
How can you best design subqueries such that they
will not fail with “ORA-01427: single-row subquery
returns more than one row” errors?

There are two common techniques: use an aggregation
so that if you do get multiple rows they will be reduced
to one, or use one of the IN, ANY, or ALL operators
so that it won’t matter if multiple rows are returned.
But these are both hacker’s solutions; the real answer
is always to use the primary key when identifying the
row to be returned, not a nonunique key.

Sometimes there is a choice between using a
subquery or using some other technique: the star
transformation is a case in point. Which is better?

It depends on the circumstances. It is not uncommon
for the different techniques to cause a different
execution method within the database. Depending
on how the instance, the database, and the data
structures within it are configured, one may be much
more efficient than another. Whenever such a choice
arises, the statements should be subjected to a tuning
analysis. Your DBA will be able to advise on this.

CERTIFICATION OBJECTIVE 8.04

Write Single-Row and Multiple-Row Subqueries
Following are examples of single- and multiple-row subqueries. They are based on
the HR demonstration schema.
How would you figure out which employees have a manager who works for
a department based in the United Kingdom? This is a possible solution, using
multiple-row subqueries:
select last_name from employees
where manager_id in
(select employee_id from employees where department_id in
(select department_id from departments where location_id in
(select location_id from locations where country_id=’UK’)));

In the preceding example, subqueries are nested three levels deep. Note that the
subqueries use the IN operator because it is possible that the queries could return
several rows.

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You have been asked to find the job with the highest average salary. This can be
done with a single-row subquery:
select job_title from jobs natural join employees group by job_title
having avg(salary) =
(select max(avg(salary)) from employees group by job_id);

The subquery returns a single value: the average salary of the department with the
highest average salary. It is safe to use the equality operator for this subquery because
the MAX function guarantees that only one row will be returned.
The ANY and ALL operators are supported syntax, but their function can be
duplicated with other more commonly used operators combined with aggregations.
For example, these two statements, which retrieve all employees whose salary is
above that of anyone in department 80, will return identical result sets:
select last_name from employees where salary > all
(select salary from employees where department_id=80);
select last_name from employees where salary >
(select max(salary) from employees where department_id=80);

The following table summarizes the equivalents for ANY and ALL:
Operator

Meaning

< ANY

less than the highest

> ANY

more than the lowest

= ANY

equivalent to IN

> ALL

more than the highest

< ALL

less than the lowest

EXERCISE 8-4
Write a Query That Is Reliable and User Friendly
In this exercise, develop a multi-row subquery that will prompt for user input. Use
either SQL*Plus or SQL Developer. All the queries should be run when connected
to the HR schema; it is assumed that the tables have the standard sets of rows.
1. Log on to your database as user HR.
2. Design a query that will prompt for a department name and list the last name
of every employee in that department:

Write Single-Row and Multiple-Row Subqueries

371

select last_name from employees where department_id =
(select department_id from departments
where department_name = ’&Department_name’);

3. Run the query in step 2 three times, when prompted supplying these values:
first time, Executive
second time, executive
third time, Executiv
The following illustration shows the result, using SQL*Plus:

4. Note the results from step 3. The first run succeeded because the value
entered was an exact match, but the other failed. Adjust the query to make it
more user friendly, so that it can handle minor variations in case or spelling:
select last_name from employees where department_id =
(select department_id from departments
where upper(department_name) like upper(’%&Department_name%’));

5. Run the query in step 4 three times, using the same values as used in step 3.
This time, the query will execute successfully.
6. Run the query in step 4 again, and this time enter the value Pu. The query
will fail, with an “ORA-01427: single-row subquery returns more than one
row” error, because the attempt to make it more user-friendly means that the
subquery is no longer guaranteed to be a single-row subquery. The string Pu
matches two departments.

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7. Adjust the query to make it resilient against the ORA-01427 error, and adjust
the output to prevent any possible confusion:
select last_name,department_name from employees join departments
on employees.department_id = departments.department_id
where departments.department_id in
(select department_id from departments
where upper(department_name) like upper(’%&Department_name%’));

The following illustration shows this final step: code that is approaching the ideal of
being both bullet proof and user friendly:

INSIDE THE EXAM
Use of Subqueries
Subqueries come in three general forms:
single-row, multiple-row, and correlated.
A special case of the single-row subquery is
the scalar subquery, a subquery that returns
exactly one value. This is a single-row singlecolumn subquery. For the first SQL OCP
exam, detailed knowledge is expected only
of scalar subqueries and single-column

multiple-row subqueries. Correlated subqueries
and multiple column subqueries are unlikely
to be examined at this level, but a general
knowledge of them may be tested.
When using subqueries in a WHERE clause,
you must be aware of which operators will
succeed with single-row subqueries and which
will succeed with multiple-row subqueries.

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373

CERTIFICATION SUMMARY
A subquery is a query embedded within another SQL statement. This statement can
be another query or a DML statement. Subqueries can be nested within each other
with no practical limits.
Subqueries can be used to generate values for the select list of a query to generate
an inline view to be used in the FROM clause, in the WHERE clause, and in the
HAVING clause. When used in the WHERE or HAVING clauses, single-row
subqueries can be used with these comparison operators: =, >, >=, <, <=, <>;
multiple-row subqueries can be used with these comparison operators: IN, NOT IN,
ANY, ALL.

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✓

Using Subqueries to Solve Problems

TWO-MINUTE DRILL
Define Subqueries
❑ A subquery is a select statement embedded within another SQL statement.
❑ Subqueries can be nested within each other.
❑ With the exception of the correlated subquery, subqueries are executed before

the outer query within which they are embedded.

Describe the Types of Problems That the Subqueries Can Solve
❑ Selecting rows from a table with a condition that depends on the data within

the table can be implemented with a subquery.
❑ Complex joins can sometimes be replaced with subqueries.
❑ Subqueries can add values to the outer query’s output that are not available in

the tables the outer query addresses.

List the Types of Subqueries
❑ Multiple-row subqueries can return several rows, possibly with several columns.
❑ Single-row subqueries return one row, possibly with several columns.
❑ A scalar subquery returns a single value; it is a single-row, single-column

subquery.
❑ A correlated subquery is executed once for every row in the outer query.

Write Single-Row and Multiple-Row Subqueries
❑ Single-row subqueries should be used with single-row comparison operators.
❑ Multiple-row subqueries should be used with multiple-row comparison operators.
❑ The ALL and ANY operators can be alternatives to use of aggregations.

Self Test

375

SELF TEST
Define Subqueries
1. Consider this generic description of a SELECT statement:
SELECT select_list
FROM table
WHERE condition
GROUP BY expression_1
HAVING expression_2
ORDER BY expression_3 ;
Where could subqueries be used? (Choose all correct answers.)
A. select_list
B. table
C. condition
D. expression_1
E. expression_2
F. expression_3
2. A query can have a subquery embedded within it. Under what circumstances could there be
more than one subquery? (Choose the best answer.)
A. The outer query can include an inner query. It is not possible to have another query within
the inner query.
B. It is possible to embed a single-row subquery inside a multiple-row subquery, but not the
other way around.
C. The outer query can have multiple inner queries, but they must not be embedded within
each other.
D. Subqueries can be embedded within each other with no practical limitations on depth.
3. Consider this statement:
select employee_id, last_name from employees where
salary > (select avg(salary) from employees);

When will the subquery be executed? (Choose the best answer.)
A. It will be executed before the outer query.
B. It will be executed after the outer query.

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C. It will be executed concurrently with the outer query.
D. It will be executed once for every row in the EMPLOYEES table.
4. Consider this statement:
select o.employee_id, o.last_name from employees o where
o.salary > (select avg(i.salary) from employees i
where i.department_id=o.department_id);

When will the subquery be executed? (Choose the best answer.)
A. It will be executed before the outer query.
B. It will be executed after the outer query.
C. It will be executed concurrently with the outer query.
D. It will be executed once for every row in the EMPLOYEES table.

Describe the Types of Problems That the Subqueries Can Solve
5. Consider the following statement:
select last_name from employees join departments
on employees.department_id = departments.department_id
where department_name=’Executive’;

and this statement:
select last_name from employees where department_id in
(select department_id from departments where department_name=’Executive’);

What can be said about the two statements? (Choose two correct answers.)
A. The two statements should generate the same result.
B. The two statements could generate different results.
C. The first statement will always run successfully; the second statement will error if there are
two departments with DEPARTMENT_NAME ‘Executive.’
D. Both statements will always run successfully, even if there are two departments with
DEPARTMENT_NAME ‘Executive.’

List the Types of Subqueries
6. What are the distinguishing characteristics of a scalar subquery? (Choose two correct answers.)
A. A scalar subquery returns one row.
B. A scalar subquery returns one column.

Self Test

377

C. A scalar subquery cannot be used in the SELECT LIST of the parent query.
D. A scalar subquery cannot be used as a correlated subquery.
7. Which comparison operator cannot be used with multiple-row subqueries?
(Choose the best answer.)
A. ALL
B. ANY
C. IN
D. NOT IN
E. All the above can be used.

Write Single-Row and Multiple-Row Subqueries
8. Consider this statement:
select last_name, (select count(*) from departments) from employees
where salary = (select salary from employees);

What is wrong with it? (Choose the best answer.)
A. Nothing is wrong—the statement should run without error.
B. The statement will fail because the subquery in the SELECT list references a table that is
not listed in the FROM clause.
C. The statement will fail if the second query returns more than one row.
D. The statement will run but is extremely inefficient because of the need to run the second
subquery once for every row in EMPLOYEES.
9. Which of the following statements are equivalent? (Choose two answers.)
A. select employee_id from employees where salary < all (select salary from employees where
department_id=10);
B. select employee_id from employees where salary < (select min(salary) from employees
where department_id=10);
C. select employee_id from employees where salary not >= any (select salary from employees
where department_id=10);
D. select employee_id from employees e join departments d on e.department_id=
d.department_id where e.salary < (select min(salary) from employees) and
d.department_id=10;

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10. Consider this statement, which is intended to prompt for an employee’s name and then find all
employees who have the same job as the first employee:
select last_name,employee_id from employees where job_id =
(select job_id from employees where last_name = ’&Name’);

What would happen if a value were given for &Name that did not match with any row in
EMPLOYEES? (Choose the best answer.)
A. The statement would fail with an error.
B. The statement would return every row in the table.
C. The statement would return no rows.
D. The statement would return all rows where JOB_ID is NULL.

LAB QUESTION
Exercise 8-3 included this query that attempted to find all employees whose salary is higher than that
of a nominated employee:
select last_name from employees where
salary > (select salary from employees where last_name=’Taylor’)
order by last_name;

The query runs successfully if last_name is unique. Two variations were given that will run
without error no matter what value is provided.
The first solution was as follows:
select last_name from employees where
salary > all (select salary from employees where last_name=’Taylor’)
order by last_name;

The second solution was as follows:
select last_name from employees where
salary > (select max(salary) from employees where last_name=’Taylor’)
order by last_name;

There are other queries that will run successfully; construct two other solutions, one using the
ANY comparison operator, the other using the MIN aggregation function. Now that you have four
solutions, do they all give the same result?
All these “solutions” are in fact just ways of avoiding error. They do not necessarily give the result
the user wants, and they may not be consistent. What change needs to be made to give a consistent,
unambiguous, result?

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379

SELF TEST ANSWERS
Define Subqueries
✓ A, B, C, D, E. Subqueries can be used at all these points.
1. ®
®
˚ F. A subquery cannot be used in the ORDER BY clause of a query.
✓ D. Subquery nesting can be done to many levels.
2. ®
®
˚ A, B, and C. A and C are wrong because subqueries can be nested. B is wrong because the
number of rows returned is not relevant to nesting subqueries, only to the operators being used.
✓ A. The result set of the inner query is needed before the outer query can run.
3. ®
®
˚ B, C, and D. B and C are not possible because the result of the subquery is needed before
the parent query can start. D is wrong because the subquery is only run once.
✓ D. This is a correlated subquery, which must be run for every row in the table.
4. ®
®
˚ A, B, and C. The result of the inner query is dependent on a value from the outer query; it
must therefore be run once for every row.

Describe the Types of Problems That the Subqueries Can Solve
✓ A, D. The two statements will deliver the same result, and neither will fail if the name
5. ®
is duplicated.
®
˚ B, C. B is wrong because the statements are functionally identical, though syntactically
different. C is wrong because the comparison operator used, IN, can handle a multiple-row subquery.

List the Types of Subqueries
✓ A, B. A scalar subquery can be defined as a query that returns a single value.
6. ®
®
˚ C, D. C is wrong because a scalar subquery is the only subquery that can be used in the
SELECT LIST. D is wrong because scalar subqueries can be correlated.
✓ E. ALL, ANY, IN, and NOT IN are the multiple-row comparison operators.
7. ®
®
˚ A, B, C, D. All of these can be used.

Write Single-Row and Multiple-Row Subqueries
✓ C. The equality operator requires a single-row subquery, and the second subquery could
8. ®
return several rows.
®
˚ A, B, D. A is wrong because the statement will fail in all circumstances except the unlikely
case where there is zero or one employees. B is wrong because this is not a problem; there
need be no relationship between the source of data for the inner and outer queries. D is wrong
because the subquery will only run once; it is not a correlated subquery.

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✓ A and B are identical.
9. ®
®
˚ C is logically the same as A and B but syntactically is not possible; it will give an error.
D will always return no rows, because it asks for all employees who have a salary lower than all
employees. This is not an error but can never return any rows. The filter on DEPARTMENTS is
not relevant.
✓ C. If a subquery returns NULL, then the comparison will also return NULL, meaning that
10. ®
no rows will be retrieved.
®
˚ A, B, D. A is wrong because this would not cause an error. B is wrong because a comparison
with NULL will return nothing, not everything. D is wrong because a comparison with NULL
can never return anything, not even other NULLs.

LAB ANSWER
The following are two possible solutions using ANY and MIN:
select last_name from employees where
salary > any (select salary from employees where last_name=’Taylor’)
order by last_name;
select last_name from employees where
salary not < (select min(salary) from employees where last_name=’Taylor’)
order by last_name;

These are just as valid as the solutions presented earlier that used ALL and MAX, but they do not
give the same result. There is no way to say that these are better or worse than the earlier solutions.
The problem is that the subquery is based on a column that is not the primary key. It would not be
unreasonable to say that all these solutions are wrong, and the original query is the best; it gives
a result that is unambiguously correct if the LAST_NAME is unique, and if LAST_NAME is not
unique, it throws an error rather than giving a questionable answer. The real answer is that the query
should be based on EMPLOYEE_ID, not LAST_NAME.

9
Using the Set
Operators

CERTIFICATION OBJECTIVES
9.01

Describe the Set Operators

9.02

Use a Set Operator to Combine Multiple
Queries into a Single Query

9.03

✓
Q&A

Control the Order of Rows Returned
Two-Minute Drill
Self Test

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

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Using the Set Operators

A

ny SQL query can be described mathematically (relational algebra is the term used
for this), but the use of set operators is what recalls school mathematics to many
programmers. As your math curriculum may not have included set theory, following is
a short summary.
All SELECT statements return a set of rows. The set operators take as their input
the results of two or more SELECT statements and from these generate a single
result set. This is known as a compound query. Oracle provides three set operators:
UNION, INTERSECT, and MINUS. UNION can be qualified with ALL. There
is a significant deviation from the ISO standard for SQL here, in that ISO SQL
uses EXCEPT where Oracle uses MINUS, but the functionality is identical. The
difference can be important when porting applications (or skills) developed for a
third-party database to the Oracle environment.

CERTIFICATION OBJECTIVE 9.01

Describe the Set Operators
The set operators used in compound queries are as follows:
Returns the combined rows from two queries, sorting them and
removing duplicates.

■ UNION

Returns the combined rows from two queries without sorting
or removing duplicates.

■ UNION ALL

Returns only the rows that occur in both queries’ result sets,
sorting them and removing duplicates.

■ INTERSECT

Returns only the rows in the first result set that do not appear in
the second result set, sorting them and removing duplicates.

■ MINUS

These commands are equivalent to the standard operators used in mathematics
set theory, often depicted graphically as Venn diagrams.

Describe the Set Operators

383

Sets and Venn Diagrams
Consider groupings of living creatures, classified as follows:
■ Creatures with two legs
■ Creatures that can fly
■ Creatures with fur

Humans, parrots, bats
Parrots, bats, bees

Bears, bats

Each classification is known as a set, and each member of the set is an element.
The union of the three sets is humans, parrots, bats, bees, and bears. This is all the
elements in all the sets, without the duplications. The intersection of the sets is all
elements that are common to all three sets, again removing the duplicates. In this
simple example, the intersection has just one element: bats. The intersection of the
two-legged set and the flying set has two elements: parrots and bats. The minus of
the sets is the elements of one set without the elements of another, so the two-legged
creatures set, minus the flying creatures set, minus the furry creatures set, results in a
single element: humans.
These sets can be represented graphically as the Venn diagram shown in Figure 9-1.
(Venn diagrams are named after John Venn, who formalized the theory at Cambridge
University in the nineteenth century.)
The circle in the top left of the figure represents the set of two-legged creatures;
the circle top right is creatures that can fly; the bottom circle is furry animals. The
unions, intersections, and minuses of the sets are immediately apparent by observing
the elements in the various parts of the circles that do or do not overlap. The diagram
FIGURE 9-1

A Venn diagram,
showing three
sets and the
universal set

HUMANS

PARROTS

BEES

BATS

BEARS

FISH

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in the figure also includes the universal set, represented by the rectangle. The universal
set is all elements that exist but are not members of the defined sets. In this case, the
universal set would be defined as all living creatures that evolved without developing
fur, two legs, or the ability to fly (such as fish).
That’s enough school maths; now proceed to the implementation within SQL.

Set Operator General Principles
All set operators make compound queries by combining the result sets from two
or more queries. If a SELECT statement includes more than one set operator (and
therefore more than two queries), they will be applied in the order the programmer
specifies: top to bottom and left to right. Although pending enhancements to ISO
SQL will give INTERSECT a higher priority than the others, there is currently
no priority of one operator over another. To override this precedence, based on
the order in which the operators appear, you can use parentheses: operators within
brackets will be evaluated before passing the results to operators outside the brackets.
Given the pending change in operator priority, it may be good practice always
to use parentheses.This will ensure that the code’s function won’t change
when run against a later version of the database.
Each query in a compound query will project its own list of selected columns.
These lists must have the same number of elements, be nominated in the same
sequence, and be of broadly similar data type. They do not have to have the same
names (or column aliases), nor do they need to come from the same tables (or
subqueries). If the column names (or aliases) are different, the result set of the
compound query will have columns named as they were in the first query.
While the selected column lists do not have
to be exactly the same data type, they must be
from the same data type group. For example,
the columns selected by one query could be of
The columns in the queries
data types DATE and NUMBER, and those
that make up a compound query can have
from the second query could be TIMESTAMP
different names, but the output result set
and INTEGER. The result set of the compound
will use the names of the columns in the
query will have columns with the higher
first query.
level of precision: in this case, they would be
TIMESTAMP and NUMBER. Other than

Describe the Set Operators

385

accepting data types from the same group, the set
operators will not do any implicit type casting.
If the second query retrieved columns of type
VARCHAR2, the compound query would throw
The corresponding
an error even if the string variables could be
columns in the queries that make up a
resolved to legitimate date and numeric values.
compound query must be of the same
UNION, MINUS, and INTERSECT will
data type group.
always combine the results sets of the input
queries, then sort the results to remove duplicate rows. The sorting is based on
all the columns, from left to right. If all the columns in two rows have the same
value, then only the first row is returned in the
compound result set. A side effect of this is that
the output of a compound query will be sorted.
If the sort order (which is ascending, based
A compound query will
on the order in which the columns happen to
by default return rows sorted across all
appear in the select lists) is not the order you
the columns, from left to right.The only
want, it is possible to put a single ORDER BY
exception is UNION ALL, where the rows
clause at the end of the compound query. It is
will not be sorted.The only place where
not possible to use ORDER BY in any of the
an ORDER BY clause is permitted is at
queries that make up the whole compound
the end of the compound query.
query, as this would disrupt the sorting that is
necessary to remove duplicates.
UNION ALL is the exception to the sorting-no-duplicates rule: the result sets of the
two input queries will be concatenated to form the result of the compound query. But
you still can’t use ORDER BY in the individual queries; it can only appear at the end of
the compound query where it will be applied to the complete result set.

EXERCISE 9-1
Describe the Set Operators
In this exercise, you will see the effect of the set operators. Either SQL*Plus or SQL
Developer can be used.
1. Connect to your database as user HR.
2. Run this query:
select region_name from regions;

Note the result, in particular the order of the rows. If the table is as originally
created, there will be four rows returned. The order will be Europe, America,
Asia, Middle East.

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3. Query the Regions table twice, using UNION:
select region_name from regions union select region_name from
regions;

The rows returned will be as for step 1 but sorted alphabetically.
4. This time, use UNION ALL:
select region_name from regions union all select region_name
from regions;

There will be double the number of rows, and they will not be sorted.
5. An intersection will retrieve rows common to two queries:
select region_name from regions intersect select region_name
from regions;

All four rows are common, and the result is sorted.
6. A MINUS will remove common rows:
select region_name from regions minus select region_name from
regions;

The second query will remove all the rows in the first query. Result: no rows left.

CERTIFICATION OBJECTIVE 9.02

Use a Set Operator to Combine Multiple
Queries into a Single Query
Compound queries are two or more queries, linked with one or more set operators.
The end result is a single result set.
The examples that follow are based on two tables, OLD_DEPT and NEW_DEPT.
The table OLD_DEPT is intended to represent a table created with an earlier version
of Oracle, when the only data type available for representing date and time data was
DATE, the only option for numeric data was NUMBER, and character data was
fixed-length CHAR. The table NEW_DEPT uses the more closely defined INTEGER
numeric data type (which Oracle implements as a NUMBER of up to 38 significant
digits but no decimal places), the more space-efficient VARCHAR2 for character

Use a Set Operator to Combine Multiple Queries into a Single Query

387

data, and the TIMESTAMP data type, which can by default store date and time values
with six decimals of precision on the seconds. There are two rows in each table.

The UNION ALL Operator
A UNION ALL takes two result sets and concatenates them together into a single
result set. The result sets come from two queries that must select the same number of
columns, and the corresponding columns of the two queries (in the order in which
they are specified) must be of the same data type group. The columns do not have to
have the same names.
Figure 9-2 demonstrates a UNION ALL operation from two tables. The UNION
ALL of the two tables converts all the values to the higher level of precision: the
dates are returned as timestamps (the less precise DATEs padded with zeros), the
character data is the more efficient VARCHAR2 with the length of the longer
input column, and the numbers (though this is not obvious due to the nature of the
data) will accept decimals. The order of the rows is the rows from the first table, in
whatever order they happen to be stored, followed by the rows from the second table
in whatever order they happen to be stored.

The UNION Operator
A UNION performs a UNION ALL and then sorts the result across all the columns
and removes duplicates. The first query in Figure 9-3 returns all four rows because there
are no duplicates. However, the rows are now in order. It may appear that the first two
rows are not in order because of the values in DATED, but they are: the DNAME
FIGURE 9-2

A UNION ALL
with data type
conversions

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in the table OLD_DEPTS is 20 bytes long
(padded with spaces), whereas the DNAME in
NEW_DEPTS, where it is a VARCHAR2, is only
A UNION ALL will return
as long as the name itself. The spaces give the
rows grouped from each query in their
row from OLD_DEPT a higher sort value, even
natural order. This behavior can be
though the date value is less.
modified by placing a single ORDER BY
The second query in Figure 9-3 removes any
clause at the end.
leading or trailing spaces from the DNAME
columns and chops off the time elements from
DATED and STARTED. Two of the rows thus become identical, and so only one
appears in the output.
Because of the sort, the order of the queries in a UNION compound query makes
no difference to the order of the rows returned.
If you know that there can be no duplicates between two tables, then always
use UNION ALL. It saves the database from doing a lot of sorting. Your DBA
will not be pleased with you if you use UNION unnecessarily.

The INTERSECT Operator
The intersection of two sets is the rows that are common to both sets, as shown in
Figure 9-4.
The first query shown in Figure 9-4 returns no rows, because every row in the
two tables is different. Next, applying functions to eliminate some of the differences
returns the one common row. In this case, only one row is returned; had there been
several common rows, they would be in order. The order in which the queries appear
in the compound query has no effect on this.
FIGURE 9-3

UNION
compound
queries

Use a Set Operator to Combine Multiple Queries into a Single Query

389

FIGURE 9-4

INTERSECT and
MINUS

The MINUS Operator
A MINUS runs both queries, sorts the results, and returns only the rows from the
first result set that do not appear in the second result set.
The third query in Figure 9-4 returns all the rows in OLD_DEPT because there
are no matching rows in NEW_DEPT. The last query forces some commonality,
causing one of the rows to removed. Because of the sort, the rows will be in order
irrespective of the order in which the queries appear in the compound query.

INSIDE THE EXAM
Using the Set Operators
A compound query is one query made up of
several queries, but they are not subqueries.
A subquery generates a result set that is used
by another query; the queries in a compound
query run independently, and a separate
stage of execution combines the result sets.

This combining operation is accomplished
by sorting the result sets and merging them
together. There is an exception to this:
UNION ALL does no processing after
running the two queries; it simply lists the
results of each.

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More Complex Examples
If two queries do not return the same number of columns, it may still be possible to
run them in a compound query by generating additional columns with NULL values.
For example, consider a classification system for animals: all animals have a name
and a weight, but the birds have a wingspan whereas the cats have a tail length. A
query to list all the birds and cats might be:
select name,tail_length,to_char(null) from cats
union all
select name,to_char(null),wing_span from birds;

Note the use of TO_CHAR(NULL) to generate the missing values.
A compound query can consist of more than two queries, in which case operator
precedence can be controlled with parentheses. Without parentheses, the set
operators will be applied in the sequence in which they are specified. Consider
the situation where there is a table PERMSTAFF with a listing of all permanent
staff members and a table CONSULTANTS with a listing of consultant staff.
There is also a table BLACKLIST of people blacklisted for one reason or another.
The following query will list all the permanent and consulting staff in a certain
geographical, removing those on the blacklist:
select name from permstaff where location = 'Germany'
union all
select name from consultants where work_area = 'Western Europe'
minus
select name from blacklist;

Note the use of UNION ALL, because is assumed that no one will be in both
the PERMSTAFF and the CONSULTANTS tables; a UNION would force an
unnecessary sort. The order of precedence for set operators is the order specified by
the programmer, so the MINUS operation will compare the BLACKLIST with the
result of the UNION ALL. The result will be all staff (permanent and consulting)
who do not appear on the blacklist. If the blacklisting could be applied only to
consulting staff and not to permanent staff, there would be two possibilities. First,
the queries could be listed in a different order:
select name from consultants where work_area = 'Western Europe'
minus
select name from blacklist
union all
select name from permstaff where location = 'Germany';

Use a Set Operator to Combine Multiple Queries into a Single Query

391

This would return consultants who are not blacklisted and then append all
permanent staff. Alternatively, parentheses could control the precedence explicitly:
select name from permstaff where location = 'Germany'
union all
(select name from consultants where work_area = 'Western Europe'
minus
select name from blacklist);

This query will list all permanent staff and then append all consultant staff who
are not blacklisted.
These two queries will return the same rows, but the order will be different
because the UNION ALL operations list the PERMSTAFF and CONSULTANTS
tables in a different sequence. To ensure that the queries return identical result sets,
there would need to be an ORDER BY clause at the foot of the compound queries.
The two preceding queries will return the same rows, but the second version
could be considered better code because the parentheses make it more selfdocumenting. Furthermore, relying on implicit precedence based on the order
of the queries works at the moment, but future releases of SQL may include
set operator precedence.

SCENARIO & SOLUTION
How can you present several tables
with similar data as one table?

This is a common problem, often caused by bad systems analysis
or perhaps by attempts to integrate systems together. Compound
queries are often the answer. By using type casting functions to
force columns to the same data type and TO_CHAR(NULL) to
generate missing columns, you can present the data as though it
were from one table.

Are there performance issues with
compound queries?

Perhaps. With the exception of UNION ALL, compound queries
have to sort data, across the full width of the rows. This may be
expensive in both memory and CPU. Also, if the two queries both
address the same table, there will be two passes through the data
as each query is run independently; if the same result could be
achieved with one (though probably more complicated) query, this
would usually be a faster solution. Compound queries are a powerful
tool but should be used with caution.

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EXERCISE 9-2
Using the Set Operators
In this exercise, you will run more complex compound queries.
1. Connect to your database as user HR.
2. Run this query to count the employees in three departments:
select department_id, count(1) from employees
where department_id in (20,30,40)
group by department_id;

3. Obtain the same result with a compound query:
select 20,count(1) from employees where department_id=20
union all
select 30,count(1) from employees where department_id=30
union all
select 40,count(1) from employees where department_id=40;

4. Find out if any managers manage staff in both departments 20 and 30, and
exclude any managers with staff in department 40:
select manager_id from employees where department_id=20
intersect
select manager_id from employees where department_id=30
minus
select manager_id from employees where department_id=40;

5. Use a compound query to report salaries subtotaled by department, by manager,
and the overall total:
select department_id,to_number(null),sum(salary) from employees
group by department_id
union
select to_number(null),manager_id,sum(salary) from employees
group by manager_id
union all
select to_number(null),to_number(null),sum(salary) from
employees;

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393

CERTIFICATION OBJECTIVE 9.03

Control the Order of Rows Returned
By default, the output of a UNION ALL compound query is not sorted at all: the
rows will be returned in groups in the order of which query was listed first and within
the groups in the order that they happen to be stored. The output of any other set
operator will be sorted in ascending order of all the columns, starting with the first
column named.
It is not syntactically possible to use an ORDER BY clause in the individual
queries that make up a compound query. This is because the execution of most
compound queries has to sort the rows, which would conflict with the ORDER BY.
It might seem theoretically possible that a UNION ALL (which does not sort the
rows) could take an ORDER BY for each query, but the Oracle implementation of
UNION ALL does not permit this.
There is no problem with placing an ORDER BY clause at the end of the
compound query, however. This will sort the entire output of the compound query.
The default sorting of rows is based on all the columns in the sequence they appear.
A specified ORDER BY clause has no restrictions: it can be based on any columns
(and functions applied to columns) in any order. For example:
SQL>
2
3
4

select deptno,trim(dname) name from old_dept
union
select dept_id,dname from new_dept
order by name;
DEPTNO NAME
---------- -------------------10 Accounts
30 Admin
20 Support

Note that the column names in the ORDER BY clause must be the name(s) (or,
in this case, the alias) of the columns in the first query of the compound query.

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EXERCISE 9-3
Control the Order of Rows Returned
In this exercise, you will tidy up the result of the final step in Exercise 2. This step
produced a listing of salaries totaled by department and then by manager, but the
results were not very well formatted.
1. Connect to your database as user HR.
2. Generate better column headings for the query:
select department_id dept,to_number(null) mgr,sum(salary)
from employees
group by department_id
union all
select to_number(null),manager_id,sum(salary) from employees
group by manager_id
union all
select to_number(null),to_number(null),sum(salary) from
employees;

3. Attempt to sort the results of the queries that subtotal by using UNION
instead of UNION ALL:
select department_id dept,to_number(null) mgr,sum(salary)
from employees
group by department_id
union
select to_number(null),manager_id,sum(salary) from employees
group by manager_id
union all
select to_number(null),to_number(null),sum(salary) from
employees;

This would be fine, except that the subtotals for staff without a department or
a manager are placed at the bottom of the output above the grand total, not
within the sections for departments and managers.
4. Generate a value to replace the NULLs for department and manager codes
and for the overall total:
select 20,count(1) from employees where department_id=20
union all
select 30,count(1) from employees where department_id=30
union all
select 40,count(1) from employees where department_id=40;

Certification Summary

395

5. Find out if any managers manage staff in both departments 20 and 30, and
exclude any managers with staff in department 40:
select manager_id from employees where department_id=20
intersect
select manager_id from employees where department_id=30
minus
select manager_id from employees where department_id=40;

6. Use a compound query to report salaries subtotaled by department, by manager,
and the overall total:
select department_id dept,to_number(null) mgr,sum(salary)
from employees
group by department_id
union
select to_number(null),manager_id,sum(salary) from employees
group by manager_id
union all
select to_number(null),to_number(null),sum(salary) from
employees;

CERTIFICATION SUMMARY
The set operators combine the result sets from two or more queries into one result
set. A query that uses a set operator is a compound query. The set operators are
UNION, UNION ALL, INTERSECT, and MINUS. They have equal precedence,
and if more than one is included in a compound query they will be executed in the
order in which they occur—though this can be controlled by using parentheses. All
the set operators except for UNION ALL rely on sorting to merge result sets and
remove duplicate rows.
The queries in a compound query must return the same number of columns.
The corresponding columns in each query must be of compatible data types. The
queries can use features of the SELECT statement with the exception of ORDER
BY; it is, however, permissible to place a single ORDER BY clause at the end of the
compound query.

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✓

Using the Set Operators

TWO-MINUTE DRILL
Describe the Set Operators
❑ UNION ALL concatenates the results of two queries.
❑ UNION sorts the results of two queries and removes duplicates.
❑ INTERSECT returns only the rows common to the result of two queries.
❑ MINUS returns the rows from the first query that do not exist in the second

query.

Use a Set Operator to Combine Multiple Queries
into a Single Query
❑ The queries in the compound query must return the same number of columns.
❑ The corresponding columns must be of compatible data type.
❑ The set operators have equal precedence and will be applied in the order they

are specified.

Control the Order of Rows Returned
❑ It is not possible to use ORDER BY in the individual queries that make a

compound query.
❑ An ORDER BY clause can be appended to the end of a compound query.
❑ The rows returned by a UNION ALL will be in the order they occur in the

two source queries.
❑ The rows returned by a UNION will be sorted across all their columns, left

to right.

Self Test

397

SELF TEST
Describe the Set Operators
1. Which of these set operators will not sort the rows? (Choose the best answer.)
A. INTERSECT
B. MINUS
C. UNION
D. UNION ALL
2. Which of these operators will remove duplicate rows from the final result? (Choose all that apply.)
A. INTERSECT
B. MINUS
C. UNION
D. UNION ALL

Use a Set Operator to Combine Multiple Queries into a Single Query
3. If a compound query contains both a MINUS and an INTERSECT operator, which will be
applied first? (Choose the best answer.)
A. The INTERSECT, because INTERSECT has higher precedence than MINUS.
B. The MINUS, because MINUS has a higher precedence than INTERSECT.
C. The precedence is determined by the order in which they are specified.
D. It is not possible for a compound query to include both MINUS and INTERSECT.
4. There are four rows in the REGIONS table. Consider the following statements and choose how
many rows will be returned for each: 0, 4, 8, or 16.
A. select * from regions union select * from regions
B. select * from regions union all select * from regions
C. select * from regions minus select * from regions
D. select * from regions intersect select * from regions
5. Consider this compound query:
select empno, hired from emp
union all
select emp_id,hired,fired from ex_emp;

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The columns EMP.EMPNO and EX_EMP.EMP_ID are integer; the column EMP.HIRED is
timestamp; the columns EX_EMP.HIRED and EX_EMP.FIRED are date. Why will the statement fail? (Choose the best answer.)
A. Because the columns EMPNO and EMP_ID have different names
B. Because the columns EMP.HIRED and EX_EMP.HIRED are different data types
C. Because there are two columns in the first query and three columns in the second query
D. For all the reasons above
E. The query will succeed.

Control the Order of Rows Returned
6. Which line of this statement will cause it to fail? (Choose the best answer.)
A. select ename, hired from current_staff
B. order by ename
C. minus
D. select ename, hired from current staff
E. where deptno=10
F. order by ename;
7. Study this statement:
select ename from emp union all select ename from ex_emp;

In what order will the rows be returned? (Choose the best answer.)
A. The rows from each table will be grouped and within each group will be sorted on ENAME.
B. The rows from each table will be grouped but not sorted.
C. The rows will not be grouped but will all be sorted on ENAME.
D. The rows will be neither grouped nor sorted.

LAB QUESTION
Working in the HR schema, design some queries that will generate reports using the set operators.
The reports required are as follows:
1. Employees have their current job (identified by JOB_ID) recorded in their EMPLOYEES row.
Jobs they have held previously (but not their current job) are recorded in JOB_HISTORY.
Which employees have never changed jobs? The listing should include the employees’
EMPLOYEE_ID and LAST_NAME.

Lab Question

399

2. Which employees were recruited into one job, then changed to a different job, but are now
back in a job they held before? Again, you will need to construct a query that compares
EMPLOYEES with JOB_HISTORY. The report should show the employees’ names and the job
titles. Job titles are stored in the table JOBS.
3. What jobs has any one employee held? This will be the JOB_ID for the employee’s current
job (in EMPLOYEES) and all previous jobs (in JOB_HISTORY). If the employee has held a
job more than once, there is no need to list it more than once. Use a replacement variable to
prompt for the EMPLOYEE_ID and display the job title(s). Employees 101 and 200 will be
suitable employees for testing.

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SELF TEST ANSWERS
Describe the Set Operators
✓ D. UNION ALL returns rows in the order that they are delivered by the two queries from
1. ®
which the compound query is made up.
®
˚ A, B, C. INTERSECT, MINUS, and UNION all use sorting as part of their execution.
✓ A, B, C. INTERSECT, MINUS, and UNION all remove duplicate rows.
2. ®
®
˚ D. UNION ALL returns all rows, duplicates included.

Use a Set Operator to Combine Multiple Queries into a Single Query
✓ C. All set operators have equal precedence, so the precedence is determined by the
3. ®
sequence in which they occur.
®
˚ A, B, D. A and B are wrong because set operators have equal precedence—though this may
change in future releases. D is wrong because many set operators can be used in one compound
query.
✓ A = 4; B = 8; C = 0; D = 4
4. ®
®
˚ Note that 16 is not used; that would be the result of a Cartesian product query.
✓ C. Every query in a compound query must return the same number of columns.
5. ®
®
˚ A, B, D, E. A is wrong because the columns can have different names. B is wrong because
the two columns are of the same data type group, which is all that was required. It therefore
follows that D and E are also wrong.

Control the Order of Rows Returned
✓ B. You cannot use ORDER BY for one query of a compound query; you may only place a
6. ®
single ORDER BY clause at the end.
®
˚ A, C, D, E, F. All these lines are legal.
✓ B. The rows from each query will be together, but there will be no sorting.
7. ®
®
˚ A, C, D. A is not possible with any syntax. C is wrong because that would be the result of a
UNION, not a UNION ALL. D is wrong because UNION ALL will return the rows from each
query grouped together.

Lab Answer

401

LAB ANSWER
1. To identify all employees who have not changed job, query the EMPLOYEES table and remove
all those who have a row in JOB_HISTORY:
select employee_id, last_name from employees
minus
select employee_id,last_name from
job_history join employees using (employee_id);

2. All employees who have changed job at least once will have a row in JOB_HISTORY; for those
who are now back in a job they have held before, the JOB_ID in EMPLOYEES will be the same
as the JOB_ID in one of their rows in JOB_HISTORY:
select last_name,job_title from employees join jobs using(job_id)
intersect
select last_name,job_title from job_history h
join jobs j on (h.job_id=j.job_id)
join employees e on (h.employee_id=e.employee_id);

3. This compound query will prompt for an EMPLOYEE_ID and then list the employee’s current
job and previous jobs,
select job_title from jobs join employees using (job_id)
where employee_id=&&Who
union
select job_title from jobs join job_history using (job_id)
where employee_id=&&Who;

Figure 9-5 shows the query and its result in SQL Developer, after giving 101 as the EMPLOYEE_ID.
FIGURE 9-5

A compound
query and the
result

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10
Manipulating Data

CERTIFICATION OBJECTIVES
10.01

Describe Each Data Manipulation
Language (DML) Statement

10.04

Delete Rows from a Table

10.02

Insert Rows into a Table

10.05

Control Transactions

✓

10.03

Update Rows in a Table

Q&A

Two-Minute Drill
Self Test

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

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Manipulating Data

T

his chapter is all about the commands that change data. Syntactically, the DML
commands are much simpler than the SELECT command that has been studied so far.
However, there is a large amount of relational database theory that comes along with
DML. As well as understanding the commands that make the changes, it is essential to understand
the theory of transaction management, which is part of the relational database paradigm. The
transaction management mechanisms provided by the Oracle database guarantee conformance
to the relational standards for transactions: they implement what is commonly referred to as the
ACID (atomicity, consistency, isolation, and durability) test.
Following detail on the DML commands, this chapter includes a full treatment of
the necessary theory: transactional integrity, record locking, and read consistency.

CERTIFICATION OBJECTIVE 10.01

Describe Each Data Manipulation
Language (DML) Statement
Strictly speaking, there are five DML commands:
■ SELECT
■ INSERT
■ UPDATE
■ DELETE
■ MERGE

In practice, most database professionals never include SELECT as part of DML.
It is considered to be a separate language in its own right, which is not unreasonable
when you consider that it has taken the eight preceding chapters to describe it.
The MERGE command is often dropped as well, not because it isn’t clearly a data
manipulation command but because it doesn’t do anything that cannot be done
with other commands. MERGE can be thought of as a shortcut for executing
either an INSERT or an UPDATE or a DELETE, depending on some condition.
A command often considered with DML is TRUNCATE. This is actually a DDL
(Data Definition Language) command, but as the effect for end users is the same as

Describe Each Data Manipulation Language (DML) Statement

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a DELETE (though its implementation is totally different), it does fit with DML.
So, the following commands are described in the next sections, with syntax and
examples:
■ INSERT
■ UPDATE
■ DELETE
■ TRUNCATE

In addition, we will discuss, for completeness:
■ MERGE

These are the commands that manipulate data.

INSERT
Oracle stores data in the form of rows in tables. Tables are populated with rows (just
as a country is populated with people) in several ways, but the most common method
is with the INSERT statement. SQL is a set-oriented language, so any one command
can affect one row or a set of rows. It follows that one INSERT statement can insert
an individual row into one table or many rows into many tables. The basic versions
of the statement do insert just one row, but more complex variations can, with one
command, insert multiple rows into multiple tables.
There are much faster techniques than INSERT for populating a table with
large numbers of rows.These are the SQL*Loader utility, which can upload
data from files produced by an external feeder system, and Datapump, which
can transfer data in bulk from one Oracle database to another—either via
disk files or through a network link.
Tables have rules defined that control the rows that may be inserted. These
rules are constraints. A constraint is an implementation of a business rule. The
business analysts who model an organization’s business processes will design a set
of rules for the organization’s data. Examples of such rules might be that every
employee must have a unique employee number, or that every employee must be
assigned to a valid department. Creating constraints is described in Chapter 11—for
now, remember that there is no way an INSERT command can insert a row that

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violates a constraint. So if you attempt to insert a row into EMPLOYEES with an
EMPLOYEE_ID that already exists in another row, or with a DEPARTMENT_ID
that does not match a row in the DEPARTMENTS table, the insert will fail.
Constraints guarantee that the data in the database conforms to the rules that define
the business procedures.
There are many possible sources for the row (or rows) inserted by an INSERT
statement. A single row can be inserted by providing the values for the row’s
columns individually. Such a statement can be constructed by typing it into
SQL*Plus or SQL Developer, or by a more sophisticated user process that presents
a form that prompts for values. This is the technique used for generating the rows
inserted interactively by end users. For inserts of multiple rows, the source of
the rows can be a SELECT statement. The output of any and all of the SELECT
statements discussed in the preceding eight chapters can be used as the input to an
INSERT statement.
The end result of any SELECT statement can be thought of as a table: a twodimensional set of rows. This “table” can be displayed to a user (perhaps in a simple
tool like SQL*Plus), or it can be passed to an INSERT command for populating
another table, defined within the database. Using a SELECT statement to construct
rows for an INSERT statement is a very
common technique. The SELECT can perform
many tasks. These typically include joining
tables and aggregations, so that the resulting
An INSERT command
rows inserted into the target table contain
can insert one row, with column values
information that is much more immediately
specified in the command, or a set of rows
comprehensible to the end users than the raw
created by a SELECT statement.
data in the source tables.

UPDATE
The UPDATE command is used to change rows that already exist—rows that have
been created by an INSERT command, or possibly by a tool such as Datapump. As
with any other SQL command, an UPDATE can affect one row or a set of rows. The
size of the set affected by an UPDATE is determined by a WHERE clause, in exactly
the same way that the set of rows retrieved by a SELECT statement is defined by a
WHERE clause. The syntax is identical. All the rows updated will be in one table; it
is not possible for a single update command to affect rows in multiple tables.
When updating a row or a set of rows, the UPDATE command specifies which
columns of the row(s) to update. It is not necessary (or indeed common) to update

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every column of the row. If the column being updated already has a value, then this
value is replaced with the new value specified by the UPDATE command. If the
column was not previously populated—which is to say, its value was NULL—then it
will be populated after the UPDATE with the new value.
A typical use of UPDATE is to retrieve one row and update one or more columns
of the row. The retrieval will be done using a WHERE clause that selects a row by
its primary key, the unique identifier that will ensure that only one row is retrieved.
Then the columns that are updated will be any columns other than the primary key
column. It is very unusual to change the value of the primary key. The lifetime of a
row begins when it is inserted, then may continue through several updates, until it is
deleted. Throughout this lifetime, it will not usually change its primary key.
To update a set of rows, use a less restrictive
WHERE clause than the primary key. To
update every row in a table, do not use any
WHERE clause at all. This set behavior can
One UPDATE statement
be disconcerting when it happens by accident.
can change rows in only one table, but it
If you select the rows to be updated with any
can change any number of rows in that
column other than the primary key, you may
table.
update several rows, not just one. If you omit
the WHERE clause completely, you will update
the whole table—perhaps millions of rows updated with just one statement—when
you meant to change just one.
An UPDATE command must honor any constraints defined for the table, just as
the original INSERT would have. For example, it will not be possible to update a
column that has been marked as mandatory to a NULL value or to update a primary
key column so that it will no longer be unique.

DELETE
Previously inserted rows can be removed from a table with the DELETE command.
The command will remove one row or a set of rows from the table, depending
on a WHERE clause. If there is no WHERE clause, every row in the table will be
removed (which can be a little disconcerting if you left out the WHERE clause by
mistake).
There are no “warning” prompts for any SQL commands. If you instruct the
database to delete a million rows, it will do so. Immediately.There is none of
that “Are you sure?” business that some environments offer.

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A deletion is all or nothing. It is not possible to nominate columns. When rows
are inserted, you can choose which columns to populate. When rows are updated,
you can choose which columns to update. But a deletion applies to the whole row—
the only choice is which rows in which table. This makes the DELETE command
syntactically simpler than the other DML commands.

MERGE
Earlier versions of SQL did not have a MERGE command. MERGE was introduced
with the SQL1999 standard, implemented by Oracle in database release 9i. Release
10g (conforming to the SQL2003 standard) provides some enhancements. Some
proprietary SQL implementations had a command called UPSERT. This rather
unpleasant sounding word describes the MERGE command rather well: it executes
either an UPDATE or an INSERT, depending on some condition. But the term
UPSERT is now definitely obsolete, because the current version of MERGE can,
according to circumstances, do a DELETE as well.
There are many occasions where you want to take a set of data (the source) and
integrate it into an existing table (the target). If a row in the source data already
exists in the target table, you may want to update the target row, or you may want to
replace it completely, or you may want to leave the target row unchanged. If a row
in the source does not exist in the target, you will want to insert it. The MERGE
command lets you do this. A MERGE passes through the source data, for each row
attempting to locate a matching row in the target. If no match is found, a row can
be inserted; if a match is found, the matching row can be updated. The release 10g
enhancement means that the target row can even be deleted, after being matched
and updated. The end result is a target table into which the data in the source has
been merged.
A MERGE operation does nothing that could not be done with INSERT,
UPDATE, and DELETE statements—but with one pass through the source data,
it can do all three. Alternative code without a MERGE would require three passes
through the data, one for each command.
MERGE may not be important for the OCP examinations, but it may be
of vital importance for coding applications that perform well and use the
database efficiently.
The source data for a MERGE statement can be a table or any subquery. The
condition used for finding matching rows in the target is similar to a WHERE
clause. The clauses that update or insert rows are as complex as an UPDATE or

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409

an INSERT command. It follows that MERGE is the most complicated of the DML
commands, which is not unreasonable, as it is (arguably) the most powerful.

TRUNCATE
The TRUNCATE command is not a DML command; it is DDL command. The
difference is enormous. When DML commands affect data, they insert, update, and
delete rows as part of transactions. Transactions are defined later in this chapter in
section 10.05, “Control Transactions.” For now, let it be said that a transaction can
be controlled, in the sense that the user has the
choice of whether to make the work done in a
transaction permanent, or whether to reverse it.
This is very useful but forces the database to do
Transactions, consisting
additional work behind the scenes that the user
of INSERT, UPDATE, and DELETE (or
is not aware of. DDL commands are not user
even MERGE) commands can be made
transactions (though within the database, they
permanent (with a COMMIT) or reversed
are in fact implemented as transactions—but
(with a ROLLBACK). A TRUNCATE
developers cannot control them), and there
command, like any other DDL command,
is no choice about whether to make them
is immediately permanent: it can never be
permanent or to reverse them. Once executed,
reversed.
they are done. However, in comparison to
DML, they are very fast.
From the user’s point of view, a truncation of a table is equivalent to executing a
DELETE of every row: a DELETE command without a WHERE clause. But whereas
a deletion may take some time (possibly hours, if there are many rows in the table)
a truncation will go through instantly. It makes no difference whether the table
contains one row or billions; a TRUNCATE will be virtually instantaneous. The
table will still exist, but it will be empty.
DDL commands, such as TRUNCATE, will fail if there is any DML command
active on the table. A transaction will block the DDL command until the DML
command is terminated with a COMMIT or a ROLLBACK.

DML Statement Failures
Commands can fail for many reasons, including the following:
■ Syntax errors
■ References to nonexistent objects or columns

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■ Access permissions
■ Constraint violations
■ Space issues

As a SQL command can affect a set of rows, there is the complication that a
command may partially succeed: the failure could occur only some way into the set.
The preceding first three classes of error also apply to SELECT statements.
One purpose of reading this book is to prevent syntax errors. When they occur
(and they will), they should be detected by the tool that is constructing the SQL to
be sent to the database: SQL*Plus or SQL Developer if the SQL is being entered
interactively, or whatever other tool is being used to generate a more sophisticated
interface. There are any number of possible syntax errors, starting with simple
spelling mistakes or transposition errors.
People who type with only one finger do not make transposition errors.
Errors of this nature will not impact the database, because the database will
never see them. The erroneous SQL is stopped at the source. The degree of
assistance the tool provides for fixing such errors will depend on the effort the tool’s
developers take.
A SQL statement may be syntactically correct but refer to objects that do not
exist. Typical problems are spelling mistakes, but there are more complex issues: a
statement could refer to a column that existed at one time but has been dropped
from the table or renamed. A statement of this nature will be sent to the database
and will fail then, before the database attempts execution. This is worse for the
database than a simple syntax error, but the statement is still stopped before it
consumes any significant database resources.
A related error has to do with type casting. SQL is a strongly typed language:
columns are defined as a certain data type, and an attempt to enter a value of a
different data type will fail. However, you may get away with this because Oracle’s
implementation of SQL will, in some circumstances, do automatic type casting.
Figure 10-1 shows several attempted executions of a statement with SQL*Plus.
In Figure 10-1, a user connects as SUE (password, SUE—not an example of good
security) and queries the EMPLOYEES table. The statement fails because of a simple
syntax error, correctly identified by SQL*Plus. Note that SQL*Plus never attempts
to correct such mistakes, even when it knows exactly what you meant to type. Some
third-party tools may be more helpful, offering automatic error correction.

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FIGURE 10-1

Some examples
of statement
failure

The second attempt to run the statement fails with an error stating that the
object does not exist. This is because it does not exist in the current user’s schema;
it exists in the HR schema. Having corrected that, the third run of the statement
succeeds—but only just. The value passed in the WHERE clause is a string,
‘21-APR-00,’ but the column HIRE_DATE is not defined in the table as a string, it
is defined as a date. To execute the statement, the database had to work out what
the user really meant and cast the string as a date. In the last example, the type
casting fails. This is because the string passed is formatted as a European-style date,
but the database has been set up as American: the attempt to match “21” to a month
fails. The statement would have succeeded if the string had been ‘04/21/2007.’
Developers should never rely on automatic type casting. It is extremely lazy
programming. They should always do whatever explicit type casting is necessary,
using appropriate functions as discussed in previous chapters. If automatic type
casting does work, at best there is a performance hit as the database has to do extra
work. This may be substantial if the type casting prevents Oracle from using indexes.
In this example, if there is an index on the HIRE_DATE column, it will be an index
of dates; there is no way the database can use it when you pass a string. At worst,
the result will be wrong. If the date string passed in were ‘04/05/2007,’ this would
succeed—but would it be the fourth of May or the fifth of April?

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Oracle will attempt to correct data type mismatches in SQL statements (DML
and SELECT) by automatic type casting, but the results are unpredictable and
no programmer should ever rely on it.
If a statement is syntactically correct and has no errors with the objects to which
it refers, it can still fail because of access permissions. If the user attempting to
execute the statement does not have the relevant permissions on the tables to which
it refers, the database will return an error identical to that which would be returned
if the object did not exist. As far as the user is concerned, it does not exist.
Errors caused by access permissions are a case where SELECT and DML
statements may return different results: it is possible for a user to have permission
to see the rows in a table, but not to insert, update, or delete them. Such an
arrangement is not uncommon; it often makes business sense. Perhaps more
confusingly, permissions can be set up in such a manner that it is possible to insert
rows that you are not allowed to see. And, perhaps worst of all, it is possible to
delete rows that you can neither see nor update. However, such arrangements are
not common.
A constraint is a business rule, implemented within the database. Typical
constraints are that a table must have a primary key: a value of one column (or
combination of columns) that can uniquely identify each row. An INSERT
command can insert several rows into a table, and for every row the database will
check whether a row already exists with the same primary key. This occurs as each
row is inserted. It could be that the first few rows (or the first few million rows) go
in without a problem, and then the statement hits a row with a duplicate value. At
this point it will return an error, and the statement will fail. This failure will trigger
a reversal of all the insertions that had already succeeded. This is part of the SQL
standard: a statement must succeed in total, or not at all. The reversal of the work is
a rollback. The mechanisms of a rollback are described in the section of this chapter
titled “Controlling Transactions.”
If a statement fails because of space problems, the effect is similar. A part of the
statement may have succeeded before the database ran out of space. The part that
did succeed will be automatically rolled back. Rollback of a statement is a serious
matter. It forces the database to do a lot of extra work and will usually take at least as
long as the statement has taken already (sometimes much longer).

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CERTIFICATION OBJECTIVE 10.02

Insert Rows into a Table
The simplest form of the INSERT statement inserts one row into one table, using
values provided in line as part of the command. The syntax is as follows:
INSERT INTO table [(column [,column…])] VALUES (value [,value…]);
For example:
insert into hr.regions
insert into hr.regions
('Australasia',11);
insert into hr.regions
insert into hr.regions

values (10,'Great Britain');
(region_name, region_id) values
(region_id) values (12);
values (13,null);

The first of the preceding commands provides values for both the columns of the
REGIONS table. If the table had a third column, the statement would fail because
it relies upon positional notation. The statement does not say which value should be
inserted into which column; it relies on the position of the values: their ordering in
the command. When the database receives a statement using positional notation, it
will match the order of the values to the order in which the columns of the table are
defined. The statement would also fail if the column order were wrong: the database
would attempt the insertion but would fail because of data type mismatches.
The second command nominates the columns to be populated and the values
with which to populate them. Note that the order in which columns are mentioned
now becomes irrelevant—as long as the order of the columns is the same as the order
of the values.
The third example lists one column, and therefore only one value. All other
columns will be left null. This statement would fail if the REGION_NAME column
were not nullable. The fourth example will produce the same result, but because
there is no column list, a value of some sort must be provided for each column—at
the least, a NULL.
It is often considered good practice not to rely on positional notation and
instead always to list the columns.This is more work but makes the code selfdocumenting (always a good idea!) and also makes the code more resilient
against table structure changes. For instance, if a column is added to a table,
all the INSERT statements that rely on positional notation will fail until they
are rewritten to include a NULL for the new column. INSERT code that names
the columns will continue to run.

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Very often, an INSERT statement will include functions to do type casting or other
editing work. Consider this statement:
insert into employees (employee_id, last_name, hire_date)
values (1000,'WATSON','03-Nov-07');

in contrast with this:
insert into employees (employee_id, last_name, hire_date)
values (1000,upper('Watson'),to_date('03-Nov-07','dd-mon-yy'));

The rows inserted with each statement would be identical. But the first will insert
exactly the literals provided. It may well be that the application relies on employee
surnames being in uppercase—without this, perhaps sort orders will be wrong and
searches on surname will give unpredictable results. Also, the insertion of the date
value relies on automatic type casting of a string to a date, which is always bad for
performance and can result in incorrect values being entered. The second statement
forces the surname into uppercase whether it was entered that way or not, and
specifies exactly the format mask of the date string before explicitly converting it
into a date. There is no question that the second statement is a better piece of code
than the first.
The following is another example of using functions:
insert into employees (employee_id,last_name,hire_date)
values (1000 + 1,user,sysdate - 7);

In the preceding statement, the EMPLOYEE_ID column is populated with the
result of some arithmetic, the LAST_NAME column is populated with the result of
the function USER (which returns the database logon name of the user), and the
HIRE_DATE column is populated with the result of a function and arithmetic: the
date seven days before the current system date.
Figure 10-2 shows the execution of the previous three insertions, followed by a
query showing the results.
Using functions to preprocess values before inserting rows can be particularly
important when running scripts with substitution variables, as they will allow the
code to correct many of the unwanted variations in data input that can occur when
users enter values interactively.
To insert many rows with one INSERT command, the values for the rows must
come from a query. The syntax is as follows:
INSERT INTO table [ (column [, column…] ) ] subquery;
Note that this syntax does not use the VALUES keyword. If the column list
is omitted, then the subquery must provide values for every column in the table.

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415

FIGURE 10-2

Using functions
with the INSERT
command

To copy every row from one table to another, if the tables have the same column
structure, a command such as this is all that is needed:
insert into regions_copy select * from regions;

This presupposes that the table REGIONS_COPY does exist (with or without
any rows). The SELECT subquery reads every row from the source table, which
is REGIONS, and the INSERT inserts them into the target table, which is
REGIONS_COPY.
There are no restrictions on the nature of the subquery. Any query returns
(eventually) a two-dimensional array of rows; if the target table (which is also a twodimensional array) has columns to receive them, the insertion will work. A common
requirement is to present data to end users in a form that will make it easy for them
to extract information and impossible for them to misinterpret it. This will usually
mean denormalizing relational tables, making aggregations, renaming columns, and
adjusting data that can distort results if not correctly processed.
Consider a simple case within the HR schema: a need to report on the salary bill
for each department. The query will need to perform a full outer join to ensure that
any employees without a department are not missed, and that all departments are
listed whether or not they have employees. It should also ensure that any null values
will not distort any arithmetic by substituting zeros or strings for nulls. This query is
perfectly straightforward for any SQL programmer, but when end users attempt to
run this sort of query they are all too likely to produce inaccurate results by omitting

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the checks. A daily maintenance job in a data warehouse that would assemble the
data in a suitable form could be a script such as this:
truncate table department_salaries;
insert into department_salaries (department,staff,salaries)
select
coalesce(department_name,'Unassigned'),
count(employee_id),
sum(coalesce(salary,0))
from employees e full outer join departments d
on e.department_id = d.department_id
group by department_name
order by department_name;

The TRUNCATE command will empty
the table, which is then repopulated from the
subquery. The end users can be let loose on this
table, and it should be impossible for them to
misinterpret the contents—a simple natural
Any SELECT statement,
join with no COALESCE functions, which
specified as a subquery, can be used as
might be all an end user would do, might be
the source of rows passed to an INSERT.
very misleading. By doing all the complex work
This enables insertion of many rows.
in the INSERT statement, users can then run
Alternatively, using the VALUES clause will
much simpler queries against the denormalized
insert one row.The values can be literals
and aggregated data in the summary table.
or prompted for as substitution variables.
Their queries will be fast, too: all the hard work
has been done already.
To conclude the description of the INSERT command, it should be mentioned
that it is possible to insert rows into several tables with one statement. This is not
part of the SQL OCP examination, but for completeness here is an example:
insert all
when 1=1 then
into emp_no_name (department_id,job_id,salary,commission_pct,hire_date)
values (department_id,job_id,salary,commission_pct,hire_date)
when department_id <> 80 then
into emp_non_sales (employee_id,department_id,salary,hire_date)
values (employee_id,department_id,salary,hire_date)
when department_id = 80 then
into emp_sales (employee_id,salary,commission_pct,hire_date)
values (employee_id,salary,commission_pct,hire_date)
select employee_id,department_id,job_id,salary,commission_pct,hire_date
from employees where hire_date > sysdate - 30;

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To read this statement, start at the bottom. The subquery retrieves all employees
recruited in the last 30 days. Then go to the top. The ALL keyword means that
every row selected will be considered for insertion into all the tables following,
not just into the first table for which the condition applies. The first condition
is 1=1, which is always true, so every source row will create a row in EMP_NO_
NAME. This is a copy of the EMPLOYEES table with the personal identifiers
removed, a common requirement in a data warehouse. The second condition is
DEPARTMENT_ID <> 80, which will generate a row in EMP_NON_SALES for
every employee who is not in the sales department; there is no need for this table
to have the COMMISSION_PCT column. The third condition generates a row
in EMP_SALES for all the salesmen; there is no need for the DEPARTMENT_ID
column, because they will all be in department 80.
This is a simple example of a multitable insert, but it should be apparent that with
one statement, and therefore only one pass through the source data, it is possible
to populate many target tables. This can take an enormous amount of strain off the
database.

EXERCISE 10-1
Use the INSERT Command
In this exercise, use various techniques to insert rows into a table.
1. Connect to the HR schema, with either SQL Developer or SQL*Plus.
2. Query the REGIONS table, to check what values are already in use for the
REGION_ID column:
select * from regions;

This exercise assumes that values above 100 are not in use. If they are, adjust
the values suggested below to avoid primary key conflicts.
3. Insert a row into the REGIONS table, providing the values in line:
insert into regions values (101,'Great Britain');

4. Insert a row into the REGIONS table, providing the values as substitution
variables:
insert into regions values (&Region_number,'&Region_name');

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When prompted, give the values 102 for the number, Australasia for the
name. Note the use of quotes around the string.
5. Insert a row into the REGIONS table, calculating the REGION_ID to be
one higher than the current high value. This will need a scalar subquery:
insert into regions values ((select max(region_id)+1
from regions), 'Oceania');

6. Confirm the insertion of the rows:
select * from regions;

7. Commit the insertions:
commit;

The following illustration shows the results of the exercise, using SQL*Plus:

Update Rows in a Table

419

CERTIFICATION OBJECTIVE 10.03

Update Rows in a Table
The UPDATE command changes column values in one or more existing rows in a
single table. The basic syntax is the following:
UPDATE table SET column=value [,column=value…] [WHERE condition];
The more complex form of the command uses subqueries for one or more of the
column values and for the WHERE condition. Figure 10-3 shows updates of varying
complexity, executed from SQL*Plus.
The first example is the simplest. One column of one row is set to a literal value.
Because the row is chosen with a WHERE clause that uses the equality predicate on
the table’s primary key, there is an absolute guarantee that at most only one row will
be affected. No row will be changed if the WHERE clause fails to find any rows at all.
The second example shows use of arithmetic and an existing column to set the new
value, and the row selection is not done on the primary key column. If the selection
is not done on the primary key, or if a nonequality predicate (such as BETWEEN) is

FIGURE 10-3

Examples of using
the UPDATE
statement

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used, then the number of rows updated may be more than one. If the WHERE clause is
omitted entirely, the update will be applied to every row in the table.
The third example in Figure 10-3 introduces the use of a subquery to define
the set of rows to be updated. A minor additional complication is the use of a
replacement variable to prompt the user for a value to use in the WHERE clause
of the subquery. In this example, the subquery (lines 3 and 4) will select every
employee who is in a department whose name includes the string ‘IT’ and increment
their current salary by 10 percent (unlikely to happen in practice).
It is also possible to use subqueries to determine the value to which a column will
be set, as in the fourth example. In this case, one employee (identified by primary
key, in line 5) is transferred to department 80 (the sales department), and then the
subquery in lines 3 and 4 set his commission rate to whatever the lowest commission
rate in the department happens to be.
The syntax of an update that uses subqueries is as follows:
UPDATE table
SET column=[subquery] [,column=subquery…]
WHERE column = (subquery) [AND column=subquery…] ;
There is a rigid restriction on the subqueries using update columns in the SET
clause: the subquery must return a scalar value. A scalar value is a single value of
whatever data type is needed: the query must return one row, with one column. If the
query returns several values, the UPDATE will fail. Consider these two examples:
update employees
set salary=(select salary from employees where employee_id=206);
update employees
set salary=(select salary from employees where last_name='Abel');

The first example, using an equality predicate on the primary key, will always
succeed. Even if the subquery does not retrieve a row (as would be the case if there
were no employee with EMPLOYEE_ID equal to 206), the query will still return
a scalar value: a null. In that case, all the rows in EMPLOYEES would have their
SALARY set to NULL—which might not be desired but is not an error as far as SQL
is concerned. The second example uses an equality predicate on the LAST_NAME,
which is not guaranteed to be unique. The statement will succeed if there is only
one employee with that name, but if there were more than one it would fail with the
error “ORA-01427: single-row subquery returns more than one row.” For code that
will work reliably, no matter what the state of the data, it is vital to ensure that the
subqueries used for setting column values are scalar.

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A common fix for making sure that queries are scalar is to use MAX or MIN.
This version of the statement will always succeed:
update employees
set salary=(select max(salary) from employees where last_name='Abel');

However, just because it will work, doesn’t necessarily mean that it does what
is wanted.
The subqueries in the WHERE clause must also be scalar, if it is using the equality
predicate (as in the preceding examples) or the greater/less than predicates. If it is
using the IN predicate, then the query can return multiple rows, as in this example
which uses IN:
update employees
set salary=10000
where department_id in (select department_id from departments
where department_name like '%IT%');

This will apply the update to all employees in a department whose name includes
the string ‘IT.’ There are several of these. But even though the query can return
several rows, it must still return only one column.

The subqueries used to SET
column values must be scalar subqueries.
The subqueries used to select the rows

must also be scalar, unless they use the IN
predicate.

EXERCISE 10-2
Use the UPDATE Command
In this exercise, use various techniques to update rows in a table. It is assumed that
the HR.REGIONS table is as seen in the illustration at the end of Exercise 10-1. If
not, adjust the values as necessary.
1. Connect to the HR schema using SQL Developer or SQL*Plus.
2. Update a single row, identified by primary key:
update regions set region_name='Scandinavia' where region_id=101;

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This statement should return the message “1 row updated.”
3. Update a set of rows, using a nonequality predicate:
update regions set region_name='Iberia' where region_id > 100;

This statement should return the message “3 rows updated.”
4. Update a set of rows, using subqueries to select the rows and to provide values:
update regions
set region_id=(region_id+(select max(region_id) from regions))
where region_id in (select region_id from regions where
region_id > 100);

This statement should return the message “3 rows updated.”
5. Confirm the state of the rows:
select * from regions;

6. Commit the changes made:
commit;

The following illustration shows the exercise, as done from SQL*Plus.

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CERTIFICATION OBJECTIVE 10.04

Delete Rows from a Table
To remove rows from a table, there are two options: the DELETE command and the
TRUNCATE command. DELETE is less drastic, in that a deletion can be rolled
back whereas a truncation cannot be. DELETE is also more controllable, in that it
is possible to choose which rows to delete, whereas a truncation always affects the
whole table. DELETE is, however, a lot slower and can place a lot of strain on the
database. TRUNCATE is virtually instantaneous and effortless.

Removing Rows with DELETE
The DELETE commands removes rows from a single table. The syntax is as follows:
DELETE FROM table [WHERE condition];
This is the simplest of the DML commands, particularly if the condition is omitted.
In that case, every row in the table will be removed with no prompt. The only
complication is in the condition. This can be a simple match of a column to a literal:
delete
delete
delete
delete

from
from
from
from

employees
employees
employees
employees

where
where
where
where

employee_id=206;
last_name like 'S%';
department_id=&Which_department;
department_id is null;

The first statement identifies a row by primary key. One row only will be removed—
or no row at all, if the value given does not find a match. The second statement uses a
nonequality predicate that could result in the deletion of many rows: every employee
whose surname begins with an uppercase “S.” The third statement uses an equality
predicate but not on the primary key. It prompts for a department number with a
substitution variable, and all employees in that department will go. The final statement
removes all employees who are not currently assigned to a department.
The condition can also be a subquery:
delete from employees where department_id in
(select department_id from departments where location_id in
(select location_id from locations where country_id in
(select country_id from countries where region_id in
(select region_id from regions where region_name='Europe')
)
)
)

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This example uses a subquery for row selection that navigates the HR
geographical tree (with more subqueries) to delete every employee who works for
any department that is based in Europe. The same rule for the number of values
returned by the subquery applies as for an UPDATE command: if the row selection
is based on an equality predicate (as in the preceding example) the subquery must be
scalar, but if it uses IN the subquery can return several rows.
If the DELETE command finds no rows to delete, this is not an error. The command
will return the message “0 rows deleted” rather than an error message because the
statement did complete successfully—it just didn’t find anything to do.

EXERCISE 10-3
Use the DELETE Command
In this exercise, use various techniques to delete rows in a table. It is assumed that
the HR.REGIONS table is as seen in the illustration at the end of Exercise 10-2.
If not, adjust the values as necessary.
1. Connect to the HR schema using SQL Developer or SQL*Plus.
2. Remove one row, using the equality predicate on the primary key:
delete from regions where region_id=204;

This should return the message “1 row deleted.”
3. Attempt to remove every row in the table by omitting a WHERE clause:
delete from regions;

This will fail, due to a constraint violation.
4. Remove rows with the row selection based on a subquery:
delete from regions where
region_id in (select region_id from regions where region_
name='Iberia');

This will return the message “2 rows deleted.”
5. Confirm that the REGIONS table now contains just the original four rows:
select * from regions;

6. Commit the deletions:
commit;

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The following illustration shows the exercise, as done from SQL*Plus:

Removing Rows with TRUNCATE
TRUNCATE is a DDL (Data Definition Language) command. It operates within
the data dictionary and affects the structure of the table, not the contents of the
table. However, the change it makes to the structure has the side effect of destroying
all the rows in the table.
One part of the definition of a table as stored
in the data dictionary is the table’s physical
location. When first created, a table is allocated
a single area of space, of fixed size, in the
TRUNCATE completely
database’s data files. This is known as an extent
empties the table.There is no concept of
and will be empty. Then, as rows are inserted,
row selection, as there is with a DELETE.
the extent fills up. Once it is full, more extents
will be allocated to the table automatically.
A table therefore consists of one or more extents, which hold the rows. As well as
tracking the extent allocation, the data dictionary also tracks how much of the space
allocated to the table has been used. This is done with the high water mark. The high
water mark is the last position in the last extent that has been used; all space below

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the high water mark has been used for rows at one time or another, and none of the
space above the high water mark has been used yet.
Note that it is possible for there to be plenty of space below the high water mark
that is not being used at the moment; this is because of rows having been removed
with a DELETE command. Inserting rows into a table pushes the high water mark
up. Deleting them leaves the high water mark where it is; the space they occupied
remains assigned to the table but is freed up for inserting more rows.
Truncating a table resets the high water mark. Within the data dictionary, the
recorded position of the high water mark is moved to the beginning of the table’s
first extent. As Oracle assumes that there can be no rows above the high water mark,
this has the effect of removing every row from the table. The table is emptied and
remains empty until subsequent insertions begin to push the high water mark back
up again. In this manner, one DDL command, which does little more than make an
update in the data dictionary, can annihilate billions of rows in a table.
A truncation is fast: virtually instantaneous, irrespective of whether the table
has many millions of rows or none. A deletion may take seconds, minutes,
hours—and it places much more strain on the database than a truncation.
But a truncation is all or nothing.
The syntax to truncate a table couldn’t be simpler:
TRUNCATE TABLE table;
Figure 10-4 shows access to the TRUNCATE command through the SQL
Developer navigation tree, but of course it can also be executed from SQL*Plus.

MERGE
The MERGE command is often ignored, because it does nothing that cannot be
done with INSERT, UPDATE, and DELETE. It is, however, very powerful, in
that with one pass through the data it can carry out all three operations. This can
improve performance dramatically. Use of MERGE is not on the OCP syllabus, but
for completeness here is a simple example:
merge into employees e using new_employees n
on (e.employee_id = n.employee_id)
when matched then
update set e.salary=n.salary
when not matched then
insert (employee_id,last_name,salary)
values (n.employee_id,n.last_name,n.salary);

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FIGURE 10-4

The TRUNCATE
command in SQL
Developer, from
the command
line and from the
menus

The preceding statement uses the contents of a table NEW_EMPLOYEES to
update or insert rows in EMPLOYEES. The situation could be that EMPLOYEES
is a table of all staff, and NEW_EMPLOYEES is a table with rows for new staff
and for salary changes for existing staff. The command will pass through NEW_
EMPLOYEES, and for each row, attempt to find a row in EMPLOYEES with the
same EMPLOYEE_ID. If there is a row found, its SALARY column will be updated
with the value of the row in NEW_EMPLOYEES. If there is not such a row, one will
be inserted. Variations on the syntax allow the use of a subquery to select the source
rows, and it is even possible to delete matching rows.

CERTIFICATION OBJECTIVE 10.05

Control Transactions
The concepts behind a transaction are a part of the relational database paradigm.
A transaction consists of one or more DML statements, followed by either a
ROLLBACK or a COMMIT command. It is possible use the SAVEPOINT

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command to give a degree of control within the transaction. Before going into the
syntax, it is necessary to review the concept of a transaction. Related topics are read
consistency; this is automatically implemented by the Oracle server, but to a certain
extent programmers can manage it by the way they use the SELECT statement.

Database Transactions
This is not the place to go into detail on the relational database transactional
paradigm—there are any number of academic texts on this, and there is not space
to cover this topic in a practical guide. Following is a quick review of some of the
principles of a relational database to which all databases (not just Oracle’s) must
conform. Other database vendors comply with the same standards with their own
mechanisms, but with varying levels of effectiveness. In brief, any relational database
must be able to pass the ACID test: it must guarantee atomicity, consistency,
isolation, and durability.

A is for Atomicity
The principle of atomicity states that all parts of a transaction must complete or none
of them. (The reasoning behind the term is that an atom cannot be split—now
well known to be a false assumption). For example, if your business analysts have
said that every time you change an employee’s salary you must also change the
employee’s grade, then the atomic transaction will consist of two updates. The
database must guarantee that both go through or neither. If only one of the updates
were to succeed, you would have an employee on a salary that was incompatible
with his grade: a data corruption, in business terms. If anything (anything at all!)
goes wrong before the transaction is complete, the database itself must guarantee
that any parts that did go through are reversed; this must happen automatically. But
although an atomic transaction sounds small (like an atom), it can be enormous.
To take another example, it is logically impossible for an accounting suite nominal
ledger to be half in August and half in September: the end-of-month rollover is
therefore (in business terms) one atomic transaction, which may affect millions of
rows in thousands of tables and take hours to complete (or to roll back, if anything
goes wrong). The rollback of an incomplete transaction may be manual (as when
you issue the ROLLBACK command), but it must be automatic and unstoppable in
the case of an error.

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C is for Consistency
The principle of consistency states that the results of a query must be consistent with the
state of the database at the time the query started. Imagine a simple query that averages
the value of a column of a table. If the table is large, it will take many minutes to pass
through the table. If other users are updating the column while the query is in progress,
should the query include the new or the old values? Should it include rows that were
inserted or deleted after the query started? The principle of consistency requires that
the database ensure that changed values are not seen by the query; it will give you an
average of the column as it was when the query started, no matter how long the query
takes or what other activity is occurring on the tables concerned. Oracle guarantees
that if a query succeeds, the result will be consistent. However, if the database
administrator has not configured the database appropriately, the query may not succeed:
there is a famous Oracle error, “ORA-1555 snapshot too old,” that is raised. This used
to be an extremely difficult problem to fix with earlier releases of the database, but with
recent versions the database administrator should always be able to prevent this.

I is for Isolation
The principle of isolation states that an incomplete (that is, uncommitted)
transaction must be invisible to the rest of the world. While the transaction is in
progress, only the one session that is executing the transaction is allowed to see the
changes; all other sessions must see the unchanged data, not the new values. The
logic behind this is first, that the full transaction might not go through (remember
the principle of atomicity and automatic or manual rollback?) and that therefore no
other users should be allowed to see changes that might be reversed. And second,
during the progress of a transaction the data is (in business terms) incoherent:
there is a short time when the employee has had his salary changed but not his
grade. Transaction isolation requires that the database must conceal transactions in
progress from other users: they will see the preupdate version of the data until the
transaction completes, when they will see all the changes as a consistent set. Oracle
guarantees transaction isolation: there is no way any session (other than that making
the changes) can see uncommitted data. A read of uncommitted data is known as
a dirty read, which Oracle does not permit (though some other databases do).

D is for Durable
The principle of durability states that once a transaction completes, it must be
impossible for the database to lose it. During the time that the transaction is in
progress, the principle of isolation requires that no one (other than the session

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concerned) can see the changes it has made so far. But the instant the transaction
completes, it must be broadcast to the world, and the database must guarantee that
the change is never lost; a relational database is not allowed to lose data. Oracle
fulfills this requirement by writing out all change vectors that are applied to data to
log files as the changes are done. By applying this log of changes to backups taken
earlier, it is possible to repeat any work done in the event of the database being
damaged. Of course, data can be lost through user error such as inappropriate DML,
or dropping or truncating tables. But as far as Oracle and the DBA are concerned,
such events are transactions like any other: according to the principle of durability,
they are absolutely nonreversible.

The Start and End of a Transaction
A session begins a transaction the moment it issues any INSERT, UPDATE, or
DELETE statement (but not a TRUNCATE—that is a DDL command, not DML).
The transaction continues through any number of further DML commands until
the session issues either a COMMIT or a ROLLBACK statement. Only then
will the changes be made permanent and become visible to other sessions (if it is
committed, rather than rolled back). It is impossible to nest transactions. The SQL
standard does not allow a user to start one transaction and then start another before
terminating the first. This can be done with PL/SQL (Oracle’s proprietary thirdgeneration language), but not with industry-standard SQL.
The explicit transaction control statements are COMMIT, ROLLBACK, and
SAVEPOINT. There are also circumstances other than a user-issued COMMIT or
ROLLBACK that will implicitly terminate a transaction:
■ Issuing a DDL or DCL statement
■ Exiting from the user tool (SQL*Plus or SQL Developer or anything else)
■ If the client session dies
■ If the system crashes

If a user issues a DDL (CREATE, ALTER, or DROP) or DCL (GRANT or
REVOKE) command, the transaction he has in progress (if any) will be committed:
it will be made permanent and become visible to all other users. This is because the
DDL and DCL commands are themselves transactions. If it were possible to see the
source code for these commands, it would be obvious. They adjust the data structures
by performing DML commands against the tables that make up the data dictionary,

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and these commands are terminated with a COMMIT. If they were not, the changes
made couldn’t be guaranteed to be permanent. As it is not possible in SQL to nest
transactions, if the user already has a transaction running, the statements the user
has run will be committed along with the statements that make up the DDL or DCL
command.
If a user starts a transaction by issuing a DML command and then exits from the
tool he is using without explicitly issuing either a COMMIT or a ROLLBACK,
the transaction will terminate—but whether it terminates with a COMMIT or
a ROLLBACK is entirely dependent on how the tool is written. Many tools will
have different behavior, depending on how the tool is exited. (For instance, in the
Microsoft Windows environment, it is common to be able to terminate a program
either by selecting the File | Exit options from a menu on the top left of the
window, or by clicking an “X” in the top right corner. The programmers who wrote
the tool may well have coded different logic into these functions.) In either case, it
will be a controlled exit, so the programmers should issue either a COMMIT or a
ROLLBACK, but the choice is up to them.
If a client’s session fails for some reason, the database will always roll back the
transaction. Such failure could be for a number of reasons: the user process can die
or be killed at the operating system level, the network connection to the database
server may go down, or the machine where the client tool is running can crash.
In any of these cases, there is no orderly issue of a COMMIT or ROLLBACK
statement, and it is up to the database to detect what has happened. The behavior
is that the session is killed, and an active transaction is rolled back. The behavior
is the same if the failure is on the server side. If the database server crashes for any
reason, when it next starts up all transactions from any sessions that were in progress
will be rolled back.

The Transaction Control Statements
A transaction begins implicitly with the first DML statement. There is no command
to explicitly start a transaction. The transaction continues through all subsequent
DML statements issued by the session. These statements can be against any number
of tables: a transaction is not restricted to one table. It terminates (barring any of
the events listed in the previous section) when the session issues a COMMIT or
ROLLBACK command. The SAVEPOINT command can be used to set markers
that will stage the action of a ROLLBACK, but the same transaction remains in
progress irrespective of the use of SAVEPOINT.

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COMMIT
Syntactically, COMMIT is the simplest SQL command. The syntax is as follows:
COMMIT;
This will end the current transaction, which has the dual effect of making
the changes both permanent and visible to other sessions. Until a transaction
is committed, it cannot be seen by any other sessions, even if they are logged
on to the database with the same username as that of the session executing the
transactions. Until a transaction is committed, it is invisible to other sessions
and can be reversed. But once it is committed, it is absolutely nonreversible. The
principle of durability applies. All database administrators dread a telephone call
along these lines:
User: I’ve just deleted a million rows instead of one.
DBA: How?
User: Because I forgot to put a WHERE clause on my DELETE statement.
DBA: Did you say COMMIT?
User: Of course.
DBA: Um…
The state of the data before the COMMIT is that the changes have been made,
but all sessions other than the one that made the changes are redirected to copies
of the data in its prechanged form. So if the session has inserted rows, other sessions
that SELECT from the table will not see them. If the transaction has deleted rows,
other sessions selecting from the table will still see them. If the transaction has made
updates, it will be the unupdated versions of the rows that are presented to other
sessions. This is in accordance with the principle of isolation: no session can be in
any way dependent on the state of an uncommitted transaction.
After the COMMIT, all sessions will immediately see the new data in any
queries they issue: they will see the new rows, they will not see the deleted rows,
they will see the new versions of the updated rows. In the case of the hypothetical
conversation just discussed, one moment other sessions’ queries will see millions of
rows in the table; the next it will be empty. This is in accordance with the principle
of durability.

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ROLLBACK
While a transaction is in progress, Oracle keeps an image of the data as it was before
the transaction. This image is presented to other sessions that query the data while
the transaction is in progress. It is also used to roll back the transaction automatically
if anything goes wrong, or deliberately if the session requests it. The syntax to request
a rollback is as follows:
ROLLBACK [TO SAVEPOINT savepoint] ;
The optional use of savepoints is detailed in the section following.
The state of the data before the rollback is that the data has been changed, but
the information needed to reverse the changes is available. This information is
presented to all other sessions, in order to implement the principle of isolation. The
rollback will discard all the changes by restoring the prechange image of the data;
any rows the transaction inserted will be deleted, rows the transaction deleted will
be inserted back into the table, and any rows that were updated will be returned to
their original state. Other sessions will not be aware that anything has happened at
all; they never saw the changes. The session that did the transaction will now see
the data as it was before the transaction started.
A COMMIT is instantaneous, because it doesn’t really have to do anything.
The work has already been done. A ROLLBACK can be very slow: it will
usually take as long (if not longer) to reverse a transaction than it took to
make the changes in the first place. Rollbacks are not good for database
performance.

EXERCISE 10-4
Use the COMMIT and ROLLBACK Commands
In this exercise, demonstrate the use of transaction control statements and transaction
isolation. It is assumed that the HR.REGIONS table is as seen in the illustration at
the end of Exercise 10-3. If not, adjust the values as necessary. Connect to the HR
schema with two sessions concurrently. These can be two SQL*Plus sessions or two
SQL Developer sessions or one of each. The following table lists steps to follow in
each session.

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Step

In your first session

In your second session

1

select * from regions;

select * from regions;

Both sessions see the same data.
2

insert into regions
values(100,'UK');

insert into regions
values(101,'GB');

3

select * from regions;

select * from regions;

Both sessions see different results: the original data, plus their own change.
4

commit;

5

select * from regions;

select * from regions;

One transaction has been published to the world, the other is still visible to only one session.
6

rollback;

rollback;

7

select * from regions;

select * from regions;

The committed transaction was not reversed because it has already been committed, but the uncommitted
one is now completely gone, having been terminated by rolling back the change.
8

delete from regions where
region_id=100;

delete from regions where
region_id=101;

9

select * from regions;

select * from regions;

Each deleted row is still visible in the session that did not delete it, until you do the following:
10

commit;

commit;

11

select * from regions;

select * from regions;

With all transactions terminated, both sessions see a consistent view of the table.

SAVEPOINT
The use of savepoints is to allow a programmer to set a marker in a transaction
that can be used to control the effect of the ROLLBACK command. Rather than
rolling back the whole transaction and terminating it, it becomes possible to
reverse all changes made after a particular point but leave changes made before
that point intact. The transaction itself remains in progress: still uncommitted, still
rollbackable, and still invisible to other sessions.
The syntax is as follows:
SAVEPOINT savepoint;

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435

This creates a named point in the transaction that can be used in a subsequent
ROLLBACK command. The following table illustrates the number of rows in a table
at various stages in a transaction. The table is a very simple table called TAB, with
one column.
Command

Rows Visible to the User

Rows Visible to Others

truncate table tab;

0

0

insert into tab values ('one');

1

0

savepoint first;

1

0

2

0

savepoint second;

2

0

insert into tab values ('three');

3

0

rollback to savepoint second;

2

0

rollback to savepoint first;

1

0

commit;

1

1

delete from tab;

0

1

rollback;

1

1

insert into tab values ('two');

The example in the table shows two transactions: the first terminated with a
COMMIT, the second with a ROLLBACK. It can be seen that the use of savepoints
is visible only within the transaction: other sessions see nothing that is not
committed.
The SAVEPOINT command is not (yet) part of the official SQL standard, so it
may be considered good practice to avoid it in production systems. It can be
very useful in development, though, when you are testing the effect of DML
statements and walking through a complex transaction step by step.

The AUTOCOMMIT in SQL*Plus and SQL Developer
The standard behavior of SQL*Plus and SQL Developer is to follow the SQL
standard: a transaction begins implicitly with a DML statement and ends explicitly
with a COMMIT or a ROLLBACK. It is possible to change this behavior in both
tools so that every DML statement commits immediately, in its own transaction. If
this is done, there is no need for any COMMIT statements, and the ROLLBACK

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statement can never have any effect: all DML statements become permanent and
visible to others as soon as they execute.
In SQL*Plus, enable the autocommit mode with the command:
SET AUTOCOMMIT ON
To return to normal:
SET AUTOCOMMIT OFF
In SQL Developer, from the Tools menu, select Preferences. Then expand
Database and Worksheet Parameters: you will see the Autocommit in SQL
Worksheet check box.
It may be hard to justify enabling the autocommit mode of the SQL*Plus and
SQL Developer tools. Perhaps the only reason is for compatibility with some
third-party products that do not follow the SQL standard. SQL scripts written
for such products may not have any COMMIT statements.

SELECT FOR UPDATE
One last transaction control statement is SELECT FOR UPDATE. Oracle, by
default, provides the highest possible level of concurrency: readers do not block
writers, and writers do not block readers. Or in plain language, there is no problem
with one session querying data that another session is updating, or one session
updating data that another session is querying. However, there are times when you
may wish to change this behavior and prevent changes to data that is being queried.
It is not unusual for an application to retrieve a set of rows with a SELECT
command, present them to a user for perusal, and prompt him for any changes.
Because Oracle is a multiuser database, it is not impossible that another session has
also retrieved the same rows. If both sessions attempt to make changes, there can be
some rather odd effects. The following table depicts such a situation.
First User

Second User

select * from regions;

select * from regions;
delete from regions
where region_id=5;
commit;

update regions set region_name='GB'
where region_id=5;

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437

This is what the first user will see, from a SQL*Plus prompt:
SQL> select * from regions;
REGION_ID REGION_NAME
---------- ------------------------5 UK
1 Europe
2 Americas
3 Asia
4 Middle East and Africa
SQL> update regions set region_name='GB' where region_id=5;
0 rows updated.

This is a bit disconcerting. One way around this problem is to lock the rows in
which one is interested:
select * from regions for update;

INSIDE THE EXAM
Understanding Transaction Isolation
All DML statements are private to the session
that makes them, until the transaction
commits. The transaction is started implicitly
with the first DML statement executed. Until
it is committed, it can be reversed with a
ROLLBACK. No other session will ever see
changes that have not been committed, but
the instant they are committed they will be
visible to all other sessions.
Transaction structure is vital for good
programming. A transaction is a logical unit
of work: the changes made by the transaction,
whether it is one statement affecting one row
in one table, or many statements affecting

any number of rows in many tables, should
be self contained. It should not be in any way
dependent on statements executed outside
the transaction, and it should not be divisible
into smaller, self-contained transactions. A
transaction should be the right size; it should
contain all the statements that cannot be
separated in terms of business logic and no
statements that can be.
The decisions on transaction structure
may be complex, but with some thought
and investigation, they can be made for any
situation. Business and systems analysts can
and should advise.

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SCENARIO & SOLUTION
Transactions, like constraints, are business rules: a
technique whereby the database can enforce rules
developed by business analysts. If the “logical unit
of work” is huge, such as an accounting suite period
rollover, should this actually be implemented as one
transaction?

Not necessarily. Such a transaction might take hours,
occupying a vast amount of database resources. In such
cases, you must discuss with your business analysts
and DBA whether it is possible to break up the
one business transaction into a number of database
transactions. Of course, if something goes wrong
partway through, you will have an accounting suite
that is partly in one period and partly in another. This
application will need to be able to sort out the mess.

Being able to do DML operations, look at the result,
then roll back and try them again can be very useful.
But is it really a good idea?

No, not really. If the application is designed so that
end users can do this, the DBA will not be happy.
He will see many transactions being rolled back,
which stresses the database. It is much better for the
application to do all such work on the client side
and only submit the work to the database when it is
ready and can be committed immediately.

The FOR UPDATE clause will place a lock on all the rows retrieved. No changes
can be made to them by any session other than that which issued the command, and
therefore the subsequent updates will succeed: it is not possible for the rows to have
been changed. This means that one session will have a consistent view of the data
(it won’t change), but the price to be paid is that other sessions will hang if they try
to update any of the locked rows (they can, of course, query them).
The locks placed by a FOR UPDATE clause will be held until the session issuing
the command issues a COMMIT or ROLLBACK. This must be done to release the
locks, even if no DML commands have been executed.

CERTIFICATION SUMMARY
There are four DML commands that affect data: INSERT, UPDATE, DELETE,
and (the optional command) MERGE. TRUNCATE is a DDL command that is
functionally equivalent to a DELETE command without a WHERE clause, but it
is far faster. All DML commands can be rolled back, either automatically in the
case of error, or manually with the ROLLBACK command—unless they have been
committed with a COMMIT. Once committed, the changes can never be reversed.
TRUNCATE, like all DDL commands, has a built-in COMMIT that is unstoppable.

Two-Minute Drill

✓

439

TWO-MINUTE DRILL
Describe Each Data Manipulation Language (DML) Statement
❑ INSERT enters rows into a table.
❑ UPDATE adjusts the values in existing rows.
❑ DELETE removes rows.
❑ MERGE can combine the functions of INSERT, UPDATE, and DELETE.
❑ Even though TRUNCATE is not DML, it does remove all rows in a table.

Insert Rows into a Table
❑ INSERT can enter one row or a set of rows.
❑ It is possible for an INSERT to enter rows into multiple tables.
❑ Subqueries can be used to generate the rows to be inserted.
❑ Subqueries and functions can be used to generate column values.
❑ An INSERT is not permanent until it is committed.

Update Rows in a Table
❑ UPDATE can affect one row or a set of rows in one table.
❑ Subqueries can be used to select the rows to be updated.
❑ Subqueries and functions can be used to generate column values.
❑ An UPDATE is not permanent until it is committed.

Delete Rows from a Table
❑ DELETE can remove one row or a set of rows from one table.
❑ A subquery can be used to select the rows to be deleted.
❑ A DELETE is not permanent until it is committed.
❑ TRUNCATE removes every row from a table.
❑ A TRUNCATE is immediately permanent: it cannot be rolled back.

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Control Transactions
❑ A transaction is a logical unit of work, possibly several DML statements.
❑ Transactions are invisible to other sessions until committed.
❑ Until committed, transactions can be rolled back.
❑ Once committed, a transaction cannot be reversed.
❑ A SAVEPOINT lets a session roll back part of a transaction.

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441

SELF TEST
The following questions will help you measure your understanding of the material presented in this
chapter. Read all the choices carefully because there might be more than one correct answer. Choose
all the correct answers for each question.

Describe Each Data Manipulation Language (DML) Statement
1. Which of the following commands can be rolled back?
A. COMMIT
B. DELETE
C. INSERT
D. MERGE
E. TRUNCATE
F. UPDATE
2. How can you change the primary key value of a row? (Choose the best answer.)
A. You cannot change the primary key value.
B. Change it with a simple UPDATE statement.
C. The row must be removed with a DELETE and reentered with an INSERT.
D. This is only possible if the row is first locked with a SELECT FOR UPDATE.
3. If an UPDATE or DELETE command has a WHERE clause that gives it a scope of several
rows, what will happen if there is an error part way through execution? The command is one of
several in a multistatement transaction. (Choose the best answer.)
A. The command will skip the row that caused the error and continue.
B. The command will stop at the error, and the rows that have been updated or deleted will
remain updated or deleted.
C. Whatever work the command had done before hitting the error will be rolled back, but
work done already by the transaction will remain.
D. The whole transaction will be rolled back.

Insert Rows into a Table
4. If a table T1 has four numeric columns, C1, C2, C3, and C4, which of these statements will
succeed? (Choose the best answer.)
A. insert into T1 values (1,2,3,null);
B. insert into T1 values (‘1’,‘2’,‘3’,‘4’);
C. insert into T1 select * from T1;

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D. All the statements (A, B, and C) will succeed.
E. None of the statements (A, B, or C) will succeed.
5. Study the result of this SELECT statement:
SQL> select * from t1;
C1
C2
C3
C4
---------- ---------- ---------- ---------1
2
3
4
5
6
7
8

If you issue this statement:
insert into t1 (c1,c2) values(select c1,c2 from t1);

why will it fail? (Choose the best answer.)
A. Because values are not provided for all the table’s columns: there should be NULLs for C3
and C4.
B. Because the subquery returns multiple rows: it requires a WHERE clause to restrict the
number of rows returned to one.
C. Because the subquery is not scalar: it should use MAX or MIN to generate scalar values.
D. Because the VALUES keyword is not used with a subquery.
E. It will succeed, inserting two rows with NULLs for C3 and C4.
6. Consider this statement:
insert into regions (region_id,region_name)
values ((select max(region_id)+1 from regions), 'Great Britain');

What will the result be? (Choose the best answer.)
A. The statement will not succeed if the value generated for REGION_ID is not unique,
because REGION_ID is the primary key of the REGIONS table.
B. The statement has a syntax error because you cannot use the VALUES keyword with
a subquery.
C. The statement will execute without error.
D. The statement will fail if the REGIONS table has a third column.

Update Rows in a Table
7. You want to insert a row and then update it. What sequence of steps should you follow?
(Choose the best answer.)
A. INSERT, UPDATE, COMMIT
B. INSERT, COMMIT, UPDATE, COMMIT

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443

C. INSERT, SELECT FOR UPDATE, UPDATE, COMMIT
D. INSERT, COMMIT, SELECT FOR UPDATE, UPDATE, COMMIT
8. If you issue this command:
update employees set salary=salary * 1.1;

what will be the result? (Choose the best answer.)
A. The statement will fail because there is no WHERE clause to restrict the rows affected.
B. The first row in the table will be updated.
C. There will be an error if any row has its SALARY column NULL.
D. Every row will have SALARY incremented by 10 percent, unless SALARY was NULL.

Delete Rows from a Table
9. How can you delete the values from one column of every row in a table? (Choose the best answer.)
A. Use the DELETE COLUMN command.
B. Use the TRUNCATE COLUMN command.
C. Use the UPDATE command.
D. Use the DROP COLUMN command.
10. Which of these commands will remove every row in a table? (Choose one or more correct
answers.)
A. A DELETE command with no WHERE clause
B. A DROP TABLE command
C. A TRUNCATE command
D. An UPDATE command, setting every column to NULL and with no WHERE clause

Control Transactions
11. User JOHN updates some rows and asks user ROOPESH to log in and check the changes before
he commits them. Which of the following statements is true? (Choose the best answer.)
A. ROOPESH can see the changes but cannot alter them because JOHN will have locked
the rows.
B. ROOPESH will not be able to see the changes.
C. JOHN must commit the changes so that ROOPESH can see them and, if necessary, roll
them back.
D. JOHN must commit the changes so that ROOPESH can see them, but only JOHN can roll
them back.

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12. User JOHN updates some rows but does not commit the changes. User ROOPESH queries the
rows that JOHN updated. Which of the following statements is true? (Choose three correct
answers.)
A. ROOPESH will not be able to see the rows because they will be locked.
B. ROOPESH will be able to see the new values, but only if he logs in as JOHN.
C. ROOPESH will see the old versions of the rows.
D. ROOPESH will see the state of the state of the data as it was when JOHN last created
a SAVEPOINT.
13. Which of these commands will terminate a transaction? (Choose three correct answers.)
A. COMMIT
B. DELETE
C. ROLLBACK
D. ROLLBACK TO SAVEPOINT
E. SAVEPOINT
F. TRUNCATE

LAB QUESTION
Carry out this exercise in the OE schema.
1. Insert a customer into CUSTOMERS, using a function to generate a unique customer number:
insert into customers
(customer_id,cust_first_name,cust_last_name)
values((select max(customer_id)+1 from customers),'John','Watson');

2. Give him a credit limit equal to the average credit limit:
update customers set
credit_limit=(select avg(credit_limit) from customers)
where cust_last_name='Watson';

3. Create another customer using the customer just created, but make sure the CUSTOMER_ID is
unique:
insert into customers
(customer_id,cust_first_name,cust_last_name,credit_limit)
select customer_id+1,cust_first_name,cust_last_name,credit_limit
from customers
where cust_last_name='Watson';

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445

4. Change the name of the second entered customer:
update customers
set cust_last_name='Ramklass',cust_first_name='Roopesh'
where customer_id=(select max(customer_id) from customers);

5. Commit this transaction:
commit;

6. Determine the CUSTOMER_IDs of the two new customers and lock the rows:
7. select customer_id,cust_last_name from customers
where cust_last_name in (‘Watson’,‘Ramklass’) for update;
From another session connected to the OE schema, attempt to update one of the locked rows:
update customers set credit_limit=0 where cust_last_name='Ramklass';

8. This command will hang. In the first session, release the locks by issuing a commit:
commit;

9. The second session will now complete its update. In the second session, delete the two rows:
delete from customers where cust_last_name in ('Watson','Ramklass');

10. In the first session, attempt to truncate the CUSTOMERS table:
truncate table customers;

11. This will fail because there is a transaction in progress against the table, which will block all
DDL commands. In the second session, commit the transaction:
commit;

12. The CUSTOMERS table will now be back in the state it was in at the start of the exercise.
Confirm this by checking the value of the highest CUSTOMER_ID:
select max(customer_id) from customers;

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SELF TEST ANSWERS
Describe Each Data Manipulation Language (DML) Statement
✓ B, C, D, F. These are the DML commands: they can all be rolled back.
1. ®

®
˚ A, E. COMMIT terminates a transaction, which can then never be rolled back.
TRUNCATE is a DDL command and includes a built-in COMMIT.
✓ B. Assuming no constraint violations, the primary key can updated like any other column.
2. ®

®
˚ A, C, D. A is wrong because there is no restriction on updating primary keys (other than
constraints). C is wrong because there is no need to do it in such a complex manner. D is wrong
because the UPDATE will apply its own lock: you do not have to lock the row first.
✓ C. This is the expected behavior: the statement is rolled back, and the rest of the
3. ®
transaction remains uncommitted.

®
˚ A, B, D. A is wrong because, while this behavior is in fact configurable, it is not enabled
by default. B is wrong because, while this is in fact possible in the event of space errors, it is
not enabled by default. D is wrong because only the one statement will be rolled back, not the
whole transaction.

Insert Rows into a Table
✓ D. A, B, and C will all succeed, even though B will force the database to do some
4. ®
automatic type casting.

®
˚ A, B, C, E. A, B, and C are wrong because each one will succeed. E is wrong because A, B,
and C will all succeed.
✓ D. The syntax is wrong: use either the VALUES keyword or a subquery, but not both.
5. ®
Remove the VALUES keyword, and it will run. C3 and C4 would be populated with NULLs.

®
˚ A, B, C, E. A is wrong because there is no need to provide values for columns not listed.
B and C are wrong because an INSERT can insert a set of rows, so there is no need to restrict
the number with a WHERE clause or by using MAX or MIN to return only one row. E is wrong
because the statement is not syntactically correct.
✓ C. The statement is syntactically correct, and the use of “MAX(REGION_ID) + 1”
6. ®
guarantees generating a unique number for the primary key column.

®
˚ A, B, D. A is wrong because the function will generate a unique value for the primary
key. B is wrong because there is no problem using a scalar subquery to generate a value for a
VALUES list. What cannot be done is to use the VALUES keyword and then a single nonscalar
subquery to provide all the values. D is wrong because if there is a third column, it will be
populated with a NULL value.

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447

Update Rows in a Table
✓ A. This is the simplest (and therefore the best) way.
7. ®

®
˚ B, C, D. All these will work, but they are all needlessly complicated: no programmer should
use unnecessary statements.
✓ D. Any arithmetic operation on a NULL returns a NULL, but all other rows will be updated.
8. ®

®
˚ A, B, C. A and B are wrong because the lack of a WHERE clause means that every row
will be processed. C is wrong because trying to do arithmetic against a NULL is not an error
(though it isn’t very useful, either).

Delete Rows from a Table
✓ C. An UPDATE, without a WHERE clause, is the only way.
9. ®

®
˚ A, B, D. A is wrong because there is no such syntax: a DELETE affects the whole row.
B is wrong because there is no such syntax: a TRUNCATE affects the whole table. D is wrong
because, while this command does exist (it is part of the ALTER TABLE command), it will
remove the column completely, not just clear the values out of it.
✓ A, C. The TRUNCATE will be faster, but the DELETE will get there too.
10. ®

®
˚ B is wrong because this will remove the table as well as the rows within it. D is wrong
because the rows will still be there—even though they are populated with NULLs.

Control Transactions
✓ B. The principle of isolation means that only JOHN can see his uncommitted transaction.
11. ®

®
˚ A, C, D. A is wrong because transaction isolation means that no other session will be able to
see the changes. C and D are wrong because a committed transaction can never be rolled back.
✓ C. Transaction isolation means that no other session will be able to see the changes until
12. ®
they are committed.

®
˚ A, B, D. A is wrong because locking is not relevant; writers do not block readers. B is
wrong because isolation restricts visibility of in-progress transactions to the session making the
changes; the schema the users are connecting to does not matter. D is wrong because savepoints
are only markers in a transaction; they do not affect publishing changes to other sessions.
✓ A, C, F. COMMIT and ROLLBACK are the commands to terminate a transaction
13. ®
explicitly; TRUNCATE will do it implicitly.

®
˚ B, D, E. B is wrong because DELETE is a DML command that can be executed within a
transaction. D and E are wrong because creating savepoints and rolling back to them leave the
transaction in progress.

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LAB ANSWER
Figure 10-5 shows the first five steps of the exercise.
Figure 10-6 shows the final seven steps of the exercise, as seen from the point of view of the first
session.

FIGURE 10-5

Steps 1 through 5
of the lab exercise

FIGURE 10-6

The final steps of
the lab exercise,
including the
TRUNCATE
error

11
Using DDL
Statements to Create
and Manage Tables

CERTIFICATION OBJECTIVES
11.01

Categorize the Main Database Objects

11.02

Review the Table Structure

11.03

List the Data Types That Are Available
for Columns

11.04

11.05

✓
Q&A

Explain How Constraints Are Created
at the Time of Table Creation
Two-Minute Drill
Self Test

Create a Simple Table

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

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T

here are several types of data objects in a database that can be addressed by users with
SQL. The most commonly used type of object is the table. Tables come in various forms,
but SQL is not aware of this. A table may also be associated with other objects such as
indexes or LOBs (a large object— a structure designed for storing big items of information, such
as video recordings) that are addressed implicitly. The statement will address only the table with
which the other objects are associated. This chapter details table creation; Chapter 12 will cover
some other object types.
When creating a table, there are certain rules that must be followed regarding
the table’s structure: its columns may only be of certain data types. There are also
rules that can be defined for the individual rows; these are known as constraints.
The structural rules and the constraint rules together restrict the data that can be
inserted into the table.

CERTIFICATION OBJECTIVE 11.01

Categorize the Main Database Objects
There are various types of objects that can exist within a database, many more with the
current release than with earlier versions. All objects have a names, and all objects
are owned by someone. The “someone” is a database user, such as HR. The objects the
user owns are their schema. An object’s name must conform to certain rules.

Object Types
This query lists the object types that happen to exist in this particular database, with
a count of how many there are:
SQL> select object_type,count(object_type) from dba_objects
2 group by object_type order by object_type;
OBJECT_TYPE
COUNT(OBJECT_TYPE)
------------------- -----------------CLUSTER
10
CONSUMER GROUP
12
CONTEXT
6
DIMENSION
5
DIRECTORY
9

Categorize the Main Database Objects

EDITION
EVALUATION CONTEXT
FUNCTION
INDEX
INDEX PARTITION
INDEXTYPE
JAVA CLASS
JAVA DATA
JAVA RESOURCE
JOB
JOB CLASS
LIBRARY
LOB
LOB PARTITION
MATERIALIZED VIEW
OPERATOR
PACKAGE
PACKAGE BODY
PROCEDURE
PROGRAM
QUEUE
RESOURCE PLAN
RULE
RULE SET
SCHEDULE
SEQUENCE
SYNONYM
TABLE
TABLE PARTITION
TRIGGER
TYPE
TYPE BODY
UNDEFINED
VIEW
WINDOW
WINDOW GROUP
XML SCHEMA
42 rows selected.

451

1
13
286
3023
342
12
22018
322
820
11
11
177
769
7
3
60
1240
1178
118
17
37
7
1
21
2
204
26493
2464
199
413
2630
231
6
4669
9
4
93

This query addresses the view DBA_OBJECTS, which has one row for every object in
the database. The numbers are low, because the database is a very small one used only for
teaching. A database used for a business application might have hundreds of thousands
of objects. You may not be able to see the view DBA_OBJECTS, depending on what
permissions your account has. Alternate views are USER_OBJECTS, which will show all
the objects owned by you, and ALL_OBJECTS, which will show all the objects to which
you have been granted access (including your own). All users have access to these.

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The objects of greatest interest to a SQL programmer are those that contain, or
give access to, data. These are
■ Tables
■ Views
■ Synonyms
■ Indexes
■ Sequences

This chapter covers tables; the others are described in Chapter 12. Briefly, a view
is a stored SELECT statement that can be addressed as though it were a table. It is
nothing more than a SELECT statement but, rather than running the statement
itself, the user issues a SELECT statement against the view instead. In effect, the
user is selecting from the result of another selection. A synonym is an alias for
a table (or a view). Users can execute SQL statements against the synonym, and
the database will map them into statements against the object to which the synonym
points. Indexes are a means of improving access times to rows in tables. If a query
requires only one row, then rather than scanning the entire table to find the row,
an index can give a pointer to the row’s exact location. Of course, the index itself
must be searched, but this is often faster than scanning the table. A sequence is
a construct that generates unique numbers. There are many cases where unique
numbers are needed. Sequences issue numbers in order, on demand: it is absolutely
impossible for the same number to be issued twice.
The remaining object types are less commonly relevant to a SQL programmer. Their
use falls more within the realm of PL/SQL programmers and database administrators.

Users and Schemas
Many people use the terms “user” and “schema” interchangeably. In the Oracle
environment, you can get away with this (though not necessarily with other
database management systems). A user is a person who can connect to the database.
The user will have a username and a password. A schema is a container for
the objects owned by a user. When a user is created, their schema is created too.
A schema is the objects owned by a user; initially, it will be empty.
Some schemas will always be empty: the user will never create any objects, because
they do not need to and (if the user is set up correctly) will not have the necessary
privileges anyway. Users such as this will have been granted permissions, either through
direct privileges or through roles, to use code and access data in other schemas, owned
by other users. Other users may be the reverse of this: they will own many objects but

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453

will never actually log on to the database. They need not even have been granted the
CREATE SESSION privilege, so the account is effectively disabled (or indeed it can be
locked)—these schemas are used as repositories for code and data accessed by others.
Schema objects are objects with an owner. The unique identifier for an object of
a particular type is not its name—it is its name, prefixed with the name of the schema
to which it belongs. Thus the table HR.REGIONS is a table called REGIONS, which
is owned by user HR. There could be another table SYSTEM.REGIONS that would
be a completely different table (perhaps different in both structure and contents)
owned by user SYSTEM and residing in his or her schema.
A number of users (and their associated schemas) are created automatically at
database creation time. Principal among these are SYS and SYSTEM. User SYS owns
the data dictionary: a set of tables (in the SYS schema) that define the database and
its contents. SYS also owns several hundred PL/SQL packages: code that is provided
for the use of database administrators and developers. Objects in the SYS schema
should never be modified with DML commands. If you were to execute DML against
the data dictionary tables, you would run the risk of corrupting the data dictionary,
with disastrous results. You update the data dictionary by running DDL commands
(such as CREATE TABLE), which provide a layer of abstraction between you and the
data dictionary itself. The SYSTEM schema stores various additional objects used for
administration and monitoring.
Depending on the options selected during database creation, there may be more
users created—perhaps up to 30 in total. These others are used for storing the code
and data required by various options. For example, the user MDSYS stores the objects
used by Oracle Spatial, an option that extends the capabilities of the Oracle database
to manage geographical information.

Naming Schema Objects
A schema object is an object that is owned by a user. All schema object names must
conform to certain rules:
■ The name may be from 1 to 30 characters long (with the exception of

database link names that may be up to 128 characters long).
■ Reserved words (such as SELECT) cannot be used as object names.
■ All names must begin with a letter from “A” through “Z.”
■ The characters in a name can only be letters, numbers, an underscore (_),

the dollar sign ($), or the hash symbol (#).
■ Lowercase letters will be converted to uppercase.

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Object names must be no
more than 30 characters.The characters
can be letters, digits, underscore, dollar,
or hash.

By enclosing the name within double quotes,
all these rules (with the exception of the
length) can be broken, but to get to the object,
subsequently, it must always be specified with
double quotes, as in the examples in Figure 11-1.
Note that the same restrictions also apply to
column names.

Although tools such as SQL*Plus and SQL Developer will automatically convert
lowercase letters to uppercase unless the name is enclosed within double quotes;
remember that object names are always case sensitive. In this example, the two
tables are completely different:
SQL> create table lower(c1 date);
Table created.
SQL> create table "lower"(col1 varchar2(2));
Table created.
SQL> select table_name from dba_tables where
2 lower(table_name) = 'lower';
TABLE_NAME
-----------------------------lower
LOWER

FIGURE 11-1

Using double
quotes to use
nonstandard
names

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455

While it is possible to use lowercase names and nonstandard characters (even
spaces), it is considered bad practice because of the confusion it can cause.

Object Namespaces
It is often said that the unique identifier for an object is the object name, prefixed
with the schema name. While this is generally true, for a full understanding of
naming, it is necessary to introduce the concept of a namespace. A namespace defines
a group of object types, within which all names must be uniquely identified,
by schema and name. Objects in different namespaces can share the same name.
These object types all share the same namespace:
■ Tables
■ Views
■ Sequences
■ Private synonyms

Thus it is impossible to create a view with the same name as a table—at least, it
is impossible if they are in the same schema. And once created, SQL statements can
address a view or a synonym as though it were a table. The fact that tables, views,
and private synonyms share the same namespace means that you can set up several
layers of abstraction between what the users see and the actual tables, which can be
invaluable for both security and for simplifying application development. Indexes
and constraints each have their own namespace. Thus it is possible for an index to
have the same name as a table, even within the same schema.

EXERCISE 11-1
Determine What Objects Are Accessible to Your Session
In this exercise, query various data dictionary views as user HR to determine what
objects are in the HR schema and what objects in other schemas HR has access to.
1. Connect to the database with SQL*Plus or SQL Developer as user HR.
2. Determine how many objects of each type are in the HR schema:
select object_type,count(*) from user_objects group by object_type;

The USER_OBJECTS view lists all objects owned by the schema to which
the current session is connected, in this case HR.

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3. Determine how many objects in total HR has permissions on:
select object_type,count(*) from all_objects group by object_type;

The ALL_OBJECTS view lists all objects to which the user has some sort
of access.
4. Determine who owns the objects HR can see:
select distinct owner from all_objects;

CERTIFICATION OBJECTIVE 11.02

Review the Table Structure
According to the relational database paradigm, a table is a two-dimensional
structure storing rows. A row is one or more columns. Every row in the table has the
same columns, as defined by the structure of the table. The Oracle database does
permit variations on this two-dimensional model. Some columns can be defined as
nested tables, which themselves have several columns. Other columns may be of an
unbounded data type such as a binary large object, theoretically terabytes big. It is
also possible to define columns as objects. The object will have an internal structure
(possibly based on columns) that is not visible as part of the table.
The systems analysis phase of the system development lifecycle will have modeled
the data structures needed to store the system’s information into third normal form,
as described in Chapter 1. The result is a set of two-dimensional tables, each with
a primary key and linked to each other with foreign keys. The system design phase
may have compromised this structure, perhaps by denormalizing the tables or by
taking advantage of Oracle-specific capabilities such as nested tables. But the end
result, as far as the SQL developer is concerned, is a set of tables.
Each table exists as a definition in the data dictionary. On creation, the table will
have been assigned a limited amount of space (known as an extent) within the database.
This may be small, perhaps only a few kilobytes or megabytes. As rows are inserted into
the table, this extent will fill. When it is full, the database will (automatically) assign
another extent to the table. As rows are deleted, space within the assigned extents
becomes available for reuse. Even if every row is deleted, the extents remain allocated
to the table. They will only be freed up and returned to the database for use elsewhere
if the table is dropped or truncated (as described in Chapter 10).

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457

EXERCISE 11-2
Investigate Table Structures
In this exercise, query various data dictionary views as user HR to determine the
structure of a table.
1. Connect to the database with SQL*Plus or SQL Developer as user HR.
2. Determine the names and types of tables that exist in the HR schema:
select table_name,cluster_name,iot_type from user_tables;

Clustered tables and index organized tables (IOTs) are advanced table
structures. In the HR schema, all tables are standard heap tables except for
COUNTRIES which is an IOT.
3. Use the DESCRIBE command to display the structure of a table:
describe regions;

4. Retrieve similar information by querying a data dictionary view:
select column_name,data_type,nullable from user_tab_columns
where table_name='REGIONS';

CERTIFICATION OBJECTIVE 11.03

List the Data Types That Are Available for Columns
When creating tables, each column must be assigned a data type, which determines
the nature of the values that can be inserted into the column. These data types are
also used to specify the nature of the arguments for PL/SQL procedures and functions.
When selecting a data type, you must consider the data that you need to store and
the operations you will want to perform upon it. Space is also a consideration: some
data types are fixed length, taking up the same number of bytes no matter what data
is actually in it; others are variable. If a column is not populated, then Oracle will not
give it any space at all. If you later update the row to populate the column, then
the row will get bigger, no matter whether the data type is fixed length or variable.

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The following are the data types for alphanumeric data:
Variable-length character data, from 1 byte to 4KB. The data
is stored in the database character set.

■ VARCHAR2

Like VARCHAR2, but the data is stored in the alternative
national language character set, one of the permitted Unicode character sets.

■ NVARCHAR2

Fixed-length character data, from 1 byte to 2KB, in the database
character set. If the data is not the length of the column, then it will be
padded with spaces.

■ CHAR

For ISO/ANSI compliance, you can specify a VARCHAR data type, but any
columns of this type will be automatically converted to VARCHAR2.
The following is the data type for binary data:
Variable-length binary data, from 1 byte to 4KB. Unlike the CHAR
and VARCHAR2 data types, RAW data is not converted by Oracle Net from
the database’s character set to the user process’s character set on SELECT or
the other way on INSERT.

■ RAW

The following are the data types for numeric data, all variable length:
Numeric data, for which you can specify precision and scale.
The precision can range from to 1 to 38, the scale can range from −84 to 127.

■ NUMBER

This is an ANSI data type, floating-point number with precision
of 126 binary (or 38 decimal). Oracle also provides BINARY_FLOAT and
BINARY_DOUBLE as alternatives.

■ FLOAT

■ INTEGER

Equivalent to NUMBER, with scale zero.

The following are the data types for date and time data, all fixed length:
This is either length zero, if the column is empty, or 7 bytes. All
DATE data includes century, year, month, day, hour, minute, and second.
The valid range is from January 1, 4712 BC to December 31, 9999 AD.

■ DATE

This is length zero if the column is empty, or up to 11 bytes,
depending on the precision specified. Similar to DATE, but with precision of
up to 9 decimal places for the seconds, 6 places by default.

■ TIMESTAMP

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459

Like TIMESTAMP, but the data
is stored with a record kept of the time zone to which it refers. The length
may be up to 13 bytes, depending on precision. This data type lets Oracle
determine the difference between two times by normalizing them to UTC,
even if the times are for different time zones.

■ TIMESTAMP WITH TIMEZONE

Like TIMESTAMP, but the
data is normalized to the database time zone on saving. When retrieved, it is
normalized to the time zone of the user process selecting it.

■ TIMESTAMP WITH LOCAL TIMEZONE

Used for recording a period in years and
months between two DATEs or TIMESTAMPs.

■ INTERVAL YEAR TO MONTH

Used for recording a period in days and
seconds between two DATEs or TIMESTAMPs.

■ INTERVAL DAY TO SECOND

The following are the large object data types:
Character data stored in the database character set, size effectively
unlimited: 4GB multiplied by the database block size.

■ CLOB

Like CLOB, but the data is stored in the alternative national
language character set, one of the permitted Unicode character sets.

■ NCLOB

Like CLOB, but binary data that will not undergo character set
conversion by Oracle Net.

■ BLOB

A locator pointing to a file stored on the operating system of the
database server. The size of the files is limited to 4GB.

■ BFILE

Character data in the database character set, up to 2GB. All the
functionality of LONG (and more) is provided by CLOB; LONGs should
not be used in a modern database, and if your database has any columns of
this type they should be converted to CLOB. There can only be one LONG
column in a table.

■ LONG

Like LONG, but binary data that will not be converted by
Oracle Net. Any LONG RAW columns should be converted to BLOBs.

■ LONG RAW

The following is the ROWID data type:
A value coded in base 64 that is the pointer to the location of a
row in a table. Encrypted. Within it is the exact physical address. ROWID is
an Oracle proprietary data type, not visible unless specifically selected.

■ ROWID

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The VARCHAR2 data type must be qualified
with a number indicating the maximum length
of the column. If a value is inserted into the
All examinees will be
column that is less than this, it is not a problem:
expected to know about these data types:
the value will only take up as much space as it
VARCHAR2, CHAR, NUMBER, DATE,
needs. If the value is longer than this maximum,
TIMESTAMP, INTERVAL, RAW, LONG,
the INSERT will fail with an error. If the value
LONG RAW, CLOB, BLOB, BFILE, and
is updated to a longer or shorter value, the
ROWID. Detailed knowledge will also
length of the column (and therefore the row
be needed for VARCHAR2, NUMBER,
itself) will change accordingly. If is not entered
and DATE.
at all or is updated to NULL, then it will take
up no space at all.
The NUMBER data type may optionally be qualified with a precision and a scale.
The precision sets the maximum number of digits in the number, and the scale is how
many of those digits are to the right of the decimal point. If the scale is negative, this
has the effect of replacing the last digits of any number inserted with zeros, which
do not count toward the number of digits specified for the precision. If the number
of digits exceeds the precision, there will be an error; if it is within the precision
but outside the scale, the number will be rounded (up or down) to the nearest value
within the scale.
The DATE data type always includes century, year, month, day, hour, minute,
and second—even if all these elements are not specified at insert time. Year, month,
and date must be specified; if the hours, minutes, and seconds are omitted they will
default to midnight. Using the TRUNC function on a date also has the effect of
setting the hours, minutes, and seconds to midnight.
Oracle provides a range of type casting functions for converting between data
types and in some circumstances will do automatic type casting. Figure 11-2
illustrates using both the manual and the automatic type casting techniques.
In the preceding example, the first INSERT uses type casting functions to convert
the character data entered to the data types specified for the table columns. The second
INSERT attempts to insert character strings into all three columns, but the insert still
succeeds because Oracle can convert data types automatically if necessary—but only if
the format of the data is suitable. Note that if the value for the date has been entered in
any format other than DD-MM-YY, such as ’18-Nov-07,’ it would have failed.
Do not rely on automatic type casting. It can impact performance and may
not always work.The Oracle environment is strongly typed, and programmers
should respect this.

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461

FIGURE 11-2

Use of type
casting functions
and automatic
type casting

EXERCISE 11-3
Investigate the Data Types in the HR schema
In this exercise, find out what data types are used in the tables in the HR schema,
using two techniques.
1. Connect to the database as user HR with SQL*Plus or SQL Developer.
2. Use the DESCRIBE command to show the data types in some tables:
describe employees;
describe departments;

3. Use a query against a data dictionary view to show what columns make up
the EMPLOYEES table, as the DESCRIBE command would:
select column_name,data_type,nullable,data_length,data_
precision,data_scale from user_tab_columns where table_
name='EMPLOYEES';

The view USER_TAB_COLUMNS shows the detail of every column in every table
in the current user’s schema.

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CERTIFICATION OBJECTIVE 11.04

Create a Simple Table
Tables can be stored in the database in several ways. The simplest is the heap table.
A heap is variable length rows in random order. There may be some correlation
between the order in which rows are entered and the order in which they are stored,
but this is a matter of luck. More advanced table structures, such as the following,
may impose ordering and grouping on the rows or force a random distribution:
■ Index organized tables

Store rows in the order of an index key.

Can denormalize tables in parent-child relationships so that
related rows from different table are stored together.

■ Index clusters

Force a random distribution of rows, which will break down
any ordering based on the entry sequence.

■ Hash clusters

Store rows in separate physical structures, the partitions,
allocating rows according to the value of a column.

■ Partitioned tables

Using the more advanced table structures has no effect whatsoever on SQL. Every
SQL statement executed against tables defined with these options will return exactly
the same results as though the tables were standard heap tables, so use of these
features will not affect code. But while their use is transparent to programmers, they
do give enormous benefits in performance.

Creating Tables with Column Specifications
To create a standard heap table, use this syntax:
CREATE TABLE [schema.]table [ORGANIZATION HEAP]
(column datatype [DEFAULT expression]
[,column datatype [DEFAULT expression]…);

As a minimum, specify the table name (it will be created in your own schema, if
you don’t specify someone else’s) and at least one column with a data type. There are
very few developers who ever specify ORGANIZATION HEAP, as this is the default
and is industry standard SQL. The DEFAULT keyword in a column definition lets
you provide an expression that will generate a value for the column when a row is
inserted if a value is not provided by the INSERT statement.

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463

Consider this statement:
CREATE TABLE SCOTT.EMP
(EMPNO NUMBER(4),
ENAME VARCHAR2(10),
HIREDATE DATE DEFAULT TRUNC(SYSDATE),
SAL NUMBER(7,2),
COMM NUMBER(7,2) DEFAULT 0.03);

This will create a table called EMP in the SCOTT schema. Either user SCOTT
himself has to issue the statement (in which case nominating the schema would
not actually be necessary), or another user could issue it if he has been granted
permission to create tables in another user’s schema. Taking the columns one by one:
■ EMPNO can be 4 digits long, with no decimal places. If any decimals are

included in an INSERT statement, they will be rounded (up or down) to
the nearest integer.
■ ENAME can store any characters at all, up to ten of them.
■ HIREDATE will accept any date, optionally with the time, but if a value is

not provided, today’s date will be entered as at midnight.
■ SAL, intended for the employee’s salary, will accept numeric values with up

to 7 digits. If any digits over 7 are to the right of the decimal point, they will
be rounded off.
■ COMM (for commission percentage) has a default value of 0.03, which will

be entered if the INSERT statement does not include a value for this column.
Following creation of the table, these statements insert a row and select the
result:
SQL> insert into scott.emp(empno,ename,sal) values(1000,'John',1000.789);
1 row created.
SQL> select * from emp;
EMPNO ENAME
HIREDATE
SAL
COMM
---------- ---------- --------- ---------- ---------1000 John
19-NOV-07
1000.79
.03

Note that values for the columns not mentioned in the INSERT statement have
been generated by the DEFAULT clauses. Had those clauses not been defined in the
table definition, the columns would have been NULL. Also note the rounding of
the value provided for SAL.

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The DEFAULT clause can be useful, but it is of limited functionality. You
cannot use a subquery to generate the default value: you can only specify
literal values or functions.

Creating Tables from Subqueries
Rather than creating a table from nothing and then inserting rows into it (as in
the previous section), tables can be created from other tables by using a subquery.
This technique lets you create the table definition and populate the table with rows
with just one statement. Any query at all can be used as the source of both the table
structure and the rows. The syntax is as follows:
CREATE TABLE [schema.]table AS subquery;
All queries return a two-dimensional set of rows; this result is stored as the new
table. A simple example of creating a table with a subquery is:
create table employees_copy as select * from employees;

This statement will create a table EMPLOYEES_COPY, which is an exact copy
of the EMPLOYEES table, identical in both definition and the rows it contains. Any
not null and check constraints on the columns will also be applied to the new table,
but any primary-key, unique, or foreign-key constraints will not be. (Constraints
are discussed in section 11.05, “Explain How Constraints Are Created at the Time
of Table Creation.”) This is because these three types of constraints require indexes
that might not be available or desired.
The following is a more complex example:
create table emp_dept as select
last_name ename,department_name dname,round(sysdate – hire_date) service
from employees natural join departments order by dname,ename;

The rows in the new table will be the result of joining the two source tables,
with two of the selected columns having their names changed. The new SERVICE
column will be populated with the result of the arithmetic that computes the
number of days since the employee was hired. The rows will be inserted in the order
specified. This ordering will not be maintained by subsequent DML but, assuming
the standard HR schema data, the new table will look like this:
SQL> select * from emp_dept where rownum < 10;
ENAME
DNAME
SERVICE
--------------- --------------- ---------Gietz
Accounting
4914

Create a Simple Table

De Haan
Executive
Kochhar
Executive
Chen
Finance
Faviet
Finance
Popp
Finance
Sciarra
Finance
Urman
Finance
Austin
IT
9 rows selected.

465

5424
6634
3705
4844
2905
3703
3545
3800

The subquery can of course include a WHERE clause to restrict the rows inserted
into the new table. To create a table with no rows, use a WHERE clause that will
exclude all rows:
create table no_emps as select * from employees where 1=2;

The WHERE clause 1=2 can never return TRUE, so the table structure will be
created ready for use, but no rows will be inserted at creation time.

Altering Table Definitions after Creation
There are many alterations that can be made to a table after creation. Those that
affect the physical storage fall into the domain of the database administrator, but
many changes are purely logical and will be carried out by the SQL developers. The
following are examples (for the most part self-explanatory):
■ Adding columns:
alter table emp add (job_id number);

■ Modifying columns:
alter table emp modify (comm number(4,2) default 0.05);

■ Dropping columns:
alter table emp drop column comm;

■ Marking columns as unused:
alter table emp set unused column job_id;

■ Renaming columns:
alter table emp rename column hiredate to recruited;

■ Marking the table as read-only:
alter table emp read only;

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All of these changes are DDL commands with the built-in COMMIT. They
are therefore nonreversible and will fail if there is an active transaction against
the table. They are also virtually instantaneous with the exception of dropping
a column. Dropping a column can be a time-consuming exercise because as each
column is dropped, every row must be restructured to remove the column’s data.
The SET UNUSED command, which makes columns nonexistent as far as SQL
is concerned, is often a better alternative, followed when convenient by
ALTER TABLE tablename DROP UNUSED COLUMNS;

which will drop all the unused columns in one pass through the table.
Marking a table as read-only will cause errors for any attempted DML commands.
But the table can still be dropped. This can be disconcerting but is perfectly logical
when you think it through. A DROP command doesn’t actually affect the table:
it affects the tables in the data dictionary that define the table, and these are not
read-only.

Dropping and Truncating Tables
The TRUNCATE TABLE command was described in Chapter 10: it has the effect
of removing every row from a table, while leaving the table definition intact. DROP
TABLE is more drastic in that the table definition is removed as well. The syntax is
as follows:
DROP TABLE [schema.]tablename ;
If schema is not specified, then the table called tablename in your currently logged
on schema will be dropped.
As with a TRUNCATE, SQL will not produce a warning before the table is
dropped, and furthermore, as with any DDL command, it includes a COMMIT.
A DROP is therefore absolutely nonreversible. But there are some restrictions: if
any session (even your own) has a transaction in progress that includes a row in
the table, then the DROP will fail, and it is also impossible to drop a table that is
referred to in a foreign key constraint defined for a another table. This table (or the
constraint) must be dropped first.

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467

EXERCISE 11-4
Create Tables
In this exercise, use SQL Developer to create a heap table, insert some rows with
a subquery, and modify the table. Do some more modifications with SQL*Plus, then
drop the table.
1. Connect to the database as user HR with SQL Developer.
2. Right-click the Tables branch of the navigation tree, and click New Table.
3. Name the new table EMPS, and use the Add Column button to set it up as
in the following illustration:

4. Click the DDL tab to see if the statement that has been constructed. It
should look like this:
CREATE TABLE EMPS
(
EMPNO NUMBER,
ENAME VARCHAR2(25),

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SALARY NUMBER,
DEPTNO NUMBER(4, 0)
)
;

Return to the Table tab (as in the preceding illustration) and click OK to
create the table.
5. Run this statement:
insert into emps select employee_id,last_name,salary,department_id
from employees;

and commit the insert:
commit;

6. Right-click the EMPS table in the SQL Developer navigator, click Column
and Add.
7. Define a new column HIRED, type DATE, as in the following illustration
below; and click Apply to create the column.

8. Connect to the database as HR with SQL*Plus.

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469

9. Define a default for HIRED column in the EMPS table:
alter table emps modify (hired default sysdate);

10. Insert a row without specifying a value for HIRED and check that the new
row does have a HIRED date but that the other rows do not:
insert into emps (empno,ename) values(99,'Newman');
select hired,count(1) from emps group by hired;

11. Tidy up by dropping the new table:
drop table emps;

CERTIFICATION OBJECTIVE 11.05

Explain How Constraints Are Created
at the Time of Table Creation
Table constraints are a means by which the database can enforce business rules, and
guarantee that the data conforms to the entity-relationship model determined by the
systems analysis that defines the application data structures. For example, the business
analysts of your organization may have decided that every customer and every invoice
must be uniquely identifiable by number, that no invoices can be issued to a customer
before that customer has been created, and that every invoice must have a valid date
and a value greater than zero. These would implemented by creating primary-key
constraints on the CUSTOMER_NUMBER column of the CUSTOMERS table and
the INVOICE_NUMBER column of the INVOICES table, a foreign-key constraint
on the INVOICES table referencing the CUSTOMERS table, a not-null constraint
on the DATE column of the INVOICES table (the DATE data type will itself ensure
that that any dates are valid automatically—it will not accept invalid dates), and
a check constraint on the AMOUNT column on the INVOICES table.
When any DML is executed against a table with constraints defined, if the DML
violates a constraint, then the whole statement will be rolled back automatically.
Remember that a DML statement that affects many rows might partially succeed
before it hits a constraint problem with a particular row. If the statement is part of
a multistatement transaction, then the statements that have already succeeded will
remain intact but uncommitted.

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The Types of Constraints
The constraint types supported by the Oracle database are as follows:
UNIQUE
NOT NULL
PRIMARY KEY
FOREIGN KEY
CHECK
Constraints have names. It is good practice to specify the names with a standard
naming convention, but if they are not explicitly named, Oracle will generate names.

Unique Constraints
A unique constraint nominates a column (or combination of columns) for which the
value must be different for every row in the table. If based on a single column, this
is known as the key column. If the constraint is composed of more than one column
(known as a composite key unique constraint) the columns do not have to be the
same data type or be adjacent in the table definition.
An oddity of unique constraints is that it is possible to enter a NULL value into
the key column(s); it is indeed possible to have any number of rows with NULL
values in their key column(s). So selecting rows on a key column will guarantee that
only one row is returned—unless you search for NULL, in which case all the rows
where the key columns are NULL will be returned.
Unique constraints are enforced by an index. When a unique constraint is
defined, Oracle will look for an index on the key column(s), and if one does not
exist it will be created. Then whenever a row is inserted, Oracle will search the
index to see if the values of the key columns are already present: if they are, it will
reject the insert. The structure of these indexes (known as B*Tree indexes) does
not include NULL values, which is why many rows with NULL are permitted: they
simply do not exist in the index. While the first purpose of the index is to enforce
the constraint, it has a secondary effect: improving performance if the key columns
are used in the WHERE clauses of SQL statements. However, selecting WHERE
key_column IS NULL cannot use the index because it doesn’t include the NULLs
and will therefore always result in a scan of the entire table.

Not Null Constraints
The not null constraint forces values to be entered into the key column. Not null
constraints are defined per column: if the business requirement is that a group of

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471

columns should all have values, you cannot define one not null constraint for the
whole group, but instead must define a not null constraint for each column.
Any attempt to insert a row without specifying values for the not null constrained
columns results in an error. It is possible to bypass the need to specify a value by
including a DEFAULT clause on the column when creating the table, as discussed in
the previous section on creating tables.

Primary Key Constraints
The primary key is the means of locating a single row in a table. The relational
database paradigm includes a requirement that every table should have a primary
key, a column (or combination of columns) that can be used to distinguish every
row. The Oracle database deviates from the paradigm (as do some other RDBMS
implementations) by permitting tables without primary keys.
Tables without primary keys are possible but not a good idea. Even if the
business rules do not require the ability to identify every row, primary keys are
often needed for maintenance work.
The implementation of a primary key constraint is, in effect, the union of a
unique constraint and a not null constraint. The key columns must have unique
values, and they may not be null. As with unique constraints, an index must exist
on the constrained column(s). If one does not exist already, an index will be created
when the constraint is defined. A table can have only one primary key. Try to
create a second, and you will get an error. A table can, however, have any number
of unique constraints and not null columns, so if there are several columns that the
business analysts have decided must be unique
and populated, one of these can be designated
the primary key and the others made unique
and not null. An example could be a table of
A primary key constraint
employees, where e-mail address, social security
is a unique constraint combined with a
number, and employee number should all be
not null constraint.
required and unique.

Foreign Key Constraints
A foreign key constraint is defined on the child table in a parent-child relationship.
The constraint nominates a column (or columns) in the child table that corresponds
to the primary key column(s) in the parent table. The columns do not have to have

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the same names, but they must be of the same data type. Foreign key constraints
define the relational structure of the database: the many-to-one relationships that
connect the table, in their third normal form.
If the parent table has unique constraints as well as (or instead of) a primary key
constraint, these columns can be used as the basis of foreign key constraints, even if
they are nullable.
Just as a unique constraint permits null values in the constrained column, so does
a foreign key constraint. You can insert rows into the child table with null foreign key
columns—even if there is not a row in the parent table with a null value. This creates
orphan rows and can cause dreadful confusion. As a general rule, all the columns in
a unique constraint and all the columns in a foreign key constraint are best defined
with not null constraints as well; this will often be a business requirement.
Attempting to inset a row in the child table for which there is no matching row
in the parent table will give an error. Similarly, deleting a row in the parent table
will give an error if there are already rows referring to it in the child table. There are
two techniques for changing this behavior. First, the constraint may be created as
ON DELETE CASCADE. This means that if a row in the parent table is deleted,
Oracle will search the child table for all the matching rows and delete them too.
This will happen automatically. A less drastic technique is to create the constraint as
ON DELETE SET NULL. In this case, if a row in the parent table is deleted, Oracle
will search the child table for all the matching rows and set the foreign key columns
to null. This means that the child rows will be orphaned, but will still exist. If the
columns in the child table also have a not null constraint, then the deletion from
the parent table will fail.
It is not possible to drop or truncate the parent table in a foreign key relationship,
even if there are no rows in the child table. This still applies if the ON DELETE
SET NULL or ON DELETE CASCADE clauses were used.
A variation on the foreign key constraint is the self-referencing foreign key constraint.
This defines a condition where the parent and child rows exist in the same table. An
example would be a table of employees, which includes a column for the employee’s
manager. The manager is himself an employee and must exist in the table. So if the
primary key is the EMPLOYEE_NUMBER
column, and the manager is identified by a
column MANAGER_NUMBER, then the
foreign key constraint will state that the value
A foreign key constraint in
of the MANAGER_NUMBER column must
a child table must reference the columns
refer back to a valid EMPLOYEE_NUMBER.
of either a unique constraint or a primary
If an employee is his own manager, then the row
key constraint in the parent table.
would refer to itself.

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Check Constraints
A check constraint can be used to enforce simple rules, such as that the value entered
in a column must be within a range of values. The rule must be an expression which
will evaluate to TRUE or FALSE. The rules can refer to absolute values entered as
literals or to other columns in the same row and may make use of some functions.
As many check constraints as you want can be applied to one column, but it is
not possible to use a subquery to evaluate whether a value is permissible or to use
functions such as SYSDATE.
The not null constraint is in fact implemented as a preconfigured check
constraint.

Defining Constraints
Constraints can be defined when creating a table or added to the table later. When
defining constraints at table creation time, the constraint can be defined in line with
the column to which it refers or at the end of the table definition. There is more
flexibility to using the latter technique. For example, it is impossible to define a
foreign key constraint that refers to two columns, or a check constraint that refers to
any column other than that being constrained if the constraint is defined in line, but
both of these are possible if the constraint is defined at the end of the table.

SCENARIO & SOLUTION
You are designing table structures for
a human resources application. The
business analysts have said that when
an employee leaves the company, his
employee record should be moved to
an archive table. Can constraints help?

Probably not. Constraints are intended to enforce simple business
rules: this may be too complicated. It may well be necessary to use
a DML trigger on the live table, which will automatically insert a
row into the archive table whenever an employee is deleted from
the live table. Triggers can do much more complicated processing
than a constraint.

Active transactions block some DDL
statements against tables. If you
want to add a constraint or rename a
column in a busy table and find the
statement always fails with “ORA00054: resource busy and acquire
with NOWAIT specified or timeout
expired,” what can you do?

Perhaps you shouldn’t be doing this sort of thing when the database
is in use, but should wait until the next period of scheduled
downtime. However, if you really need to make the change in a
hurry, ask the database administrator to quiesce the database: this is
a process that will freeze all user sessions. If you are very quick, you
can make the change then unquiesce the database before end users
complain.

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For the constraints that require an index (the unique and primary key constraints),
the index will be created with the table if the constraint is defined at table
creation time.
Consider these two table creation statements (to which line numbers have
been added):
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

create table dept(
deptno number(2,0) constraint dept_deptno_pk primary key
constraint dept_deptno_ck check (deptno between 10 and 90),
dname varchar2(20) constraint dept_dname_nn not null);
create table emp(
empno number(4,0) constraint emp_empno_pk primary key,
ename varchar2(20) constraint emp_ename_nn not null,
mgr number (4,0) constraint emp_mgr_fk references emp (empno),
dob date,
hiredate date,
deptno number(2,0) constraint emp_deptno_fk references dept(deptno)
on delete set null,
email varchar2(30) constraint emp_email_uk unique,
constraint emp_hiredate_ck check (hiredate >= dob + 365*16),
constraint emp_email_ck
check ((instr(email,'@') > 0) and (instr(email,'.') > 0)));

Taking these statements line by line:
1. The first table created is DEPT, intended to have one row for each department.
2. DEPTNO is numeric, 2 digits, no decimals. This is the table’s primary key.
The constraint is named DEPT_DEPTNO_PK.
3. A second constraint applied to DEPTNO is a check limiting it to numbers in
the range 10 to 90. The constraint is named DEPT_DEPTNO_CK.
4. The DNAME column is variable length characters, with a constraint DEPT_
DNAME_NN making it not nullable.
5. The second table created is EMP, intended to have one row for every employee.
6. EMPNO is numeric, up to 4 digits with no decimals. Constraint EMP_
EMPNO_PK marks this as the table’s primary key.
7. ENAME is variable length characters, with a constraint EMP_ENAME_NN
making it not nullable.
8. MGR is the employee’s manager, who must himself be an employee. The column is defined in the same way as the table’s primary key column of EMPNO.
The constraint EMP_MGR_FK defines this column as a self-referencing

Explain How Constraints Are Created at the Time of Table Creation

475

foreign key, so any value entered must refer to an already extant row in EMP
(though it is not constrained to be not null, so can be left blank).
9. DOB, the employee’s birthday, is a date and not constrained.
10. HIREDATE is the date the employee was hired and is not constrained. At
least, not yet.
11. DEPTNO is the department with which the employee is associated. The
column is defined in the same way as the DEPT table’s primary key column
of DEPTNO, and the constraint EMP_DEPTNO_FK enforces a foreign key
relationship: it is not possible to assign an employee to a department that
does not exist. Though this is nullable.
12. The EMP_DEPTO_FK constraint is further defined as ON DELETE SET
NULL, so if the parent row in DEPT is deleted, all matching child rows in
EMPNO will have DEPTNO set to NULL.
13. EMAIL is variable length character data, and must be unique if entered
(though it can be left empty).
14. This defines an additional table level constraint EMP_HIREDATE_CK. The
constraint checks for use of child labor by rejecting any rows where the date
of hiring is not at least 16 years later than the birthday. This constraint could
not be defined in line with HIREDATE, because the syntax does not allow
references to other columns at that point.
15. An additional constraint EMP_EMAIL_CK is added to the EMAIL column,
which makes two checks on the e-mail address. The INSTR functions search
for the at symbol (@ and dot (.) characters (which will always be present in
a valid e-mail address); if it can’t find both of them, the check condition will
return FALSE and the row will be rejected.
The preceding examples show several possibilities for defining constraints at table
creation time. The following are further possibilities not covered:
■ Controlling the index creation for the unique and primary key constraints
■ Defining whether the constraint should be checked at insert time (which it is

by default) or later on when the transaction is committed
■ Stating whether the constraint is in fact being enforced at all (which is the

default) or is disabled
It is possible to create tables with no constraints and then to add them later with
an ALTER TABLE command. The end result will be the same, but this technique

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does make the code less self documenting, as the complete table definition will then
be spread over several statements rather than being in one.

EXERCISE 11-5
Work with Constraints
Use SQL*Plus or SQL Developer to create tables, add constraints, and demonstrate
their use.
1. Connect to the database as user HR.
2. Create a table EMP as a copy of some columns from EMPLOYEES:
create table emp as
select employee_id empno, last_name ename, department_id deptno
from employees;

3. Create a table DEPT as a copy of some columns from DEPARTMENTS:
create table dept as
select department_id deptno, department_name dname from departments;

4. Use DESCRIBE to describe the structure of the new tables. Note that
the not null constraint on ENAME and DNAME has been carried over from
the source tables.
5. Add a primary key constraint to EMP and to DEPT and a foreign key constraint
linking the tables:
alter table emp add constraint emp_pk primary key (empno);
alter table dept add constraint dept_pk primary key (deptno);
alter table emp add constraint
dept_fk foreign key (deptno) references dept on delete set null;

The preceding last constraint does not specify which column of DEPT to
reference; this will default to the primary key column.
6. Demonstrate the effectiveness of the constraints by trying to insert data that
will violate them:
insert into dept values(10,'New Department');
insert into emp values(9999,'New emp',99);
truncate table dept;

Certification Summary

477

INSIDE THE EXAM
Using DDL Statements to Create and Manage Tables
The CREATE TABLE statement can be very
complex indeed. The SQL Reference volume
of the Oracle Database 11g documentation
set devotes 69 pages to it, with another 85
pages for ALTER TABLE (by contrast, DROP
TABLE takes only four pages). For examination

purposes, only knowledge of the simplest table
structure—the heap table—is required, with
knowledge of the most basic data types. Understanding, defining, and using constraints
is needed but not the methods for controlling
when (or if) they are enforced.

7. Tidy up by dropping the tables. Note that this must be done in the correct order:
drop table emp;
drop table dept;

CERTIFICATION SUMMARY
Tables are two-dimensional structures consisting of rows of columns of defined data
types. The Oracle database does permit user-defined data types, but for the most part
you will be using the built-in data types.
Tables can be created from scratch: defining every column and then inserting
rows. Alternatively, tables can be created using the output of a query. This latter
technique can define the table and insert rows in one command, but there are
limitations: it will be necessary to add primary key and unique constraints later,
whereas they can be defined at table creation time using the former technique.
To assist with enforcing business rules, constraints can be defined for columns.
These will maintain data integrity by making it absolutely impossible to insert data
that breaks the rules.

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✓

Using DDL Statements to Create and Manage Tables

TWO-MINUTE DRILL
Categorize the Main Database Objects
❑ Some objects contain data, principally tables and indexes.
❑ Programmatic objects such as stored procedures and functions are executable

code.
❑ Views and synonyms are objects that give access to other objects.

Review the Table Structure
❑ Tables are two-dimensional structures, storing rows defined with columns.
❑ Tables exist within a schema. The schema name with the table name make

a unique identifier.

List the Data Types that Are Available for Columns
❑ The most common character data types are VARCHAR2, NUMBER,

and DATE.
❑ There are many other data types.

Create a Simple Table
❑ Tables can be created from nothing or with a subquery.
❑ After creation, column definitions can be added, dropped, or modified.
❑ The table definition can include default values for columns.

Explain How Constraints Are Created
at the Time of Table Creation
❑ Constraints can be defined at table creation time or added later.
❑ A constraint can be defined inline with its column or at the table level after

the columns.
❑ Table-level constraints can be more complex than those defined inline.
❑ A table may only have one primary key but can have many unique keys.
❑ A primary key is functionally equivalent to unique plus not null.
❑ A unique constraint does not stop insertion of many null values.
❑ Foreign key constraints define the relationships between tables.

Self Test

479

SELF TEST
The following questions will help you measure your understanding of the material presented in this
chapter. Read all the choices carefully because there might be more than one correct answer. Choose
all the correct answers for each question.

Categorize the Main Database Objects
1. If a table is created without specifying a schema, in which schema will it be? (Choose the best
answer.)
A. It will be an orphaned table, without a schema.
B. The creation will fail.
C. It will be in the SYS schema.
D. It will be in the schema of the user creating it.
E. It will be in the PUBLIC schema.
2. Several object types share the same namespace, and therefore cannot have the same name in
the same schema. Which of the following object types is not in the same namespace as the
others? (Choose the best answer.)
A. Index
B. PL/SQL stored procedure
C. Synonym
D. Table
E. View
3. Which of these statements will fail because the table name is not legal? (Choose two answers.)
A. create table “SELECT” (col1 date);
B. create table “lowercase” (col1 date);
C. create table number1 (col1 date);
D. create table 1number (col1 date);
E. create table update (col1 date);

Review the Table Structure
4. What are distinguishing characteristics of heap tables? (Choose two answers.)
A. A heap can store variable length rows.
B. More than one table can store rows in a single heap.
C. Rows in a heap are in random order.

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D. Heap tables cannot be indexed.
E. Tables in a heap do not have a primary key.

List the Data Types that Are Available for Columns
5. Which of the following data types are variable length? (Choose all correct answers.)
A. BLOB
B. CHAR
C. LONG
D. NUMBER
E. RAW
F. VARCHAR2
6. Study these statements:
create table tab1 (c1 number(1), c2 date);
alter session set nls_date_format='dd-mm-yy';
insert into tab1 values (1.1,’31-01-07’);

Will the insert succeed? (Choose the best answer)
A. The insert will fail because the 1.1 is too long.
B. The insert will fail because the ’31-01-07’ is a string, not a date.
C. The insert will fail for both reasons A and B.
D. The insert will succeed.
7. Which of the following is not supported by Oracle as an internal data type? (Choose the best
answer.)
A. CHAR
B. FLOAT
C. INTEGER
D. STRING

Create a Simple Table
8. Consider this statement:
create table t1 as select * from regions where 1=2;

What will be the result? (Choose the best answer.)

Self Test

A.
B.
C.
D.

481

There will be an error because of the impossible condition.
No table will be created because the condition returns FALSE.
The table T1 will be created but no rows inserted because the condition returns FALSE.
The table T1 will be created and every row in REGIONS inserted because the condition
returns a NULL as a row filter.

9. When a table is created with a statement such as the following:
create table newtab as select * from tab;

will there be any constraints on the new table? (Choose the best answer.)
A. The new table will have no constraints, because constraints are not copied when creating
tables with a subquery.
B. All the constraints on TAB will be copied to NEWTAB.
C. Primary key and unique constraints will be copied but not check and not null constraints.
D. Check and not null constraints will be copied but not unique or primary key.
E. All constraints will be copied, except foreign key constraints.

Explain How Constraints Are Created at the Time of Table Creation
10. Which types of constraint require an index? (Choose all that apply.)
A. CHECK
B. NOT NULL
C. PRIMARY KEY
D. UNIQUE
11. A transaction consists of two statements. The first succeeds, but the second (which updates
several rows) fails partway through because of a constraint violation. What will happen?
(Choose the best answer.)
A. The whole transaction will be rolled back.
B. The second statement will be rolled back completely, and the first will be committed.
C. The second statement will be rolled back completely, and the first will remain uncommitted.
D. Only the one update that caused the violation will be rolled back; everything else will be
committed.
E. Only the one update that caused the violation will be rolled back; everything else will
remain uncommitted.

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LAB QUESTION
Consider this simple analysis of a telephone billing system:
A subscriber is identified by a customer number and also has a name and possibly one or more
telephones.
A telephone is identified by its number, which must be a 7-digit integer beginning with 2 or 3, and
also has a make, an activation date, and a flag for whether it is active. Inactive telephones are not
assigned to a subscriber; active telephones are.
For every call, it is necessary to record the time it started and the time it finished.
Create tables with constraints and defaults that can be used to implement this system.

Self Test Answers

483

SELF TEST ANSWERS
Categorize the Main Database Objects
✓ D. The schema will default to the current user.
1. ®
®
˚ A, B, C, E. A is wrong because all tables must be in a schema. B is wrong because the
creation will succeed. C is wrong because the SYS schema is not a default schema. E is wrong
because while there is a notional user PUBLIC, he does not have a schema at all.
✓ A. Indexes have their own namespace.
2. ®
®
˚ B, C, D, E. Stored procedures, synonyms, tables, and views exist in the same namespace.
✓ D, E. D violates the rule that a table name must begin with a letter, and E violates the rule
3. ®
that a table name cannot be a reserved word. Both rules can be bypassed by using double quotes.
®
˚ A, B, C. These are wrong because all will succeed (though A and B are not exactly sensible).

Review the Table Structure
✓ A, C. A heap is a table of variable length rows in random order.
4. ®
®
˚ B, D, E. B is wrong because a heap table can only be one table. D and E are wrong because
a heap table can (and usually will) have indexes and a primary key.

List the Data Types that Are Available for Columns
✓ A, C, D, E, F. All these are variable length data types.
5. ®
®
˚ B. CHAR columns are fixed length.
✓ D. The number will be rounded to 1 digit, and the string will cast as a date.
6. ®
®
˚ A, B, C. Automatic rounding and type casting will correct the “errors,” though ideally they
would not occur.
✓ D. STRING is not an internal data type.
7. ®
®
˚ A, B, C. CHAR, FLOAT, and INTEGER are all internal data types, though not as widely
used as some others.

Create a Simple Table
✓ C. The condition applies only to the rows selected for insert, not to the table creation.
8. ®
®
˚ A, B, D. A is wrong because the statement is syntactically correct. B is wrong because the
condition does not apply to the DDL, only to the DML. D is wrong because the condition will
exclude all rows from selection.

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✓ D. Check and not null constraints are not dependent on any structures other than the table
9. ®
to which they apply and so can safely be copied to a new table.
®
˚ A, B,C, E. A is wrong because not null and check constraint will be applied to the new
table. B, C, and E are wrong because these constraints need other objects (indexes or a parent
table) and so are not copied.

Explain How Constraints Are Created at the Time of Table Creation
✓ C, D. Unique and primary key constraints are enforced with indexes.
10. ®
®
˚ A, B. Check and not null constraints do not rely on indexes.
✓ C. A constraint violation will force a roll back of the current statement but nothing else.
11. ®
®
˚ A, B, D, E. A is wrong because all statements that have succeeded remain intact. B and D
are wrong because there is no commit of anything until it is specifically requested. E is wrong
because the whole statement will be rolled back, not just the failed row.

LAB ANSWER
A possible solution:
■

The SUBSCRIBERS table:
create table subscribers
(id number(4,0) constraint sub_id_pk primary key,
name varchar2(20) constraint sub_name_nn not null);

■

The TELEPHONES table:
create table telephones
(telno number (7,0) constraint tel_telno_pk primary key
constraint tel_telno_ck check (telno between 1000000 and 9999999),
activated date default sysdate,
active varchar2(1) constraint tel_active_nn not null
constraint tel_active_ck check(active='Y' or active='N'),
subscriber number(4,0) constraint tel_sub_fk references subscribers,
constraint tel_active_yn check((active='Y' and subscriber is not null)
or (active='N' and subscriber is null))
);

Lab Answer

485

This table has a constraint on the ACTIVE column that will permit only Y or N, depending
on whether there is a value in the SUBSCRIBER column. This constraint is too complex to
define in line with the column, because it references other columns. SUBSCRIBER, if not
null, must match a row in SUBSCRIBERS.
■

The CALLS table:
create table calls
(telno number (7,0) constraint calls_telno_fk references telephones,
starttime date constraint calls_start_nn not null,
endtime date constraint calls_end_nn not null,
constraint calls_pk primary key(telno,starttime),
constraint calls_endtime_ck check(endtime > starttime));

Two constraints are defined at the table level, not the column level. The primary key cannot be
defined in line with a column because it is based on two columns: one telephone can make many
calls, but not two that begin at the same time (at least, not with current technology). The final
constraint compares the start and end times of the call and so cannot be defined in line.

This page intentionally left blank

12
Creating Other
Schema Objects

CERTIFICATION OBJECTIVES
12.01

Create Simple and Complex Views

12.02

Retrieve Data from Views

12.03

Create Private and Public Synonyms

12.04

Create, Maintain, and Use Sequences

12.05

✓
Q&A

Create and Maintain Indexes
Two-Minute Drill
Self Test

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

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S

o far, most of the SQL statements used in this book have addressed tables. This seems
reasonable: SQL commands interact with data, and data is stored in tables. In practice,
most users never issue SQL that addresses tables: they issue SQL that addresses views or
synonyms.Views and synonyms do not themselves store data; they provide a layer of abstraction
between the users and the data. As tables, views, and synonyms share the same namespace, users
need never be aware of which they are addressing.
Sequences are a mechanism for issuing unique numbers. In many databases,
primary key values are defined as a unique number. A sequence can issue such
numbers on demand, without programmers needing to worry about whether they
really are unique. The structure of a sequence means that hundreds of numbers can,
if necessary, be issued every second without any performance issues.
Index management is sometimes said to lie in the database administration
domain, not the SQL developer’s domain. For this reason, the treatment here is
very brief. But whether or not index management is part of the developer’s job, all
developers must understand the purpose and design of indexes. In many cases the
indexing strategy is crucial to adequate performance: get it right, and the benefits
will be enormous; get it wrong, and the results may be disastrous. Use of indexes is
transparent, but if the programmers are aware of them they can ensure that their
code will be able to make the best use of what indexes are available and advise on
what indexes should be created.

CERTIFICATION OBJECTIVE 12.01

Create Simple and Complex Views
To the user, a view looks like a table: a two-dimensional structure of rows of
columns, against which the user can run SELECT and DML statements. The
programmer knows the truth: a view is just a SELECT statement. Any SELECT
statement returns a two-dimensional set of rows. If the SELECT statement is saved
as a view, then whenever the users query or update rows in the view (under the
impression that it is table), the statement runs, and the result is presented to users
as though it were a table. The SELECT statement on which a view is based can be
anything. It can join tables, perform aggregations, or do sorts; absolutely anything
that is legal in the SELECT command can be used. However, if the view is complex,

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489

then only SELECT statements can be run against it. A simple view is one which can
be addressed by DML statements as well as SELECT. As might be expected, simple
views are based on relatively simple SELECT statements; complex views are based
on more complicated statements.

Views share the same
namespace as tables: anywhere that a
tablename can be used, a viewname is also

syntactically correct. But DML operations
will not always succeed.

Why Use Views at All?
It is usually not a good idea to let end users loose on tables. Possible reasons include:
■ Security.
■ Simplifying user SQL.
■ Preventing error.
■ Making data comprehensible. Table and column names are often long and

pretty meaningless. The view and its columns can be much more obvious.
■ Performance.

Examples of views for each of these purposes follow.

Views to Enforce Security
It may be that users should only see certain rows or columns of a table. There
are several ways of enforcing this, but a view is often the simplest. Consider the
HR.EMPLOYEES table. This includes personal details that should not be visible to
staff outside the personnel department. But finance staff will need to be able to see
the costing information. This view will depersonalize the data:
create view hr.emp_fin as select
hire_date,job_id,salary,commission_pct,department_id from hr.employees;

Note the use of schema qualifiers for the table as the source of the data (often
referred to as either the base or the detail table) and the view: views are schema
objects and can draw their data from tables in the same schema or in other schemas.
If the schema is not specified, it will of course be in the current schema.

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Finance staff can then be given permission to see the view but not the table and
can issue statements such as this:
select * from emp_fin where department_id=50;

They will see only the five columns that make up the view, not the remaining
columns of EMPLOYEES with the personal information. The view can be joined to
other tables or aggregated as though it were a table:
select department_name, sum(salary) from departments natural join emp_fin
group by department_name;

A well constructed set of views can implement a whole security structure within
the database, giving users access to data they need to see while concealing data they
do not need to see.

Views to Simplify User SQL
It will be much easier for users to query data if the hard work (such as joins or
aggregations) is done for them by the code that defines the view. In the last example,
the user had to write code that joined the EMP_FIN view to the DEPARTMENTS
table and summed the salaries per department. This could all be done in a view:
create view dept_sal as
select d.department_name, sum(e.salary) from
departments d left outer join employees e on d.department_
id=e.department_id
group by department_name order by department_name;

Then the users can select from DEPT_SAL without needing to know anything
about joins, or even how to sort the results:
select * from dept_sal;

In particular, they do not need to know how to make sure that all departments
are listed, even those with no employees. The example in the previous section would
have missed these.

Views to Prevent Errors
It is impossible to prevent users making errors, but well constructed views can
prevent some errors arising from a lack of understanding of how data should be
interpreted. The previous section already introduced this concept by constructing a
view that will list all departments, irrespective of whether they currently have staff
assigned to them.

Create Simple and Complex Views

491

A view can help to present data in a way that is unambiguous. For example, many
applications never actually delete rows. Consider this table:
create table emp(empno number constraint emp_empno_pk primary key,
ename varchar2(10),deptno number,active varchar2(1) default 'Y');

The column ACTIVE is a flag indicating that the employee is currently employed
and will default to ‘Y’ when a row is inserted. When a user, through the user
interface, “deletes” an employee, the underlying SQL statement will be an update
that sets ACTIVE to ‘N’. If users who are not aware of this query the table, they may
severely misinterpret the results. It will often be better to give them access to a view:
create view current_staff as select * from emp where active='Y';

Queries addressed to this view cannot possibly see “deleted” staff members.

Views to Make Data Comprehensible
The data structures in a database will be normalized tables. It is not reasonable to
expect users to understand normalized structures. To take an example from the
Oracle E-Business suite, a “customer” in the Accounts Receivable module is in fact
an entity consisting of information distributed across the tables HZ_PARTIES,
HZ_PARTY_SITES, HZ_CUST_ACCTS_ALL, and many more. All these tables
are linked by primary key–to–foreign key relationships, but these are not defined
on any identifiers visible to users (such as a customer number): they are based on
columns the users never see that have values generated internally from sequences.
The forms and reports used to retrieve customer information never address these
tables directly; they all work through views.
As well as presenting data to users in a comprehensible form, the use of views
to provide a layer of abstraction between the objects seen by users and the objects
stored within the database can be invaluable for maintenance work. It becomes
possible to redesign the data structures without having to recode the application. If
tables are changed, then adjusting the view definitions may make any changes to the
SQL and PL/SQL code unnecessary. This can be a powerful technique for making
applications portable across different databases.

Views for Performance
The SELECT statement behind a view can be optimized by programmers, so that
users don’t need to worry about tuning code. There may be many possibilities for
getting the same result, but some techniques can be much slower than others. For
example, when joining two tables there is usually a choice between the nested loop

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join and the hash join. A nested loop join uses an index to get to individual rows;
a hash join reads the whole table into memory. The choice between the two will be
dependent on the state of the data and the hardware resources available.
Theoretically, one can always rely on the Oracle optimizer to work out the
best way to run a SQL statement, but there are cases where it gets it wrong. If the
programmers know which technique is best, they can instruct the optimizer. This
example forces use of the hash technique:
create view dept_emp as
select /*+USE_HASH (employees departments)*/ department_name, last_name
from departments natural join employees;

Whenever users query the DEPT_EMP view, the join will be performed by
scanning the detail tables into memory. The users need not know the syntax for
forcing use of this join method. You do not need to know it, either: this is beyond
the scope of the OCP SQL examination, but the concept of tuning with view design
should be known.

Simple and Complex Views
For practical purposes, classification of a view as simple or complex is related to
whether DML statements can be executed against it: simple views can accept DML
statements, complex views can’t. The strict definitions are as follows:
■ A simple view draws data from one detail table, uses no functions, and does

no aggregation.
■ A complex view can join detail tables, use functions, and perform aggregations.

Applying these definitions shows that of the four views used as examples in
the previous section, the first and third are simple and the second and fourth are
complex.
It is often not possible to execute INSERT, UPDATE, or DELETE commands
against a complex view. The mapping of the rows in the view back to the rows in
the detail table(s) cannot always be established on a one-to-one basis, which is
necessary for DML operations. It is usually possible to execute DML against a simple
view but not always. For example, if the view does not include a column that has a
NOT NULL constraint, then an INSERT through the view cannot succeed (unless
the column has a default value). This can produce a disconcerting effect because
the error message will refer to a table and a column that are not mentioned in the
statement, as demonstrated in the first example in Figure 12-1.

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FIGURE 12-1

DML against
simple and
complex views

The first view in the figure, RNAME_V, does conform to the definition of a
simple view, but an INSERT cannot be performed through it because it is missing a
mandatory column. The second view, RUPPERNAME_V, is a complex view because
it includes a function. This makes an INSERT impossible, because there is no way
the database can work out what should actually be inserted: it can’t reverse engineer
the effect of the UPPER function in a deterministic fashion. But the DELETE
succeeds, because that is not dependent on the function.
As a rule, simple views accept DML, complex views don’t. But there are
exceptions.

CREATE VIEW, ALTER VIEW, and DROP VIEW
The syntax to create a view is as follows:
CREATE [OR REPLACE] [FORCE | NOFORCE] VIEW
[schema.]viewname [(alias [,alias]…)]
AS subquery
[WITH CHECK OPTION [CONSTRAINT constraintname]]
[WITH READ ONLY [CONSTRAINT constraintname]] ;
Note that views are schema objects. There is no reason not to have a view owned
by one user referencing detail tables owned by another user. By default, the view will

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be created in the current schema. The optional keywords, none of which have been
used in the examples so far, are as follows:
■ OR REPLACE

If the view already exists, it will be dropped before being

created.
The FORCE keyword will create the view even if
the detail table(s) in the subquery does not exist. NOFORCE is the default
and will cause an error if the detail table does not exist.

■ FORCE or NOFORCE

This is to do with DML. If the subquery
includes a WHERE clause, then this option will prevent insertion of rows
that wouldn’t be seen in the view or updates that would cause a row to
disappear from the view. By default, this option is not enabled, which can
give disconcerting results.

■ WITH CHECK OPTION

■ WITH READ ONLY

Prevents any DML through the view.

■ CONSTRAINT constraintname

This can be used to name the WITH
CHECK OPTION and WITH READ ONLY restrictions so that error messages
when the restrictions cause statements to fail, will be more comprehensible.

In addition, a set of alias names can be provided for the names of the view’s
columns. If not provided, the columns will be named after the table’s columns or
with aliases specified in the subquery.
Figure 12-2 first demonstrates the effect of creating a view, R1, which permits
insertion of rows that cannot be seen. The second view, LOC1800, uses WITH
CHECK OPTION to prevent this from happening.
FIGURE 12-2

Use of WITH
CHECK OPTION
to prevent
seemingly
anomalous
behavior

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The main use of the ALTER VIEW command is to compile the view. A view must
be compiled successfully before it can be used. When a view is created, Oracle will
check that the detail tables and the necessary columns on which the view is based do
exist. If they do not, the compilation fails and the view will not be created—unless
you use the FORCE option. In that case, the view will be created but will be unusable
until the tables or columns to which it refers are created and the view is successfully
compiled. When an invalid view is queried, Oracle will attempt to compile it
automatically. If the compilation succeeds because the problem has been fixed, the
user won’t know there was ever a problem—except that his query may take a little
longer than usual. Generally speaking, you should manually compile views to make
sure they do compile successfully, rather than having users discover errors.
Figure 12-3 shows the cycle of error and correction.
In the figure, the first statement creates a view EX_STAFF that references a
column EMPLOYEES.LEFT_DATE, which does not exist. The statement does
not fail because the creation is forced, but it does give a warning. An attempt to
select from the view fails. After adding the necessary column to the detail table, the
SELECT succeeds. This is because the view was compiled successfully automatically.
Finally, there is a manual compilation of the view. It would have been better practice
to do this immediately after adding the missing column.
It is not possible to adjust a view’s column definitions after creation in the way
that a table’s columns can be changed. The view must be dropped and re-created.
The DROP command is as follows:
DROP VIEW [schema.]viewname ;
By using the OR REPLACE keywords with the CREATE VIEW command, the
view will be automatically dropped (if it exists at all) before being created.
FIGURE 12-3

How to correct
errors and
compile a view

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EXERCISE 12-1
Create Views
In this exercise, you will create some simple and complex views, using data in the
HR schema. Either SQL*Plus or SQL developer can be used.
1. Connect to your database as user HR.
2. Create views on the EMPLOYEES and DEPARTMENT tables that remove
all personal information:
create view emp_anon_v as
select hire_date, job_id,salary,commission_pct,department_id from
employees;
create view dept_anon_v as
select department_id,department_name,location_id from departments;

3. Create a complex view that will join and aggregate the two simple views.
Note that there is no reason not to have views of views.
create view dep_sum_v as
select e.department_id,count(1) staff, sum(e.salary) salaries,
d.department_name from emp_anon_v e join dept_anon_v d
on e.department_id=d.department_id
group by e.department_id,d.department_name;

4. Confirm that the view works by querying it, as in the following illustration:

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CERTIFICATION OBJECTIVE 12.02

Retrieve Data from Views
Queries can be addressed to views exactly as though to tables. Views and tables
share the same namespace, so syntactically there is absolutely no difference in the
SELECT statements. The user need have no knowledge of which type of object he is
addressing.
Consider this view:
create view dept_emp as
select department_name, last_name from
departments join employees using (department_id);

The following query:
select * from dept_emp where department_name='Marketing';

when parsed by Oracle for execution will become:
select * from
(select department_name, last_name
from departments join employees using (department_id))
where department_name='Marketing';

You rarely get anything for nothing (even when dealing with Oracle Corporation),
and use of views certainly does not in any way reduce the work that the database has
to do. Views can, however, reduce substantially the work that users have to do.

EXERCISE 12-2
Use Views
In this exercise, you will use the views created in Exercise 12-1 for SELECT and
DML statements. Either SQL*Plus or SQL developer can be used.
1. Connect to your database as user HR.
2. Insert a new department through the DEPT_ANON_V view and attempt to
insert an employee through EMP_ANON_V:
insert into DEPT_ANON_V values(99,'Temp Dept',1800);
insert into emp_anon_v values(sysdate,'AC_MGR',10000,0,99);

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The insert into EMP_ANON_V will fail because of missing mandatory
values. You will, however, be able to do this update through it:
update emp_anon_v set salary=salary*1.1;

Then roll back the changes:
rollback;

3. Find out the average salary of the department with the highest average salary,
by querying the EMPLOYEES table:
select max(avg_sal) from
(select AVG(SALARY) avg_sal from employees group by department_id);

and find the same information from the DEP_SUM_V view, which is a much
simpler query:
select max(salaries / staff) from dep_sum_v;

CERTIFICATION OBJECTIVE 12.03

Create Private and Public Synonyms
A synonym is an alternative name for an object. If synonyms exist for objects,
then any SQL statement can address the object either by its actual name, or by its
synonym. This may seem trivial. It isn’t. Use of synonyms means that an application
can function for any user, irrespective of which schema owns the views and tables or
even in which database the tables reside. Consider this statement:
select * from hr.employees@prod;

The user issuing the statement must know that the employees table is owned by
the HR schema in the database identified by the database link PROD (do not worry
about database links—they are a means of accessing objects in a database other than
that onto which you are logged). If a public synonym has been created with this
statement:
create public synonym emp for hr.employees@prod;

then all the user (any user!) need enter is the following:
select * from emp;

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This gives both data independence and location transparency. Tables and views
can be renamed or relocated without ever having to change code; only the synonyms
need to be adjusted.
As well as SELECT statements, DML statements can address synonyms as though
they were the object to which they refer.
Private synonyms are schema objects. Either they must be in your own schema, or
they must be qualified with the schema name. Public synonyms exist independently
of a schema. A public synonym can be referred to by any user to whom permission
has been granted to see it without the need to qualify it with a schema name. Private
synonyms must be a unique name within their schema. Public synonyms can have
the same name as schema objects. When executing statements that address objects
without a schema qualifier, Oracle will first look for the object in the local schema,
and only if it cannot be found will it look for a public synonym. Thus, in the
preceding example, if the user happened to own a table called EMP it would be this
that he would see—not the table pointed to by the public synonym.
The syntax to create a synonym is as follows:
CREATE [PUBLIC] SYNONYM synonym FOR object ;
A user will need to have been granted permission to create private synonyms
and further permission to create public synonyms. Usually, only the database
administrator can create (or drop) public synonyms. This is because their presence
(or absence) will affect every user.

The “public” in “public
synonym” means that it is not a schema
object and cannot therefore be prefixed

with a schema name. It does not mean that
everyone has permissions against it.

To drop a synonym:
DROP [PUBLIC] SYNONYM synonym ;
If the object to which a synonym refers (the table or view) is dropped, the synonym
continues to exist. Any attempt to use it will return an error. In this respect, synonyms
behave in the same way as views. If the object is recreated, the synonym must be
recompiled before use. As with views, this will happen automatically the next time the
synonym is addressed, or it can be done explicitly with
ALTER SYNONYM synonym COMPILE;

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EXERCISE 12-3
Create and Use Synonyms
In this exercise, you will create and use private synonyms, using objects in the HR
schema. Either SQL*Plus or SQL developer can be used.
1. Connect to your database as user HR.
2. Create synonyms for the three views created in Exercise 12-1:
create synonym emp_s for emp_anon_v;
create synonym dept_s for dept_anon_v;
create synonym dsum_s for dep_sum_v;

3. Confirm that the synonyms are identical to the underlying object:
describe emp_s;
describe emp_anon_v;

4. Confirm that the synonyms work (even to the extent of producing the same
errors) by running the statements in Exercises 12-1 and 12-2 against the
synonyms instead of the views:
select * from dsum_s;
insert into dept_s values(99,'Temp Dept',1800);
insert into emp_s values(sysdate,'AC_MGR',10000,0,99);
update emp_s set salary=salary*1.1;
rollback;
select max(salaries / staff) from dsum_s;

5. Drop two of the views:
drop view emp_anon_v;
drop view dept_anon_v;

6. Query the complex view that is based on the dropped views:
select * from dep_sum_v;

Note that the query fails.
7. Attempt to recompile the broken view:
alter view dep_sum_v compile;

This will fail as well.
8. Drop the DEP_SUM_V view:
drop view dep_sum_v;

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501

9. Query the synonym for a dropped view:
select * from emp_s;

This will fail.
10. Recompile the broken synonym:
alter synonym emp_s compile;

Note that this does not give an error, but rerun the query from step 9. It is
definitely still broken.
11. Tidy up by dropping the synonyms:
drop synonym emp_s;
drop synonym dept_s;
drop synonym dsum_s;

CERTIFICATION OBJECTIVE 12.04

Create, Maintain, and Use Sequences
A sequence is a structure for generating unique integer values. Only one session can
read the next value and thus force it to increment. This is a point of serialization, so
each value generated will be unique.
Sequences are an invaluable tool for generating primary keys. Many applications
will need automatically generated primary key values. Examples in everyday business
data processing are invoice numbers or order numbers: the business analysts will
have stated that every invoice and order must have a unique number, which should
continually increment. Other applications may not have such a requirement in
business terms, but it will be needed to enforce relational integrity. Consider a
telephone billing system: in business terms the unique identifier of a telephone is the
telephone number (which is a string) and that of a call will be the source telephone
number and the time the call began (which is a timestamp). These data types are
unnecessarily complex to use as primary keys for the high volumes that go through a
telephone switching system. While this information will be recorded, it will be much
faster to use simple numeric columns to define the primary and foreign keys. The
values in these columns can be sequence based.

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FIGURE 12-4

Use of a
sequence by
two sessions
concurrently

The sequence mechanism is independent of tables, the row locking mechanism,
and commit or rollback processing. This means that a sequence can issue thousands
of unique values a minute—far faster than any method involving selecting a column
from a table, updating it, and committing the change.
Figure 12-4 shows two sessions selecting values from a sequence SEQ1.
Note that in the figure each selection of SEQ1.NEXTVAL generates a unique
number. The numbers are issued consecutively in order of the time the selection was
made, and the number increments globally, not just within one session.

Creating Sequences
The full syntax for creating a sequence is as follows:
CREATE SEQUENCE [schema.]sequencename
[INCREMENT BY number]
[START WITH number]
[MAXVALUE number | NOMAXVALUE]
[MINVALUE number | NOMINVALUE]
[CYCLE | NOCYCLE]
[CACHE number | NOCACHE]
[ORDER | NOORDER] ;

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503

It can be seen that creating a sequence can be very simple. For example, the
sequence used in Figure 12-4 was created with
create sequence seq1;

The options are shown in the following table.
INCREMENT BY

How much higher (or lower) than the last number issued should the
next number be? Defaults to +1 but can be any positive number (or
negative number for a descending sequence).

START WITH

The starting point for the sequence: the number issued by the first
selection. Defaults to 1 but can be anything.

MAXVALUE

The highest number an ascending sequence can go to before
generating an error or returning to its START WITH value. The
default is no maximum.

MINVALUE

The lowest number a descending sequence can go to before
generating an error or returning to its START WITH value. The
default is no minimum.

CYCLE

Controls the behavior on reaching MAXVALUE or MINVALUE.
The default behavior is to give an error, but if CYCLE is specified
the sequence will return to its starting point and repeat.

CACHE

For performance, Oracle can preissue sequence values in batches
and cache them for issuing to users. The default is to generate and
cache the next 20 values.

ORDER

Only relevant for a clustered database: ORDER forces all instances
in the cluster to coordinate incrementing the sequence, so that
numbers issued are always in order even when issued to sessions
against different instances.

Appropriate settings for INCREMENT BY, START WITH, and MAXVALUE or
MINVALUE will come from your business analysts.
It is very rare for CYCLE to be used because it lets the sequence issue duplicate
values. If the sequence is being used to generate primary key values, CYCLE only
makes sense if there is a routine in the database that will delete old rows faster than
the sequence will reissue numbers.
Caching sequence values is vital for performance. Selecting from a sequence is a
point of serialization in the application code: only one session can do this at once.
The mechanism is very efficient: it is much faster than locking a row, updating the
row, and then unlocking it with a COMMIT. But even so, selecting from a sequence
can be a cause of contention between sessions. The CACHE keyword instructs
Oracle to pregenerate sequence numbers in batches. This means that they can be
issued faster than if they had to be generated on demand.

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The default number of values to cache is only 20. Experience shows that this
is not enough. If your application selects from the sequence 10 times a second,
then set the cache value to 50 thousand. Don’t be shy about this.

Using Sequences
To use a sequence, a session can select either the next value with the NEXTVAL
pseudo column, which forces the sequence to increment, or the last (or “current”)
value issued to that session with the CURRVAL pseudo column. The NEXTVAL
will be globally unique: each session that selects it will get a different, incremented,
value for each SELECT. The CURRVAL will be constant for one session until it
selects NEXTVAL again. There is no way to find out what the last value issued
by a sequence was: you can always obtain the next value by incrementing it with
NEXTVAL, and you can always recall the last value issued to your session with
CURRVAL, but you cannot find the last value issued.

The CURRVAL of a sequence
is the last value issued to the current
session, not necessarily the last value issued.

You cannot select the CURRVAL until after
selecting the NEXTVAL.

A typical use of sequences is for primary key values. This example uses a sequence
ORDER_SEQ to generate unique order numbers and LINE_SEQ to generate unique
line numbers for the line items of the order. First create the sequences, which is a
one-off operation:
create sequence order_seq start with 10;
create sequence line_seq start with 10;

Then insert the orders with their lines as a single transaction:
insert into orders (order_number,order_date,customer_number)
values (order_seq.nextval,sysdate,'1000');
insert into order_lines (order_number,line_number,item_number,quantity)
values (order_seq.currval,line_seq.nextval,'A111',1);
insert into order_lines (order_number,line_number,item_number,quantity)
values (order_seq.currval,line_seq.nextval,'B111',1);
commit;

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505

The first INSERT statement raises an order with a unique order number drawn
from the sequence ORDER_SEQ for customer number 1000. The second and third
statements insert the two lines of the order, using the previously issued order number
from ORDER_SEQ as the foreign key to connect the lines to the order, and the next
values from LINE_SEQ to generate a unique identifier for each line. Finally, the
transaction is committed.
A sequence is not tied to any one table. In the preceding example, there would be
no technical reason not to use one sequence to generate values for the primary keys
of the order and of the lines.
A COMMIT is not necessary to make the increment of a sequence permanent:
it is permanent and made visible to the rest of the world the moment it happens. It
can’t be rolled back, either. Figure 12-5 demonstrates this.
In the figure, the second insert is rolled back. But as the final query shows, the
sequence was incremented in spite of this. Sequence updates occur independently
of the transaction management system. For this reason, there will always be gaps in
the series. The gaps will be larger if the database has been restarted and the CACHE
clause was used. All numbers that have been generated and cached but not yet
issued will be lost when the database is shut down. At the next restart, the current
value of the sequence will be the last number generated, not the last issued. So, with
the default CACHE of 20, every shutdown/startup will lose up to 20 numbers.
FIGURE 12-5

A sequence
increment cannot
be rolled back.

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Some accountants get very upset at the idea of gaps in numbering. Ask them
why it matters. It usually doesn’t.
If the business analysts have stated that there must be no gaps in a sequence,
then another means of generating unique numbers must be used. For the preceding
example of raising orders, the current order number could be stored in this table and
initialized to 10:
create table current_on(order_number number);
insert into current_on values(10);
commit;

Then the code to create an order would have to become:
update current_on set order_number=order_number + 1;
insert into orders (order_number,order_date,customer_number)
values ((select order_number from current_on),sysdate,'1000');
commit;

This will certainly work as a means of generating unique order numbers, and
because the increment of the order number is within the transaction that inserts the
order, it can be rolled back with the insert if necessary: there will be no gaps in order
numbers, unless an order is deliberately deleted. But it is far less efficient than using
a sequence, and code like this is famous for causing dreadful contention problems. If
many sessions try to lock and increment the one row containing the current number,
the whole application will hang as they queue up to take their turn. After creating
and using a sequence, it can be modified. The syntax is as follows:
ALTER SEQUENCE sequencename
[INCREMENT BY number]
[START WITH number]
[MAXVALUE number | NOMAXVALUE]
[MINVALUE number | NOMINVALUE]
[CYCLE | NOCYCLE]
[CACHE number | NOCACHE]
[ORDER | NOORDER] ;
This ALTER command is the same as the CREATE command, with one exception:
there is no way to set the starting value. If you want to restart the sequence, the only

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507

SCENARIO & SOLUTION
You are involved in designing a database to
be used for online order entry and offline
financial reporting. What should you
consider with regard to views, synonyms,
and indexes?

The data must be normalized for storage but should be
denormalized into views for retrieval. The dual use of the
database will be a problem: too many indexes will slow down
the order entry, too few will impact adversely on reporting.
An alternative approach would be create a separate set of
tables (possibly denormalized and aggregated like the views)
for reporting and update these with batch routines. Using
synonyms at all times would make it easier to switch between
implementations with and without separate reporting tables.

Should sequences always be used for
primary keys?

In some cases, the business analysts may have defined a
primary key based on attributes of the data. For example, a list
of telephone subscribers could be keyed on telephone number:
this is a formatted string that may have information such as
geographical location (for a land line) or network operator (for
a mobile) embedded within it. But usually a primary key does
not have meaning, and in these cases a sequence is always the
best way.

way is to drop it and re-create it. To adjust the cache value from default to improve
performance of the preceding order entry example:
alter sequence order_seq cache 1000;

However, if you want to reset the sequence to its starting value, the only way is to
drop it:
drop sequence order_seq;

and create it again.

EXERCISE 12-4
Create and Use Sequences
In this exercise, you will create some sequences and use them. You will need two
concurrent sessions, either SQL Developer or SQL*Plus.
1. Log on to your database twice, as HR in separate sessions. Consider one to be
your A session and the other to be your B session.

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2. In your A session, create a sequence as follows:
create sequence seq1 start with 10 nocache maxvalue 15 cycle;

The use of NOCACHE is deleterious for performance. If MAXVALUE
is specified, then CYCLE will be necessary to prevent errors when
MAXVALUE is reached.
3. Execute the following commands in the appropriate session in the correct
order to observe the use of NEXTVAL and CURRVAL and the cycling of the
sequence:
In Your A Session
1st

select seq1.nextval from dual;

2nd
3rd

select seq1.nextval from dual;
select seq1.nextval from dual;

4th
5th

select seq1.nextval from dual;
select seq1.currval from dual;

6th
7th

select seq1.nextval from dual;
select seq1.nextval from dual;

8th
9th

In Your B Session

select seq1.currval from dual;
select seq1.nextval from dual;

10th

select seq1.nextval from dual;

4. Create a table with a primary key:
create table seqtest(c1 number,c2 varchar2(10));
alter table seqtest add constraint seqtest_pk primary key (c1);

5. Create a sequence to generate primary key values:
create sequence seqtest_pk_s;

6. In your A session, insert a row into the new table and commit:
insert into seqtest values(seqtest_pk_s.nextval,'first');
commit;

7. In your B session, insert a row into the new table and do not commit it:
insert into seqtest values(seqtest_pk_s.nextval,'second');

Create and Maintain Indexes

509

8. In your A session, insert a third row and commit:
insert into seqtest values(seqtest_pk_s.nextval,'third');
commit;

9. In your B session, roll back the second insertion:
rollback;

10. In your B session, see the contents of the table:
select * from seqtest;

This demonstrates that sequences are incremented and the next value
published immediately, outside the transaction control mechanism.
11. Tidy up:
drop table seqtest;
drop sequence seqtest_pk_s;
drop sequence seq1;

CERTIFICATION OBJECTIVE 12.05

Create and Maintain Indexes
Indexes have two functions: to enforce primary key and unique constraints and to
improve performance. An application’s indexing strategy is critical for performance.
Theoretically, the SQL developer does not need to be aware of the existence (or
otherwise) of indexes, but if the developer does know what indexes exist and how
they are structured, the developer can write code designed to exploit them.
There is no clear demarcation of whose domain index management lies within.
When the business analysts specify business rules that will be implemented as
constraints, they are in effect specifying indexes. The database administrators
will be monitoring the execution of code running in the database and will make
recommendations for indexes. The developer, who should have the best idea of
what is going on in the code and the nature of the data, will also be involved in
developing the indexing strategy.

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What Indexes Are For
Indexes are part of the constraint mechanism. If a column (or a group of columns)
is marked as a table’s primary key, then every time a row is inserted into the table,
Oracle must check that a row with the same value in the primary key does not
already exist. If the table has no index on the column(s), the only way to do this is
to scan right through the table, checking every row. While this might be acceptable
for a table of only a few rows, for a table with thousands or millions (or billions) of
rows, this is not feasible. An index gives (near) immediate access to key values, so
the check for existence can be made virtually instantaneously. When a primary key
constraint is defined, Oracle will automatically create an index on the primary key
column(s), if one does not exist already.
A unique constraint also requires an index. The difference from a primary key
constraint is that the column(s) of the unique constraint can be left null, perhaps in
many rows. This does not affect the creation and use of the index: nulls do not go
into the B*Tree indexes, as described in the next section, “Types of Index.”
Foreign key constraints are enforced by indexes, but the index must exist on the
parent table, not necessarily on the table for which the constraint is defined. A
foreign key constraint relates a column in the child table to the primary key or to a
unique key in the parent table. When a row is inserted in the child table, Oracle will
do a lookup on the index on the parent table to confirm that there is a matching
row before permitting the insert. However, you should always create indexes on the
foreign key columns within the child table for performance reasons: a DELETE on
the parent table will be much faster if Oracle can use an index to determine whether
there are any rows in the child table referencing the row that is being deleted.
Indexes are critical for performance. When executing any SQL statement that
includes a WHERE clause, Oracle has to identify which rows of the table are to
be selected or modified. If there is no index on the column(s) referenced in the
WHERE clause, the only way to do this is with a full table scan. A full table scan
reads every row of the table, in order to find the relevant rows. If the table has
billions of rows, this can take hours. If there is an index on the relevant column(s),
Oracle can search the index instead. An index is a sorted list of key values,
structured in a manner that makes the search very efficient. With each key value is
a pointer to the row in the table. Locating relevant rows via an index lookup is far
faster than using a full table scan, if the table is over a certain size and the proportion
of the rows to be retrieved is below a certain value. For small tables, or for a WHERE
clause that will retrieve a large fraction of the table’s rows, a full table scan will

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511

be quicker: you can (usually) trust Oracle to make the correct decision, based on
information the database gathers about the tables and the rows within them.
A second circumstance where indexes can be used is for sorting. A SELECT
statement that includes the ORDER BY, GROUP BY, or UNION keywords (and a
few others) must sort the rows into order—unless there is an index, which can return
the rows in the correct order without needing to sort them first.
A third circumstance when indexes can improve performance is when tables are
joined, but again Oracle has a choice: depending on the size of the tables and the
memory resources available, it may be quicker to scan tables into memory and join
them there, rather than use indexes. The nested loop join technique passes through
one table using an index on the other table to locate the matching rows: this is
usually a disk-intensive operation. A hash join technique reads the entire table
into memory, converts it into a hash table, and uses a hashing algorithm to locate
matching rows; this is more memory and CPU intensive. A sort merge join sorts the
tables on the join column then merges them together: this is often a compromise
between disk, memory, and CPU resources. If there are no indexes, then Oracle is
severely limited in the join techniques available.

Types of Index
Oracle supports several types of index, which have several variations. The two index
types of concern here are the B*Tree index, which is the default index type, and the
bitmap index. The only variation worthy of consideration for examination purposes
applies to B*Tree indexes: these can be either unique or nonunique. A unique index
will not permit insertion of two rows with the same key values; a nonunique index
will permit as many rows as you want with the same values. Nonunique is the default.
As a general rule, indexes will improve performance for data retrieval but reduce
performance for DML operations. This is because indexes must be maintained. Every
time a row is inserted into a table, a new key must be inserted into every index on
the table, which places an additional strain on the database. For this reason, on
transaction processing systems it is customary to keep the number of indexes as low
as possible (perhaps no more than those needed for the constraints) and on a queryintensive system, such as a data warehouse, to create as many as might be helpful.

B*Tree Indexes
A B*Tree index (the “B” stands for “balanced”) is a tree structure. The root node of
the tree points to many nodes at the second level, which can point to many nodes at

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the third level, and so on. The necessary depth of the tree will be largely determined
by the number of rows in the table and the length of the index key values.
The B*Tree structure is very efficient. If the depth is greater then three or four,
than either the index keys are very long or the table has billions of rows. If
neither if these is the case, then the index is in need of a rebuild.
The leaf nodes of the index tree store the rows’ keys, in order, each with a pointer
that identifies the physical location of the row. So to retrieve a row with an index
lookup, if the WHERE clause is using an equality predicate on the primary key column,
Oracle navigates down the tree to the leaf node containing the desired key value,
then uses the pointer to find the row. If the WHERE clause is using a nonequality
predicate (such as any of the operators LIKE, BETWEEN, >, or < ) then Oracle can
navigate down the tree to find the first matching key value and then navigate across
the leaf nodes of the index to find all the other matching values. As it does so, it will
retrieve the rows from the table, in order.
The pointer to the row is the rowid. The rowid is an Oracle proprietary pseudocolumn
that every row in every table has. Encrypted within it is the physical address of the
row. As rowids are not part of the SQL standard, they are never visible to a normal
SQL statement, but you can see them and use them if you want. This is demonstrated
in Figure 12-6.
A row’s rowid is globally unique. Every row in every table in the whole database
will have a different rowid. The rowid encryption gives the physical address of the
row: from it, Oracle can calculate which operating system file and where in the file
the row is, and go straight to it.
FIGURE 12-6

Displaying and
using rowids

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513

B*Tree indexes are a very efficient way of retrieving rows if the number of rows
needed is low in proportion to the total number of rows in the table and if the table
is large. Consider this statement:
select count(*) from employees where last_name between 'A%' and 'Z%';

This WHERE clause is sufficiently broad that it will include every row in the
table. It would be much slower to such the index to find the rowids and then use the
rowids to find the rows than to scan the whole table. After all, it is the whole table
that is needed. Another example would be if the table were small enough that one
disk read could scan it in its entirety: there would be no point in reading an index
first.
It is often said that if the query is going to retrieve more than 2 to 4 percent
of the rows, then a full table scan will be quicker. A major exception to this is if
the value specified in the WHERE clause is NULL. NULLs do not go into B*Tree
indexes, so a query such as this:
select * from employees where last_name is null;

will always result in a full table scan. There is little value in creating a B*Tree
index on a column with few unique values, as it will not be selective: the proportion
of the table that will be retrieved for each distinct key value will be too high. In
general, B*Tree indexes should be used if:
■ The cardinality (the number of distinct values) in the column is high, and
■ The number of rows in the table is high, and
■ The column is used in WHERE clauses or JOIN conditions

Bitmap Indexes
In many business applications, the nature of the data and the queries is such that B*Tree
indexes are not of much use. Consider the table of sales for a chain of supermarkets,
storing one year of historical data, which can be analyzed in several dimensions.
Figure 12-7 shows a simple entity-relationship diagram, with just four of the dimensions.
The cardinality of each dimension could be quite low. Make these assumptions:
SHOP

There are four shops

PRODUCT

There are 200 products

DATE

There are 365 days

CHANNEL

There are two channels (walk-in and delivery)

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FIGURE 12-7

Shop

A fact table with
four dimensions
Channel

Sales

Product

Date

Assuming an even distribution of data, only two of the dimensions (PRODUCT
and DATE) have a selectivity better than the commonly used criterion of 2 percent
to 4 percent that makes an index worthwhile. But if queries use range predicates
(such as counting sales in a month, or of a class of ten or more products) then not
even these will qualify. This is a simple fact: B*Tree indexes are often useless in a
data warehouse environment. A typical query might want to compare sales between
two shops to walk-in customers of a certain class of product in a month. There may
well be B*Tree indexes on the relevant columns, but Oracle will ignore them as
being insufficiently selective. This is what bitmap indexes are designed for.
A bitmap index stores the rowids associated with each key value as a bitmap. The
bitmaps for the CHANNEL index might look like this:
WALKIN
DELIVERY

11010111000101011101011101…..
00101000111010100010100010…..

This indicates that the first two rows were sales to walk-in customers, the third
sale was a delivery, the fourth sale was a walk-in, and so on.
The bitmaps for the SHOP index might be:
LONDON
OXFORD
READING
GLASGOW

11001001001001101001010000…..
00100010011000010001001000…..
00010001000100000100100010…..
00000100100010000010000101…..

This indicates that the first two sales were in the London shop, the third was in
Oxford, the fourth in Reading, and so on. Now if this query is received:
select count(*) from sales where channel='WALKIN' and shop='OXFORD';

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Oracle can retrieve the two relevant bitmaps and add them together with a
Boolean AND operation:
WALKIN
OXFORD
WALKIN & OXFORD

11010111000101011101011101…..
00100010011000010001001100…..
00000010000000010000000000…..

The result of the AND operation shows that only the seventh and sixteenth
rows qualify for selection. This combining of bitmaps is very fast and can be used
to implement complex Boolean algebra operations with many conditions on many
columns using any combination of AND, OR, and NOT operators. A particular
advantage that bitmap indexes have over B*Tree indexes is that they include
NULLs. As far as the bitmap index is concerned, NULL is just another distinct
value, which will have its own bitmap.
In general, bitmap indexes should be used if:
■ The cardinality (the number of distinct values) in the column is low (such as

male/female), and
■ The number of rows in the table is high, and
■ The column is used in Boolean algebra (AND/OR/NOT) operations

If you knew in advance what the queries would be then you could build
B*Tree indexes that would work, such as a composite index on SHOP and
CHANNEL. But usually you don’t know, which is where the dynamic merging
of bitmaps gives great flexibility.

Creating and Using Indexes
Indexes are created implicitly when primary key and unique constraints are defined,
if an index on the relevant column(s) does not already exist. The basic syntax for
creating an index explicitly is as follows:
CREATE [UNIQUE | BITMAP] INDEX [ schema.]indexname
ON [schema.]tablename (column [, column…] ) ;
The default type of index is a nonunique B*Tree index. It is not possible to create
a unique bitmap index (and you wouldn’t want to if you could: think about the
cardinality issue). Indexes are schema objects, and it is possible to create an index
in one schema on a table in another, but most people would find this somewhat
confusing. A composite index is an index on several columns. Composite indexes can

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be on columns of different data types, and the columns do not have to be adjacent in
the table.
Many database administrators do not consider it good practice to rely on
implicit index creation. If the indexes are created explicitly, the creator has
full control over the characteristics of the index, which can make it easier for
the DBA to manage subsequently.
Consider this example of creating tables and indexes and then defining constraints:
create table dept(deptno number,dname varchar2(10));
create table emp(empno number, surname varchar2(10), forename varchar2(10), dob
date, deptno number);
create unique index dept_i1 on dept(deptno);
create unique index emp_i1 on emp(empno);
create index emp_i2 on emp(surname,forename);
create bitmap index emp_i3 on emp(deptno);
alter table dept add constraint dept_pk primary key (deptno);
alter table emp add constraint emp_pk primary key (empno);
alter table emp add constraint emp_fk foreign key (deptno) references
dept(deptno);

The first two indexes created are flagged as UNIQUE, meaning that it will not
be possible to insert duplicate values. This is not defined as a constraint at this
point but is true nonetheless. The third index is not defined as UNIQUE and will
therefore accept duplicate values; this is a composite index on two columns. The
fourth index is defined as a bitmap index, because the cardinality of the column is
likely to be low in proportion to the number of rows in the table.
When the two primary key constraints are defined, Oracle will detect the precreated indexes and use them to enforce the constraints. Note that the index
on DEPT.DEPTNO has no purpose for performance because the table will in all
likelihood be so small that the index will never be used to retrieve rows (a scan
will be quicker), but it is still essential to have an index to enforce the primary key
constraint.

A unique and primary key
constraint can be enforced by indexes
that are either unique or nonunique: in

the latter case, it will be a nonunique
index that happens to have only unique
values.

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517

Once created, use of indexes is completely transparent and automatic. Before
executing a SQL statement, the Oracle server will evaluate all the possible ways of
executing it. Some of these ways may involve using whatever indexes are available,
others may not. Oracle will make use of the information it gathers on the tables
and the environment to make an intelligent decision about which (if any) indexes
to use.
The Oracle server should make the best decision about index use, but if it
gets it wrong it is possible for a programmer to embed instructions, known as
optimizer hints, in code that will force the use (or not) of certain indexes.

Modifying and Dropping Indexes
There is a command, ALTER INDEX…but it cannot be used to change any of the
characteristics described in this chapter: the type (B*Tree or bitmap) of the index,
the columns, or whether it is unique or nonunique. The ALTER INDEX command
lies in the database administration domain and would typically be used to adjust
the physical properties of the index, not the logical properties that are of interest
to developers. If it is necessary to change any of these properties, the index must be
dropped and recreated. Continuing the example in the previous section, to change
the index EMP_I2 to include the employees’ birthdays:
drop index emp_i2;
create index emp_i2 on emp(surname,forename,dob);

This composite index now includes columns with different data types. The columns
happen to be listed in the same order that they are defined in the table, but this is by
no means necessary.
When a table is dropped, all the indexes and constraints defined for the table are
dropped as well. If an index was created implicitly by creating a constraint, then
dropping the constraint will also drop the index. If the index had been created
explicitly and the constraint created later, then if the constraint is dropped the
index will survive.

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INSIDE THE EXAM
Creating Other Schema Objects
In many applications, users never see tables.
The table data is abstracted into views, and
the views further abstracted by synonyms.
Use of views and synonyms is completely
transparent to SQL: they all share the same
namespace, and so code will run against any
of them. Views can make complex relational
structures much easier to use, and synonyms
provide data ownership and location
transparency.
Use of indexes is also transparent. They
are required to enforce unique and primary
key constraints (and strongly advised for
foreign key constraints) and may optionally
be created to enhance performance. Indexes
come in two forms: B*Tree and bitmap.
Arriving at an appropriate indexing strategy,
which creates indexes of the right type on
the right columns, is critical for performance.

B*Tree indexes are suitable for highly
selective columns with a high cardinality;
bitmap indexes are suitable for columns
with only a few distinct values. The Oracle
server will determine at statement execution
time whether or not to use indexes, though
programmers can control this by embedding
hints in their code.
Sequences are a facility for generating
unique numbers. It is absolutely impossible
for two sessions to read the same value from
a sequence. Sequences and their values exist
outside the transaction processing structure.
If a sequence is incremented, this increment
cannot be rolled back and is immediately
visible to all other sessions without any
commit. Sequences bypass the mechanisms for
atomic transactions and transaction isolation.

EXERCISE 12-5
Creating Indexes
In this exercise, create indexes on a copy of the EMPLOYEES table in the HR
schema. Either SQL*Plus or SQL Developer can be used.
1. Connect to your database as user HR.
2. Create a table that is a copy of EMPLOYEES:
create table emps as select * from employees;

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519

This table will have neither indexes nor primary, unique, or foreign key
constraints, because these are not copied by a CREATE TABLE AS command.
The NOT NULL constraints will have been copied. Confirm this by describing
the table:
describe emps;

3. Create an index to be used for the primary key constraint:
create unique index emps_empid_i on emps(employee_id);

4. Demonstrate that a unique index cannot accept duplicates, even before a
constraint is defined:
insert into emps(employee_id,last_name,email,hire_date,job_id)
values(198,'Watson','jw@bplc.co.za',sysdate,'IT_PROG');

This will return an error because the index cannot insert a second employee_
id 198. Index uniqueness is an attribute of the index that can exist without a
constraint but should not be relied upon to enforce data integrity.
5. Create additional indexes on columns that are likely to be used in WHERE
clauses, using B*Tree for columns of high cardinality and bitmap for columns
of low cardinality:
create
create
create
create

index emps_name_i on emps(last_name,first_name);
index emps_tel_i on emps(phone_number);
bitmap index emps_mgr_i on emps(manage_id);
bitmap index emps_dept_i on emps(department_id);

6. Define some constraints:
alter table emps add constraint emps_pk primary key (employee_id);
alter table emps add constraint emps_email_uk unique(email);
alter table emps add constraint emps_tel_uk unique(phone_number);

7. Display the index names and their type:
select index_name,index_type,uniqueness from user_indexes
where table_name='EMPS';

The view USER_INDEXES shows details of all indexes in your current schema.
Note that in addition to the five indexes explicitly created in steps 3 and 5,
there is also an index created implicitly with the name of the constraint defined
on EMAIL. Note also that the unique constraint on PHONE_NUMBER

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is being enforced with a nonunique index; this is perfectly possible, because
although the constraint mechanism uses indexes, it is independent of the
structure of the index.
8. Tidy up by dropping the EMPS table, and confirm that all the indexes have
also gone:
drop table emps;
select index_name from user_indexes where table_name='EMPS';

CERTIFICATION SUMMARY
Views are select statements, stored in the data dictionary. They can be queried
as though they were tables, and in some cases they can be the object of DML
statements. A simple view is columns from one table, with no aggregations or
functions; a complex view can join tables, aggregate, and use functions. As a general
rule, it is possible to do DML through a simple view but not through a complex
view—but there are exceptions.
Synonyms are aliases that can be used to access views and tables. Synonyms can
simplify code by making it unnecessary to specify schema qualifiers or database link
names: they let code run without needing to know about data ownership or location.
Tables, views, and synonyms share the name namespace: within a schema, they must
all have different names and can be used interchangeably.
Sequences generate unique numbers, typically for use as primary key values.
Indexes have a dual purpose: enforcing constraints and enhancing performance.
If an index is available when defining a constraint, Oracle will use it; otherwise, an
index will be implicitly created. When creating indexes for performance, consider
which columns are used for joining tables or in WHERE clauses and create B*Tree
indexes on columns with high cardinality and bitmap indexes on columns with
relatively few distinct values.

Two-Minute Drill

✓

521

TWO-MINUTE DRILL
Create Simple and Complex Views
❑ A simple view has one detail (or base) table and uses neither functions nor

aggregation.
❑ A complex view can be based on any SELECT statement, no matter how

complicated.
❑ Views are schema objects. To use a view in another schema, the view name

must be qualified with the schema name.

Retrieve Data from Views
❑ A view can be queried exactly as though it were a table.
❑ Views can be joined to other views or to tables, they can be aggregated, and

in some cases they can accept DML statements.
❑ Views exist only as data dictionary constructs. Whenever you query a view,

the underlying SELECT statement must be run.

Create Private and Public Synonyms
❑ A synonym is an alternative name for a view or a table.
❑ Private synonyms are schema objects; public synonyms exist outside user

schemas and can be used without specifying a schema name as a qualifier.
❑ Synonyms share the same namespace as views and tables and can therefore be

used interchangeably with them.

Create, Maintain, and Use Sequences
❑ A sequence generates unique values—unless either MAXVALUE or

MINVALUE and CYCLE have been specified.
❑ Incrementing a sequence need not be committed and cannot be rolled back.
❑ Any session can increment the sequence by reading its next value. It is possible

to obtain the last value issued to your session but not the last value issued.

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Create and Maintain Indexes
❑ Indexes are required for enforcing unique and primary key constraints.
❑ NULLs are not included in B*Tree indexes but are included in bitmap indexes.
❑ B*Tree indexes can be unique or nonunique, which determines whether they

can accept duplicate key values.
❑ B*Tree indexes are suitable for high cardinality columns, bitmap indexes for

low cardinality columns.
❑ Compound indexes have a key consisting of several columns, which can be of

different data types.

Self Test

523

SELF TEST
The following questions will help you measure your understanding of the material presented in this
chapter. Read all the choices carefully because there might be more than one correct answer. Choose
all the correct answers for each question.

Create Simple and Complex Views
1. Which of these is a defining characteristic of a complex view, rather than a simple view?
(Choose one or more correct answers.)
A. Restricting the projection by selecting only some of the table’s columns
B. Naming the view’s columns with column aliases
C. Restricting the selection of rows with a WHERE clause
D. Performing an aggregation
E. Joining two tables
2. Consider these three statements:
create view v1 as select department_id,department_name,last_name from
departments join employees using (department_id);
select department_name,last_name from v1 where department_id=20;
select d.department_name,e.last_name from departments d, employees e
where d.department_id=e.department_id and
d.department_id=20;

The first query will be quicker than the second because (choose the best answer):
A. The view has already done the work of joining the tables.
B. The view uses ISO standard join syntax, which is faster than the Oracle join syntax used in
the second query.
C. The view is precompiled, so the first query requires less dynamic compilation than the second
query.
D. There is no reason for the first query to be quicker.
3. Study this view creation statement:
create view dept30 as
select department_id,employee_id,last_name from employees
where department_id=30 with check option;

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What might make the following statement fail? (Choose the best answer.)
update dept30 set department_id=10 where employee_id=114;

A.
B.
C.
D.

Unless specified otherwise, views will be created as WITH READ ONLY.
The view is too complex to allow DML operations.
The WITH CHECK OPTION will reject any statement that changes the DEPARTMENT_ID.
The statement will succeed.

Retrieve Data from Views
4. There is a simple view SCOTT.DEPT_VIEW on the table SCOTT.DEPT. This insert fails with
an error:
SQL> insert into dept_view values('SUPPORT','OXFORD');
insert into dept_view values('SUPPORT','OXFORD')
*
ERROR at line 1:
ORA-01400: cannot insert NULL into ("SCOTT"."DEPT"."DEPTNO")

What might be the problem? (Choose the best answer.)
A. The INSERT violates a constraint on the detail table.
B. The INSERT violates a constraint on the view.
C. The view was created as WITH READ ONLY.
D. The view was created as WITH CHECK OPTION.

Create Private and Public Synonyms
5. What are distinguishing characteristics of a public synonym rather than a private synonym?
(Choose two correct answers.)
A. Public synonyms are always visible to all users.
B. Public synonyms can be accessed by name without a schema name qualifier.
C. Public synonyms can be selected from without needing any permissions.
D. Public synonyms can have the same names as tables or views.
6. Consider these three statements:
create synonym s1 for employees;
create public synonym s1 for departments;
select * from s1;

Self Test

525

Which of the following statements is correct? (Choose the best answer.)
A. The second statement will fail because an object S1 already exists.
B. The third statement will show the contents of EMPLOYEES.
C. The third statement will show the contents of DEPARTMENTS.
D. The third statement will show the contents of the table S1, if such a table exists in
the current schema.
7. A view and a synonym are created as follows:
create view dept_v as select * from dept;
create synonym dept_s for dept_v;

Subsequently the table DEPT is dropped. What will happen if you query the synonym DEPT_S ?
(Choose the best answer.)
A. There will not be an error because the synonym addresses the view, which still exists, but
there will be no rows returned.
B. There will not be an error if you first recompile the view with the command ALTER VIEW
DEPT_V COMPILE FORCE;
C. There will be an error because the synonym will be invalid.
D. There will be an error because the view will be invalid.
E. There will be an error because the view will have been dropped implicitly when the table
was dropped.

Create, Maintain, and Use Sequences
8. A sequence is created as follows:
create sequence seq1 maxvalue 50;

If the current value is already 50, when you attempt to select SEQ1.NEXTVAL what will
happen? (Choose the best answer.)
A. The sequence will cycle and issue 0.
B. The sequence will cycle and issue 1.
C. The sequence will reissue 50.
D. There will be an error.
9. You create a sequence as follows:
create sequence seq1 start with 1;

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After selecting from it a few times, you want to reinitialize it to reissue the numbers already
generated. How can you do this? (Choose the best answer.)
A. You must drop and re-create the sequence.
B. You can’t. Under no circumstances can numbers from a sequence be reissued once they
have been used.
C. Use the command ALTER SEQUENCE SEQ1 START WITH 1; to reset the next value to 1.
D. Use the command ALTER SEQUENCE SEQ1 CYCLE; to reset the sequence to its
starting value.
10. Study the following exhibit:

Assuming that the sequence SEQ1 was created with the option ORDER and INCREMENT BY
set to 1, what value will be returned by the final SELECT statement? (Choose the best answer.)
A. 2
B. 3
C. 4
D. It will depend on whether any other sessions are selecting from the sequence while the
statements in the exhibit are being run.

Create and Maintain Indexes
11. A UNIQUE constraint on a column requires an index. Which of the following scenarios is
correct? (Choose one or more correct answers.)
A. If a UNIQUE index already exists on the column, it will be used.
B. If a NONUNIQUE index already exists it will be used.

Lab Question

527

C. If a NONUNIQUE index already exists on the column, a UNIQUE index will be created
implicitly.
D. If any index exists on the column, there will be an error as Oracle attempts to create
another index implicitly.
12. This statement will fail:
create unique bitmap index on employees(department_id,hire_date);

Why? (Choose the best answer.)
A. Bitmap indexes cannot be unique.
B. The two columns are of different data types.
C. A bitmap index can be on only one column.
D. There is already a B*Tree index on DEPARTMENT_ID.
13. You have created an index with this statement:
create index ename_i on employees(last_name,first_name);

How can you adjust the index to include the employees’ birthdays, which is a date type column
called DOB? (Choose the best answer.)
A. Use ALTER INDEX ENAME_I ADD COLUMN DOB;.
B. You can’t do this because of the data type mismatch.
C. You must drop the index and re-create it.
D. This can only be done if the column DOB is NULL in all existing rows.

LAB QUESTION
Figure 12-7 in this chapter shows an entity-relationship diagram for a simple system designed to store
and analyze sales. The columns for the fact table SALES are as follows:
System-generated primary key

■

SALE_ID

■

CHANNEL_ID

Foreign key to CHANNELS

■

PRODUCT_ID

Foreign key to PRODUCTS

■

SHOP_ID

■

DAY_ID

■

QUANTITY

Foreign key to SHOPS
Foreign key to DAYS
The quantity of the product sold

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It is expected that there will be several million SALES rows per year. The dimension tables are as
follows:
PRODUCTS A list of all products, including price. Cardinality of a few hundred
CHANNEL Possible sales methods, such as walk-in, Internet, and telephone
SHOPS Details of all the shops—no more that a couple of dozen
DAYS Dates for which sales are being stored: 365, identified by day number
Write code to create the tables; create indexes; create constraints. Create sequences to be used for
primary keys where necessary. Create some views that will present the data in an easy-to-understand
fashion.

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529

SELF TEST ANSWERS
Create Simple and Complex Views
✓ D, E. Aggregations and joins make a view complex and make DML impossible.
1. ®
®
˚ A, B, C. Selection and projection or renaming columns does not make the view complex.
✓ D. Sad but true. Views do not help performance.
2. ®
®
˚ A is wrong because a view is only a SELECT statement; it doesn’t prerun the query. B is
wrong because the Oracle optimizer will sort out any differences in syntax. C is wrong because,
although views are precompiled, this doesn’t affect the speed of compiling a user’s statement.
✓ C. The WITH CHECK OPTION will prevent DML that would cause a row to disappear
3. ®
from the view.
®
˚ A, B, D. A is wrong because views are by default created read/write. B is wrong because
the view is a simple view. D is wrong because the statement cannot succeed because the check
option will reject it.

Retrieve Data from Views
✓ A. There is a NOT NULL or PRIMARY KEY constraint on DEPT.DEPTNO.
4. ®
®
˚ B, C, D. B is wrong because constraints are enforced on detail tables, not on views. C and
D are wrong because the error message would be different.

Create Private and Public Synonyms
✓ B, D. Public synonyms are not schema objects and so can only be addressed directly. They
5. ®
can have the same names as schema objects.
®
˚ A, C. These are wrong because users must be granted privileges on a public synonym before
they can see it or select from it.
✓ B. The order of priority is to search the schema namespace before the public namespace, so
6. ®
it will be the private synonym (to EMPLOYEES) that will be found.
®
˚ A, C, D. A is wrong because a synonym can exist in both the public namespace and the
schema namespace. C is wrong because the order of priority will find the private synonym first.
D is wrong because it would not be possible to have a table and a private synonym in the same
schema with the same name.

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Creating Other Schema Objects

✓ D. The synonym will be fine, but the view will be invalid. Oracle will attempt to recompile
7. ®
the view, but this will fail.
®
˚ A, B, C, E. A is wrong because the view will be invalid. B is wrong because the FORCE
keyword can only be applied when creating a view (and it would still be invalid, even so). C is
wrong because the synonym will be fine. E is wrong because views are not dropped implicitly
(unlike indexes and constraints).

Create, Maintain, and Use Sequences
✓ D. The default is NOCYCLE, and the sequence cannot advance further.
8. ®
®
˚ A, B, C. A and B are wrong because CYCLE is disabled by default. If it were enabled,
the next number issued would be 1 (not zero) because 1 is the default for START WITH. C is
wrong because under no circumstances will a sequence issue repeating values.
✓ A. It is not possible to change the next value of a sequence, so you must re-create it.
9. ®
®
˚ B, C, D. B is wrong because, while a NOCYCLE sequence can never reissue numbers,
there is no reason why a new sequence (with the same name) cannot do so. C is wrong because
START WITH can only be specified at creation time. D is wrong because this will not force an
instant cycle; it will only affect what happens when the sequence reaches its MAXVALUE or
MINVALUE.
✓ D. If the sequence is being used by other sessions, there is no knowing how many
10. ®
increments may have taken place between the first and the second INSERT statements.
®
˚ A, B, C. The answer would be 4, C, except that there could have been increments forced
by other sessions. A and B are wrong because the ROLLBACK will not reverse the sequence
increments.

Create and Maintain Indexes
✓ A, B. Either a UNIQUE or a NONUNIQUE index can be used to enforce a UNIQUE
11. ®
constraint.
®
˚ C, D. C is wrong because there is no need to create another index (in fact, you can’t index
the same column twice even if you want to). D is wrong because if an index exists, Oracle won’t
attempt to create another.
✓ A. The keywords BITMAP and UNIQUE are mutually exclusive. And you wouldn’t want
12. ®
to do this, anyway.

Lab Answer

531

®
˚ B, C, D. B and C are wrong because a bitmap index can be composite, with columns of
different data types. D is wrong because the bitmap index is not on DEPARTMENT_ID alone,
which would not be possible.
✓ C. It is not possible to change an index’s columns after creation.
13. ®
®
˚ A, B, D. B is wrong because the data type is not the problem. A and D are wrong because
an index’s columns are fixed at creation time.

LAB ANSWER
This is a possible solution:
/*create the tables*/
create table sales (sale_id number, channel_id number, product_id number, shop_id number,
day_id number, quantity number);
create table products (product_id number, pname varchar2(20),price number);
create table channels (channel_id number, cname varchar2(20));
create table shops (shop_id number,address varchar2(20));
create table days (day_id number, day date);
/*pre-create indexes to be used for constraints*/
create unique index prod_pk on products(product_id);
create unique index chan_pk on channels(channel_id);
create unique index shop_pk on shops(shop_id);
create unique index day_id on days(day_id);
create unique index sales_pk on sales(sale_id);
/*create bitmap indexes on the dimension columns of the fact table*/
create bitmap index sales_chan on sales(channel_id);
create bitmap index sales_prod on sales(product_id);
create bitmap index sales_shop on sales(shop_id);
create bitmap index sales_date on sales(day_id);
/*add the primary key constraints/*
alter table products add constraint prod_pk primary key (product_id);
alter table channels add constraint chan_pk primary key (channel_id);
alter table shops add constraint shop_pk primary key (shop_id);
alter table days add constraint day_pk primary key (day_id);
alter table sales add constraint sales_pk primary key(sale_id);
/*add the foreign key constraints*/
alter table sales add constraint sales_prod_fk foreign key (product_id) references products;
alter table sales add constraint sales_chan_fk foreign key (channel_id) references channels;
alter table sales add constraint sales_shop_fk foreign key (shop_id) references shops;
alter table sales add constraint sales_day_fk foreign key (day_id) references days;
/*create the sequences for primary keys:
cache many values for the fact table, but don't pre-issue values for the largely static

532

Chapter 12:

Creating Other Schema Objects

dimension tables. This will save some memory*/
create sequence sales_seq cache 1000;
create sequence product_seq nocache;
create sequence channel_seq nocache;
create sequence shop_seq nocache;
create sequence day_seq nocache;
/*create a view to analyze sales in several dimensions*/
create view sales_analysis as
select cname,pname,address,day,sum(quantity) total
from sales,channels,products,shops,days
where sales.channel_id=channels.channel_id
and sales.product_id=products.product_id
and sales.shop_id=shops.shop_id
and sales.day_id=days.day_id
group by grouping sets(
(cname,pname,address,day),
(address,pname),
(pname,day));

A
About the CD

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

534

Appendix: About the CD

T

he CD-ROM included with this book comes complete with MasterExam and an electronic
version of the book. The software is easy to install on any Windows 98/NT/2000/XP/Vista
computer and must be installed to access the MasterExam feature.You may, however,
browse the electronic book directly from the CD without installation. To register for a second bonus
MasterExam, simply click the Online Training link on the Main Page and follow the directions to the
free online registration.

System Requirements
Software requires Windows 98 or higher and Internet Explorer 5.0 or above and
20MB of hard disk space for full installation. The electronic book requires Adobe
Acrobat Reader.

Installing and Running MasterExam
If your computer CD-ROM drive is configured to auto run, the CD-ROM will
automatically start up upon inserting the disk. From the opening screen you may install
MasterExam by pressing the MasterExam button. This will begin the installation
process and create a program group named “LearnKey.” To run MasterExam, navigate
to Start | Programs | LearnKey. If the auto run feature did not launch your CD, browse
to the CD and click on the LaunchTraining.exe icon.

MasterExam
MasterExam provides you with a simulation of the actual exam. The number of
questions, the type of questions, and the time allowed are intended to be an accurate
representation of the exam environment. You have the option to take an open book
exam, including hints, references, and answers; a closed book exam; or the timed
MasterExam simulation.
When you launch MasterExam, a digital clock display will appear in the upper
left-hand corner of your screen. The clock will continue to count down to zero unless
you choose to end the exam before the time expires.

Technical Support

535

Electronic Book
The entire contents of the Study Guide are provided in PDF. Adobe’s Acrobat Reader
has been included on the CD.

Help
A help file is provided through the help button on the main page in the lower left-hand
corner. An individual help feature is also available through MasterExam.

Removing Installation(s)
MasterExam is installed to your hard drive. For best results for removal of programs
use the Start | Programs | LearnKey | Uninstall options to remove MasterExam.

Technical Support
For questions regarding the technical content of the electronic book or MasterExam,
please visit www.osborne.com or e-mail customer.service@mcgraw-hill.com. For
customers outside the 50 United States, e-mail international_cs@mcgraw-hill.com.

LearnKey Technical Support
For technical problems with the software (installation, operation, removing
installations), please visit www.learnkey.com or e-mail techsupport@learnkey.com.

This page intentionally left blank

Glossary

A
ACID Atomicity, consistency, isolation, and durability. Four characteristics that
a relational database must be able to maintain for transactions.
ADDM Automatic Database Diagnostic Monitor. A tool that generates performance
tuning reports based on snapshots in the AWR.
AES Advanced Encryption Standard. A widely used data encryption method.
AL16UTF16 A Unicode fixed-width 2-byte character set, commonly specified
for the NLS character set used for NVARCHAR2, NCHAT, and NCLOB data types.
alias In Oracle Net, a pointer to a connect string. An alias must be resolved into
the address of a listener and the name of a service or instance.
ANSI American National Standards Institute. A U.S. body that defines a number
of standards relevant to computing.

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

538

OCA Oracle Database 11g: SQL Fundamentals I Exam

API Application Programming Interface. A defined method for manipulating data,
typically implemented as a set of PL/SQL procedures in a package.
ASCII American Standard Code for Information Interchange. A standard (with
many variations) for coding letters and other characters as bytes.
ASM Automatic Storage Management. An LVM provided with the Oracle database.
attribute One element of a tuple (aka a column).
AVG A function that divides the sum of a column or expression by the number of
nonnull rows in a group.

B
background process

A process that is part of the instance: launched at startup.

BFILE A large object data type that is stored as an operating system file. The value
in the table column is a pointer to the file.
bind variable A value passed from a user process to a SQL statement at statement
execution time.
BLOB Binary Large Object. A LOB data type for binary data, such as photographs
and video clips.
block The units of storage into which data files are formatted. The size can be
2KB, 4KB, 8KB, 16KB, 32KB, or 64KB. Some platforms will not permit all these sizes.

C
Cartesian product Sometimes called a cross join. A mathematical term that
refers to the set of data created by merging the rows from two or more tables.
CET Central European Time. A time zone used in much of Europe (though not
Great Britain) that is one hour ahead of UTC with daylight saving time in effect during
the summer months.

Glossary

539

character set The encoding system for representing data within bytes. Different
character sets can store different characters and may not be suitable for all languages.
Unicode character sets can store any character.
check constraint A simple rule enforced by the database that restricts the values
that can be entered into a column.
client-server architecture A processing paradigm where the application is
divided into client software that interacts with the user and server software that
interacts with the data.
CLOB Character Large Object. A LOB data type for character data, such as text
documents, stored in the database character set.
cluster A hardware environment where more than one computer shares access to
storage. A RAC database consists of several instances on several computers opening
one database on the shared storage.
cluster segment A segment that that can contain one or more tables, denormalized
into a single structure.
COALESCE A function that returns the first nonnull value from its parameter
list. If all its parameters are null, then a null value is returned.
column An element of a row: tables are two-dimensional structures, divided
horizontally into rows and vertically into columns.
commit To make permanent a change to data.
complete recovery Following a restore of damaged database files, this applies
each redo to bring the database up to date with no loss of data.
connect identifier An Oracle Net alias.
connect string The database connection details needed to establish a session:
the address of the listener and the service or instance name.
consistent backup A backup made while the database is closed.

540

OCA Oracle Database 11g: SQL Fundamentals I Exam

constraint A mechanism for enforcing rules on data: that a column value must
be unique or may only contain certain values. A primary key constraint specifies
that the column must be both unique and not null.
control file The file containing pointers to the rest of the database, critical sequence
information, and the RMAN repository.
CPU Central Processing Unit. The chip that provides the processing capability of
a computer, such as an Intel Pentium or a Sun SPARC.

D
data blocks

The units into which data files are formatted.

data dictionary The tables owned by SYS in the SYSTEM tablespace that define
the database and the objects within it.
data dictionary views Views on the data dictionary tables that let the DBA
investigate the state of the database.
data guard A facility whereby a copy of the production database is created and
updated (possibly in real time) with all changes applied to the production database.
data pump A facility for transferring large amounts of data at high speed into,
out of, or between databases.
database buffer cache An area of memory in the SGA used for working on
blocks copied from data files.
database link A connection from one database to another, based on a username
and password and a connect string.
data file The disk-based structure for storing data.
DBA Database Administrator. The person responsible for creating and managing
Oracle databases—this could be you.

Glossary

541

DBA role A preseeded role provided for backward compatibility that includes all the
privileges needed to manage a database, except that needed to start up or shut down.
DBCA The Database Configuration Assistant. A GUI tool for creating, modifying,
and dropping instances and databases.
DBMS Database Management System. Often used interchangeably with RDBMS.
DBWn or DBWR The Database Writer. The background process responsible for
writing changed blocks from the database buffer cache to the data files. An instance
may have up to 10 database writer processes, DBW0 through DBW9.
DDL Data Definition Language. The subset of SQL commands that change object
definitions within the data dictionary: CREATE, ALTER, DROP, and TRUNCATE.
deadlock A situation where two sessions block each other, such that neither can
do anything. Deadlocks are detected and resolved automatically by the database.
DECODE A function that implements if-then-else conditional logic by testing
two terms for equality and returning the third term if they are equal or, optionally,
returning some other term if they are not.
direct path A method of I/O on data files that bypasses the database buffer cache.
directory object An Oracle directory: an object within the database that points
to an operating system directory.
DML Data Manipulation Language. The subset of SQL commands that change data
within the database: INERT, UPDATE, DELETE, and MERGE.
DHCP Dynamic Host Configuration Protocol. The standard for configuring
the network characteristics of a computer, such as its IP address, in a changing
environment where computers may be moved from one location to another.
DNS Domain Name Service. The TCP mechanism for resolving network names
into IP addresses.

542

OCA Oracle Database 11g: SQL Fundamentals I Exam

domain The set of values an attribute is allowed to take. Terminology: tables
have rows; rows have columns with values. Or: relations have tuples; tuples have
attributes with values taken from their domain.
DSS Decision Support System. A database, such as a data warehouse, optimized
for running queries as against OLTP work.

E
easy connect A method of establishing a session against a database by specifying
the address on the listener and the service name without using an Oracle Net alias.
EBCDIC Extended Binary Coded Decimal Interchange Code. A standard
developed by IBM for coding letters and other characters in bytes.
environment variable A variable set in the operating system shell which can
be used by application software and by shell scripts.
equijoin A join condition using an equality operator.

F
fact table The central table in a star schema, with columns for values relevant to
the row and columns used as foreign keys to the dimension tables.
FGA Fine Grained Auditing. A facility for tracking user access to data based on
the rows that are seen or manipulated.
full backup A backup containing all blocks of the files backed up, not only those
blocks changed since the last backup.

G
GMT Greenwich Mean Time. Now referred to as UTC, this is the time zone of
the meridian through Greenwich Observatory in London.

Glossary

543

grid computing An architecture where the delivery of a service to end users is
not tied to certain server resources but can be provided from anywhere in a pool of
resources.
GROUP BY A clause that specifies the grouping attribute rows must have in
common for them to be clustered together.
GUI Graphical User Interface. A layer of an application that lets users work with
the application through a graphical terminal, such as a PC with a mouse.

H
HTTP Hypertext Transfer Protocol. The protocol that enables the World Wide
Web (both invented at the European Organization for Nuclear Research in 1989),
this is a layered protocol that runs over TCP/IP.
HWM High water mark. This is the last block of a segment that has ever been
used—blocks above this are part of the segment but are not yet formatted for use.

I
I/O Input/output. The activity of reading from or writing to disks—often the slowest
point of a data processing operation.
IBM International Business Machines. A well known computer hardware,
software, and services company.
inconsistent backup A backup made while the database was open.
INITCAP
title case.

A function that accepts a string of characters and returns each word in

incremental backup A backup containing only blocks that have been changed
since the last backup was made.

544

OCA Oracle Database 11g: SQL Fundamentals I Exam

inner join When equijoins and nonequijoins are performed, rows from the source
and target tables are matched. These are referred to as inner joins.
INSTR A function that returns the positional location of the nth occurrence of
a specified string of characters in a source string.
instance recovery The automatic repair of damage caused by a disorderly
shutdown of the database.
IOT Index Organized Table. A table type where the rows are stored in the leaf
blocks of an index segment.
IP Internet Protocol. Together with the Transmission Control Protocol, TCP/IP:
the de facto standard communication protocol used for client/server communication
over a network.
IPC Interprocess Communications Protocol. The platform-specific protocol,
provided by your OS vendor, used for processes running on the same machine to
communicate with each other.
ISO International Organization for Standardization. A group that defines many
standards, including SQL.

J
J2EE Java 2 Enterprise Edition. The standard for developing Java applications.
JOIN…ON A clause that allows the explicit specification of join columns
regardless of their column names. This provides a flexible joining format.
JOIN…USING A syntax that allows a natural join to be formed on specific
columns with shared names.
joining Involves linking two or more tables based on common attributes. Joining
allows data to be stored in third normal form in discrete tables, instead of in one
large table.

Glossary

545

JVM Java Virtual Machine. The run-time environment needed for running code
written in Java. Oracle provides a JVM within the database, and there will be one
provided by your operating system.

L
LAST_DAY
date item.

A function used to obtain the last day in a month given any valid

LDAP Lightweight Directory Access Protocol. The TCP implementation of the
X25 directory standard, used by the Oracle Internet Directory for name resolution,
security, and authentication. LDAP is also used by other software vendors, including
Microsoft and IBM.
LENGTH A function that computes the number of characters in a string including
spaces and special characters.
LGWR Log writer. The background process responsible for flushing change vectors
from the log buffer in memory to the online redo logfiles on disk.
library cache A memory structure within the shared pool, used for caching SQL
statements parsed into their executable form.
listener The server-side process that listens for database connection requests from
user processes and launches server processes to establish sessions.
LOB Large Object. A data structure that is too large to store within a table. LOBs
(Oracle supports several types) are defined as columns of a table but are physically
stored in a separate segment.
log switch The action of closing one online logfile group and opening another;
triggered by the LGWR process filling the first group.
LVM Logical Volume Manager. A layer of software that abstracts the physical storage
within your computer from the logical storage visible to an application.

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OCA Oracle Database 11g: SQL Fundamentals I Exam

M
MMON Manageability Monitor. The background process responsible for gathering
performance monitoring information and raising alerts.
MOD The modulus operation, a function that returns the remainder of a division
operation.
MONTHS_BETWEEN A function that computes the number of months
between two given date parameters and is based on a 31-day month.
mounted database A situation where the instance has opened the database
control file but not the online redo logfiles or the data files.
MTBF Mean time between failure. A measure of the average length of running
time for a database between unplanned shutdowns.
MTTR Mean time to recover. The average time it takes to make the database
available for normal use after a failure.
multiplexing

To maintain multiple copies of files.

N
namespace A logical grouping of objects within which no two objects may have
the same name.
natural join A join performed using the NATURAL JOIN syntax when the
source and target tables are implicitly equijoined using all identically named columns.
NCLOB National Character Large Object. A LOB data type for character data,
such as text documents, stored in the alternative national database character set.
NETBEUI NETBIOS Extended User Interface. An enhanced version of NETBIOS.
NETBIOS Network Basic Input Output System. The network communications
protocol that was burnt onto the first network card that IBM ever produced.

Glossary

547

NLS National Language Support. The capability of the Oracle database to support
many linguistic, geographical, and cultural environments—now usually referred to as
globalization.
node A computer attached to a network.
nonequijoin Performed when the values in the join columns fulfill the join
condition based on an inequality expression.
null The absence of a value, indicating that the value is not known, missing,
or inapplicable.
NULLIF A function that tests two terms for equality. If they are equal, the function
returns null; else it returns the first of the two terms tested.
NVL A function that returns either the original item unchanged or an alternative
item if the initial term is null.
NVL2 A function that returns a new if-null item if the original item is null or an
alternative if-not-null item if the original term is not null.

O
OC4J Oracle Containers for J2EE. The control structure provided by the Oracle
Application Server for running Java programs.

OCA
Oracle Certified Associate.

OCI Oracle Call Interface. An API, published as a set of C libraries, that
programmers can use to write user processes that will use an Oracle database.
OCP Oracle Certified Professional. The qualification you are working toward.
ODBC Open Database Connectivity. A standard developed by Microsoft for
communicating with relational databases. Oracle provides an ODBC driver that will
allow clients running Microsoft products to connect to an Oracle database.

548

OCA Oracle Database 11g: SQL Fundamentals I Exam

offline backup A backup made while the database is closed.
OLAP Online Analytical Processing. Select intensive work involving running
queries against a (usually) large database. Oracle provides OLAP capabilities as an
option, in addition to the standard query facilities.
OLTP Online Transaction Processing. A pattern of activity within a database
typified by a large number of small, short, transactions.
online backup A backup made while the database is open.
online redo log

The files to which change vectors are streamed by the LGWR.

OS Operating system. Typically, in the Oracle environment, this will be a version
of Unix (perhaps Linux) or Microsoft Windows.
Oracle Net Oracle’s proprietary communications protocol, layered on top of an
industry standard protocol.
ORACLE_BASE
ORACLE_HOME

The root directory into which Oracle products are installed.
The root directory of any one Oracle product.

outer join A join performed when rows, which are not retrieved by an inner join,
are included for retrieval.

P
parse An action that converts SQL statements into a form suitable for execution.
PGA Program Global Area. The variable sized block of memory used to maintain
the state of a database session. PGAs are private to the session and controlled by the
session’s server process.
PL/SQL Procedural Language/Structured Query Language. Oracle’s proprietary
programming language, which combines procedural constructs, such as flow control,
and user interface capabilities with SQL.

Glossary

549

PMON Process Monitor. The background process responsible for monitoring
the state of user’s sessions against an instance.
primary key The column (or combination of columns) whose value(s) can be used
to identify each row in a table.
projection The restriction of columns selected from a table. Using projection, you
retrieve only the columns of interest and not every possible column.

R
RAC Real Application Clusters. Oracle’s clustering technology, which allows several
instances in different machines to open the same database for scalability, performance,
and fault tolerance.
RAID Redundant Array of Inexpensive Disks. Techniques for enhancing
performance and/or fault tolerance by using a volume manager to present a number
of physical disks to the operating system as a single logical disk.
raw device

An unformatted disk or disk partition.

RDBMS Relational Database Management System. Often used interchangeably
with DBMS.
RAM Random Access Memory. The chips that make up the real memory in your
computer hardware, as opposed to the virtual memory presented to software by the
operating system.
referential integrity A rule defined on a table specifying that the values in
a column (or columns) must map onto those of a row in another table.
relation A two-dimensional structure consisting of tuples with attributes
(aka a table).
REPLACE A function that substitutes each occurrence of a search item in
the source string with a replacement term and returns the modified source string.

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OCA Oracle Database 11g: SQL Fundamentals I Exam

RMAN Recovery Manager. Oracle’s backup and recovery tool.
rowid The unique identifier of every row in the database, used as a pointer to
the physical location of the row.

S
schema The objects owned by a database user.
SCN System Change Number. The continually incrementing number used to track
the sequence and exact time of all events within a database.
segment A database object, within a schema, that stores data.
selection The extraction of rows from a table. Selection includes the further
restriction of the extracted rows based on various criteria or conditions. This allows
you to retrieve only the rows that are of interest and not every row in the table.
self-join A join required when the join columns originate from the same table.
Conceptually, the source table is duplicated and a target table is created. The self-join
then works as a regular join between two discrete tables.
sequence
numbers.

A database object, within a schema, that can generate consecutive

service name A logical name registered by an instance with a listener, which
can be specified by a user process when it issues a connect request.
session A user process and a server process, connected to the instance.
SGA System Global Area. The block of shared memory that contains the memory
structures that make up an Oracle instance.
SID (1) System Identifier. The name of an instance, which must be unique on
the computer the instance is running on. (2) Session Identifier. The number used to
identify uniquely a session logged on to an Oracle instance.

Glossary

551

SMON System Monitor. The background process responsible for opening a database
and monitoring the instance.
spfile Server parameter file. The file containing the parameters used to build an
instance in memory.
SQL Structured Query Language. An international standard language for extracting
data from and manipulating data in relational databases.
SSL Secure Sockets Layer. A standard for securing data transmission, using
encryption, checksumming, and digital certificates.
SUBSTR A function that extracts and returns a segment from a given source
string.
SUM A function that returns an aggregated total of all the nonnull numeric
expression values in a group.
synonym

An alternative name for a database object.

sysdba The privilege that lets a user connect with operating system or password
file authentication and create, start up, and shut down a database.
sysoper The privilege that lets a user connect with operating system or password
file authentication and start up and shut down (but not create) a database.
system A preseeded schema used for database administration purposes.

T
table A logical two-dimensional data storage structure, consisting of rows and
columns.
tablespace The logical structure that abstracts logical data storage in tables from
physical data storage in data files.

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OCA Oracle Database 11g: SQL Fundamentals I Exam

TCP Transmission Control Protocol. Together with the Internet Protocol,
TCP/IP: the de facto standard communication protocol used for client/server
communication over a network.
TCPS TCP with SSL. The secure sockets version of TCP.
tempfile The physical storage that makes up a temporary tablespace, used for
storing temporary segments.
TNS Transparent Network Substrate. The heart of Oracle Net, a proprietary
layered protocol running on top of whatever underlying network transport protocol
you choose to use—probably TCP/IP.
TO_CHAR A function that performs date-to-character and number-to-character
data type conversions.
TO_DATE

A function that explicitly transforms character items into date values.

TO_NUMBER

A function that changes character items into number values.

transaction A logical unit of work that will complete in total or not at all.
tuple A one-dimensional structure consisting of attributes (aka a row).

U
UGA User Global Area. That part of the PGA that is stored in the SGA for sessions
running through shared servers.
UI User interface. The layer of an application that communicates with end
users—nowadays, frequently graphical: a GUI.
URL Uniform Resource Locator. A standard for specifying the location of an object
on the Internet consisting of a protocol, a host name and domain, an IP port number,
a path and filename, and a series of parameters.

Glossary

553

UTC Coordinated Universal Time. Previously known as Greenwich Mean Time
(GMT), UTC is the global standard time zone; all others relate to it as offsets, ahead
or behind.

X
X-Windows The standard GUI environment used on most computers, except
those that run Microsoft Windows.
XML Extensible Markup Language. A standard for data interchange using
documents, where the format of the data is defined by tags within the document.

This page intentionally left blank

INDEX

SYMBOL
% (percentage symbol), 120–121
& (ampersand). See ampersand substitution
() (parenthesis), 384
; (semicolons), 65
_ (underscore character), 120–122
|| (double pipe symbols). See character
concatenation operator
" (double quotes), 454
' (single quotes), 77–81
3GLs (third-generation languages), 10

A
access permissions, 412
ACID test
atomicity, 428
consistency, 429
defined, 537
durability, 429–430
isolation, 429
ad hoc queries, 147
ADD_MONTHS function, 173, 208–211
ADDM (Automatic Database Diagnostic
Monitor), 537
Advanced Encryption Standard (AES), 537
Advanced option, 36
AES (Advanced Encryption Standard), 537
aggregate reporting, 275
aggregation functions. See group functions
AL16UTF16 character set, 537
aliases, 61, 74, 537
ALL operator
group functions, 275
subqueries, 365, 370

alphanumeric data types, 458
ALTER command, 506–507
ALTER INDEX command, 517
ALTER TABLE command, 475–476
alter user commands, 42
ALTER VIEW command, 493–496
American National Standards Institute (ANSI),
23, 537
American Standard Code for Information
Interchange (ASCII), 538
ampersand substitution
for column names, 147
DEFINE command, 149–154
double, 145–147
for expressions, 147–149
overview, 142–143
single, 143–145
for text, 147–149
UNDEFINE command, 149–154
VERIFY command, 149–150, 154–156
AND operator, 126–128
ANSI (American National Standards Institute),
23, 537
ANY operator, 365, 370
Application Programming Interface (API), 538
Application server, 5–7
Application Server Control tool, 8
application tier, 6
architecture, 3–5
arithmetic operators, 70–73
AS keyword, 75–76
ascending sorting, 137–138
ASCII (American Standard Code for Information
Interchange), 538
ASM (Automatic Storage Management), 538
atomicity, 428

Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

556

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

attributes, defined, 538. See also columns
AUTOCOMMIT command, 435–436
Automatic Database Diagnostic Monitor
(ADDM), 537
Automatic Storage Management (ASM), 538
AVG function, 281–282, 538

B
B*Tree indexes, 470, 511–513
background processes, 538
Basic option, 36
BETWEEN operator, 113, 116–119, 331
BFILE data type, 459, 538
binary data type, 458
Binary Large Object (BLOB) data type, 55, 459, 538
bind variables, 538
bitmap indexes, 513–515
black box calculations, 170–171
blind queries, 60, 63
BLOB (Binary Large Object) data type, 55, 459, 538
blocks, defined, 538
Boolean operators
AND, 126–128
versus comparison operators, 135
HAVING clause, 298
NOT, 131–132
and BETWEEN operator, 117–118
OR, 128–131
and rows, 104
bracketed expressions, 70, 132

C
CACHE option, 503
Cartesian products
defined, 310, 538
overview, 314
of two or more tables, 342–346
case conversion functions. See character case
conversion functions
CASE expression, 254, 257–261

CD-ROM included with book
help file, 535
LearnKey software, 535
MasterExam feature, 534
overview, 533–534
removing installation, 535
Study Guide, 535
system requirements, 534
technical support, 535
CDs, companion, 43
Central European Time (CET), 538
Central Processing Units (CPUs), 540
CET (Central European Time), 538
CHAR data type, 55, 458
character case conversion functions
INITCAP, 179–182, 543
LOWER, 177–179
overview, 171
UPPER, 179
character concatenation (||) operator
character manipulations functions, 183
and expressions, 70
and NULL values, 84
overview, 75–76
character conversion
converting dates to, 235–240
converting numbers to, 232–235
to dates, 231, 241–242
to numbers, 230, 242–244
Character Large Object (CLOB) data type,
459, 539
character literal data, 65
character manipulations functions
CONCAT, 183–184
INSTR, 189–190
LENGTH, 184–185
LPAD, 186–187
overview, 171
REPLACE, 193–196
RPAD, 186–187
SUBSTR, 190–193
TRIM, 187–189

Index

character sets, 539
character strings
converting numbers into, 235
implicit conversions, 229
character version, SQL*Plus, 28
character-based conditions, 107–110
check constraints, 473, 539
client tier
SQL, 4
web applications, 6
client tools
SQL Developer
AUTOCOMMIT command in, 435–436
database connections, 35–37
installing and launching, 33–34
User Interface, 34–35
SQL*Plus
AUTOCOMMIT command in, 435–436
database connections, 30–32
on Linux, 27–28
overview, 26
on Windows, 28–30
client-server architecture, 4, 539
CLOB (Character Large Object) data type, 459, 539
cluster segments, 539
clusters, 6, 539
COALESCE function, 247, 253–254, 539
column aliasing, 74–75, 384
column names
ampersand substitution for, 147
qualifying ambiguous, 317–318
columns. See also names of specific columns
creating tables with specifications, 462–464
data types available for, 457–461
defined, 456, 539
grouping by multiple, 291–294
on HR schema, 40–41
COMM column, 463
COMMIT command, 409, 427, 431–432
committing, defined, 539
companion CDs, 43

comparison operators. See also WHERE clause
BETWEEN, 116–119, 331
versus Boolean operators, 135
equality, 113–116
IN, 119–120, 365, 369
inequality, 113–116
IS NULL, 126
LIKE, 120–125, 182
multiple-row subqueries, 365–366
overview, 113
single-row subqueries, 365–366
comparison parameters, 255–256
complete recovery, 539
complex views, 492–493
composite indexes, 515–516
composite inequality operators, 115
composite key unique constraints, 470
composite keys, 18
composite sorting, 139–142
compound queries, 382
CONCAT function, 171, 183–184
concatenation operator. See character
concatenation operator
conditional clauses, 113
conditional expressions
conditional functions
CASE expression, 257–261
DECODE function, 254–257
general functions
COALESCE, 253–254
NULLIF, 249–252
NVL, 247–248
NVL2, 248–249
nesting functions, 245–246
conditional functions
CASE expression, 257–261
DECODE function, 254–257
conditional operators
character-based expressions, 109–110
and NOT operator, 131
WHERE clause, 105

557

558

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

connect identifiers, 539
connect strings, 539
connection pooling, 7
Connection Type radio buttons, 36
connections, indirect, 4
consistency, 429
consistent backups, 539
CONSTRAINT keyword, 494
constraints
defined, 405, 412, 450, 540
defining, 473–477
overview, 469
types of
check, 473, 539
foreign key, 471–472
not null, 470–471
primary key, 471
unique, 470
control files, 540
control transactions
database transactions
ACID test, 428–430
start and end of, 430–431
overview, 427–428
transaction control statements
AUTOCOMMIT command, 435–436
COMMIT command, 432
overview, 431
ROLLBACK command, 433–434
SAVEPOINT command, 434–435
SELECT FOR UPDATE command,
436–438
conversion functions
overview, 174, 228–229
TO_CHAR
converting dates to characters using,
235–240
converting numbers to characters using,
232–235
defined, 228, 552
overview, 231–232

TO_DATE
converting characters to dates using,
205–206, 241–242
defined, 228
overview, 231–232
TO_NUMBER
converting characters to numbers using,
243–244
defined, 228, 552
overview, 231–232
types of
explicit data type conversion, 231
implicit data type conversion, 229–230
Coordinated Universal Time (UTC), 553
correlated subqueries, 366–369
COUNT function, 275, 279–280
COUNTRIES table, 41, 55–56
CPUs (Central Processing Units), 540
CREATE command, 506–507
CREATE SCHEMA command, 38
CREATE TABLE statement, 477
CREATE USER command, 38
CREATE VIEW command, 493–496
cross joins
creating Cartesian products with, 342–346
defined, 310, 538
overview, 314
crow’s feet, 19
CURRVAL pseudo column, 504
CYCLE option, 503

D
data blocks, 540
Data Control Language (DCL) commands, 24
data, defined, 53
Data Definition Language (DDL)
commands, 24
defined, 541
tables
altering definitions, 465–466

Index

constraints, 469–477
creating from subqueries, 464–465
creating with column specifications,
462–464
data types available for columns, 457–461
dropping, 466–469
main database objects, 450–456
table structure, 456–457
truncating, 466–469
data dictionaries, 53, 540
data dictionary views, 540
data files, defined, 540
data guards, 540
Data Manipulation Language (DML)
defined, 541
DELETE command, 407–408
failures of, 409–412
generating rows to be passed to statements,
363–364
INSERT command, 405–406
MERGE command, 408–409
overview, 24, 404–405
TRUNCATE command, 409
UPDATE command, 406–407
and views, 492–493
data models, 22
data normalization, 14–22
data pumps, 540
data retrieval. See SELECT statement
data sorting, 137–142
data tier, 6
data types, 457–46. See also names of specific data types,
Database Administrators (DBAs), 540–541
database buffer caches, 540
Database Configuration Assistant (DBCA), 541
database connections
indirect, 4
SQL Developer, 35–37
SQL*Plus, 30–32
Database Control tool, 8
database links, 498, 540
database logon, 29

559

Database Management System (DBMS), 541
database tier, 6
database transactions
ACID test, 428–430
start and end of, 430–431
Database Writer (DBWn or DBWR), 541
Datapump tool, 405–406
date built-in functions
date arithmetic, 205–206
date storage, 202–204
SYSDATE, 204–205
date conversion
to characters, 229, 235–240
converting characters to, 241–242
DATE data type, 55, 458, 460
date format masks, 236–238
date literals, 190
date manipulation functions
ADD_MONTHS, 208–211
LAST_DAY, 213–214
MONTHS_BETWEEN, 207–208
NEXT_DAY, 211–213
overview, 173, 206
ROUND, 214–216
TRUNC, 216–217, 460
date storage, 202–204
date-based conditions, 110–113
day of the week parameter, 211–212
DBAs (Database Administrators), 540–541
DBCA (Database Configuration Assistant), 541
DBMS (Database Management System), 541
DBWn or DBWR (Database Writer), 541
DCL (Data Control Language) commands, 24
DDL. See Data Definition Language
DD-MON-RR format mask, 203
deadlocks, defined, 541
Debug button, 35
decimal precision parameter, 197, 199
Decision Support Systems (DSS), 14, 542
DECODE function, 254–257, 541
DEFAULT clause, 464
default date format, 111

560

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

DEFAULT keyword, 462
DEFINE command, 149–154
DELETE command
deleting rows from tables with, 423–425
overview, 407–408
demonstration schemas
creating, 42–43
HR, 38–42
OE, 38–42
overview, 37
DEPARTMENTS table, 41, 55
descending sorting, 137–138
DESCRIBE command, 53–57, 228
detail tables, 489
DHCP (Dynamic Host Configuration Protocol), 541
direct paths, 541
directory objects, 541
dirty reads, 429
displaying data from multiple tables
Cartesian products, 342–346
JOIN…ON clause, 322–326
JOIN...USING clause, 321–322
NATURAL JOIN clause, 319–321
nonequijoins, 329–331
N-way joins, 327–329
outer joins, 334–342
overview, 310–311
qualifying ambiguous column names, 317–318
self-joins, 331–334
SQL:1999 syntax, 316–317
types of joins, 311–316
DISTINCT keyword, 61, 105, 279
dividend parameter, 200
divisor parameter, 200
DML. See Data Manipulation Language
Domain Name Service (DNS), 541
domains, defined, 542
dot notation, 317–318
double ampersand substitution, 145–147
double pipe symbols (||). See character
concatenation operator

double quotes ("), 454
DROP command, 466
DROP VIEW command, 493–496
dropping tables, 466–469
DSS (Decision Support Systems), 14, 542
DUAL table, 54, 77, 248
durability, 429–430
Dynamic Host Configuration Protocol (DHCP), 541

E
easy connects, 542
EBCDIC (Extended Binary Coded Decimal
Interchange Code), 542
Edit button, 35
EMPLOYEES table, 41, 55
EMPNO column, 463
ENAME column, 463
end-user sessions, 6–7
Enterprise Manager, 544
entities. See tables
entity-relationship diagram, 19
environment variables, 542
equality operators, 105, 113–116, 370. See also
WHERE clause
equijoins, 310–311, 542
error messages
database connection, 32
“ORA-00904: invalid identifier”, 108
“ORA-00918:column ambiguously defined”, 317
“ORA-00923: FROM keyword not found
where expected”, 73, 75
“ORA-00932: inconsistent data types”, 253
“ORA-00934: group function is not allowed
here”, 290
“ORA-00935: group function is nested too
deeply”, 287
“ORA-00937: not a single-group group
function”, 290
“ORA-00942: table or view does not exist”, 73
“ORA-01427: single-row subquery returns more
than one row”, 367, 371, 420

Index

“ORA-1555 snapshot too old”, 429
“ORA-25154:column part of USING clause
cannot have qualifier”, 317
ESCAPE identifier, 123–124
explicit data type conversion, 231
expressions. See also conditional expressions
ampersand substitution for, 147–149
CASE, 254, 257–261
column aliasing and, 74–75
DUAL table, 77
literal values, 77
Extended Binary Coded Decimal Interchange Code
(EBCDIC), 542
Extensible Markup Language (XML), 553
extents, 425, 456

F
fact tables, 542
FGA (Fine Grained Auditing), 542
fields. See columns
File button, 35
fill mode (fm) operator, 236, 238, 240–241
Fine Grained Auditing (FGA), 542
first normal form, 15–16
FLOAT data type, 458
fm (fill mode) operator, 236, 238, 240–241
FOR UPDATE clause, 438
FORCE keyword, 494
foreign key constraints, 471–472
foreign keys, 18, 85–88
formal names, 194–195
format masks, 203–204
format parameter, 232
FROM clause, 358
full backups, 542
full outer joins, 338–342
functions. See also names of specific functions
multiple-row, 175–176
operating on character data, 171–172
operating on date information, 173
operating on numeric data, 172–173

561

G
general functions
COALESCE, 253–254, 539
NULLIF, 249–252, 547
NVL, 247–248, 547
NVL2, 248–249, 547
overview, 247
GMT (Greenwich Mean Time), 542
good practice, 66–69
Graphical User Interfaces (GUIs), 543
graphical version, SQL*Plus, 28
Greenwich Mean Time (GMT), 542
grid computing, 8–9, 543
Grid Control tool, 8
GROUP BY clause
creating groups of data, 287–288
defined, 543
grouping by multiple columns, 291–294
and HAVING clause, 298
nesting group functions, 285
overview, 288–291
group functions
AVG, 281–282, 538
COUNT, 275, 279–280
defined, 274–275
GROUP BY clause
creating groups of data, 287–288
grouping by multiple columns, 291–294
overview, 288–291
HAVING clause
overview, 296–299
restricting group results, 294–296
MAX, 282–285
MIN, 282–285
nesting, 285–287
overview, 175
versus single-row functions, 213
SUM, 280–281, 551
syntax of, 275–278
types of, 275–278

562

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

grouped rows
overview, 296–299
restricting group results, 294–296
grouping attribute, 288, 291
group-level results, 295
GUIs (Graphical User Interfaces), 543

H
hash clusters, 462
hash joins, 492, 511
HAVING clause
overview, 296–299
restricting group results, 274, 294–296
heap tables, 462
high water mark (HWM), 425–426, 543
HIREDATE column, 463
HR (Human Resources) demonstration schema,
38–42
HTTP (Hypertext Transfer Protocol), 543
Human Resources (HR) demonstration
schema, 38–42
HWM (high water mark), 425–426, 543
Hypertext Transfer Protocol (HTTP), 543

I
IBM (International Business Machines), 543
if-then-else logic, 228, 254
implicit data type conversion, 229–230
implicit precedence, 391
IN operator, 113, 119–120, 365, 369
inconsistent backups, 543
INCREMENT BY option, 503
incremental backups, 543
indentation, 66–69
index clusters, 462
Index Organized Tables (IOTs), 462, 544

indexes
creating, 515–517
defined, 452
dropping, 517–520
modifying, 517–520
overview, 509
purpose of, 510–511
types of, 511–515
using, 515–517
indirect database connections, 4
inequality operators, 113–116
INITCAP function, 171, 179–182, 543
inline views, 362
inner joins
defined, 544
versus outer, 335–336
inner queries, 358
input parameters, 170, 176
input/output (I/O), defined, 543
INSERT command, 5, 405–406
insertion anomalies, 15–16
installing SQL Developer, 33–34
instance recovery, 544
instances, 3–4
INSTR function (in-string), 172, 189–190, 544
INTEGER data type, 386, 458
International Business Machines (IBM), 543
International Organization for Standardization
(ISO), 23, 544
Internet Protocol (IP), defined, 544
Interprocess Communications Protocol (IPC), 544
INTERSECT set operator, 382, 388–389
INTERVAL DAY TO SECOND data type, 459
INTERVAL YEAR TO MONTH data type, 459
I/O (input/output), defined, 543
IOTs (Index Organized Tables), 462, 544
IP (Internet Protocol), defined, 544
IPC (Interprocess Communications Protocol), 544
IS NULL operator, 113, 126

Index

ISO (International Organization for
Standardization), 23, 544
isolation, 429

563

K
key columns, 470

J

L

Java 2 Enterprise Edition (J2EE), 6, 544
Java Runtime Environment (JRE), 33
Java Virtual Machine (JVM), 545
JOB_HISTORY table, 41, 55
JOBS table, 41, 55
JOIN…ON clause
defined, 544
joining tables using, 331–334
overview, 313, 322–326, 328
joins
cross
creating Cartesian products with, 342–346
defined, 310, 538
overview, 314
defined, 58, 544
equijoins, 310–311, 542
hash, 492, 511
inner, 335–336, 544
natural, 312–314, 546
nested loop, 491–492, 511
nonequijoins, 310, 329–331, 547
N-way, 327–329
outer
defined, 310, 548
full, 338–342
versus inner, 335–336
left, 336–337
overview, 314, 334–335
right, 337–338
overview, 311–312
self-joins, 310, 331–334, 346, 550
sort merge, 511
syntax of, 314–316
JOIN...USING clause, 313, 321–322, 544
JRE (Java Runtime Environment), 33
JVM (Java Virtual Machine), 545

Large Object (LOB) data types, 450, 459, 545
LAST_DAY function, 173, 213–214, 545
LD_LIBRARY_PATH variable, 27
LDAP (Lightweight Directory Access Protocol), 545
LearnKey software, 534, 535
left outer joins, 336–337
LENGTH function, 171, 184–185, 545
LGWR (log writer) process, 545
library caches, 545
Lightweight Directory Access Protocol (LDAP), 545
LIKE operator, 113, 120–125, 182
Linux, SQL*Plus on, 27–28
listeners, 545
literal values, 65, 77, 190
LOB (Large Object) data types, 450, 459, 545
LOCATIONS table, 41, 55–56
log switch action, 545
log writer (LGWR) process, 545
logical storage, 13
Logical Volume Manager (LVM), 545
log-on/off cycle, 4
LONG data type, 459
LONG RAW data type, 459
LOWER function, 171, 177–179
lowercase, writing SQL statements in, 64–65
LPAD function, 172, 186–187
LVM (Logical Volume Manager), 545

M
main database objects
namespaces, 455–456
naming schema objects, 453–455
schemas, 452–453
types of, 450–452
users, 452–453

564

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

Manageability Monitor (MMON), 546
manipulating data
control transactions
database transactions, 428–431
overview, 427–428
transaction control statements, 431–438
deleting rows from tables
DELETE command, 423–425
MERGE command, 426–427
TRUNCATE command, 425–426
DML statements
DELETE command, 407–408
failures of, 409–412
INSERT command, 405–406
MERGE command, 408–409
overview, 404–405
TRUNCATE command, 409
UPDATE command, 406–407
inserting rows into tables, 413–418
updating rows in tables, 419–422
many-to-many relationship, 20–21
many-to-one relationship, 18
master-detail relationship, 335
MasterExam feature, 534
MAX function, 282–285
MAXVALUE option, 503
mean time between failure (MTBF), 546
mean time to recover (MTTR), 546
MERGE command
deleting rows from tables with, 426–427
overview, 404, 408–409
metadata, defined, 53
middle tier, 6
Migrate button, 35
MIN function, 282–285
MINUS set operator, 382, 389
MINVALUE option, 503
MMON (Manageability Monitor), 546
MOD function (modulus), 173, 199–202, 546
MONTHS_BETWEEN function, 173, 207–208, 546
mounted databases, 546

MTBF (mean time between failure), 546
MTTR (mean time to recover), 546
multiple-row functions. See group functions
multiple-row subqueries, 365–366, 369–372
multiplexing, 546

N
namespaces, defined, 546
National Character Large Object (NCLOB) data
type, 459, 546
National Language Support (NLS), 116,
232–233, 547
NATURAL JOIN clause, 312, 319–321
natural joins, 312–314, 546
NATURAL keyword, 323, 346
Navigate button, 35
NCLOB (National Character Large Object) data
type, 459, 546
negative start position parameter, 192
nested loop joins, 491–492, 511
nesting
functions
CONCAT, 183
overview, 245–246
single-row, 174, 287
subqueries, 358
NETBEUI (NETBIOS Extended User
Interface), 546
NETBIOS (Network Basic Input Output
System), 546
NETBIOS Extended User Interface
(NETBEUI), 546
Network Basic Input Output System
(NETBIOS), 546
NEXT_DAY function, 173, 211–213
NEXTVAL pseudo column, 504
NLS (National Language Support),
116, 232–233, 547
NLS_CURRENCY value, 232

Index

NLS_DATE_LANGUAGE database parameter,
211–212
NLS_SESSION_PARAMETERS view, 232
nodes, defined, 547
NOFORCE keyword, 494
nonequijoins, 310, 329–331, 547
nonunique indexes, 511
normalization, 10–11, 22
NOT IN operator, 361, 365
not null constraints, 470–471
NOT operator, 131–132
null comparison, 113, 126
null values
defined, 55, 547
foreign keys and nullable columns, 85–88
NOT NULL columns, 82–84
NULLABLE columns, 82–84
overview, 81–82
NULLIF function, 247, 249–252, 547
NULLS FIRST keywords, 137
NULLS LAST keywords, 137
number conversion
to characters, 229, 232–235
converting characters to, 242–244
NUMBER data type, 458, 460
number parameter, 232
numeric columns, 54
numeric data types, 458
numeric format masks, 234
numeric functions
MOD, 199–202, 546
overview, 172–173
ROUND, 196–198
TRUNC, 198–199
numeric literals, 190
numeric-based conditions, 105–107
NVARCHAR2 data type, 458
NVL function, 247–248, 547
NVL2 function, 247–249, 547
N-way joins, 327–329

565

O
object namespaces, 455–456
OC4J (Oracle Containers for J2EE), 547
OCAs (Oracle Certified Associates), 547
OCI (Oracle Call Interface), 10, 547
OCPs (Oracle Certified Professionals), 547
ODBC (Open Database Connectivity), 547
OE (Order Entry) demonstration schema, 38–42
offline backups, 548
OLAP (Online Analytical Processing), 548
OLTP (Online Transaction Processing) systems,
14, 548
ON DELETE CASCADE constraint, 472
ON DELETE SET NULL constraint, 472
ON keyword, 323, 346
one-legged rows, 310
Online Analytical Processing (OLAP), 548
online backups, 548
Online Transaction Processing (OLTP) systems,
14, 548
Open Database Connectivity (ODBC), 547
operating systems (OS), defined, 548
operators. See also names of specific operators
arithmetic, 70–73
Boolean
AND, 126–128
versus comparison operators, 135
HAVING clause, 298
NOT, 131–132
and BETWEEN operator, 117–118
OR, 128–131
and rows, 104
character concatenation
character manipulations functions, 183
and expressions, 70
and NULL values, 84
overview, 75–76
comparison
BETWEEN, 116–119, 331
versus Boolean operators, 135

566

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

operators (Cont.)
equality, 113–116
IN, 119–120, 365, 369
inequality, 113–116
IS NULL, 126
LIKE, 120–125, 182
multiple-row subqueries, 365–366
overview, 113
single-row subqueries, 365–366
conditional
character-based expressions, 109–110
and NOT operator, 131
WHERE clause, 105
set
combining multiple queries into single,
386–392
controlling order of rows returned,
393–395
general principles of, 384–386
overview, 382–383
Venn diagrams and, 383–384
optimizer hints, 517
OR operator, 128–131
OR REPLACE keywords, 494
“ORA-00904: invalid identifier” error
message, 108
“ORA-00918: column ambiguously defined” error
message, 317
“ORA-00923: FROM keyword not found where
expected” error message, 73, 75
“ORA-00932: inconsistent data types” error
message, 253
“ORA-00934: group function is not allowed here”
error message, 290
“ORA-00935: group function is nested too deeply”
error message, 287
“ORA-00937: not a single-group group function”
error message, 290
“ORA-00942: table or view does not exist”
error message, 73
“ORA-01427: single-row subquery returns more
than one row” error message, 367, 371, 420

“ORA-1555 snapshot too old” error message, 429
“ORA-25154: column part of USING clause
cannot have qualifier” error message, 317
Oracle Application server, 5–7
Oracle Call Interface (OCI), 10, 547
Oracle Certified Associates (OCAs), 547
Oracle Certified Professionals (OCPs), 547
Oracle Containers for J2EE (OC4J), 547
Oracle Enterprise Manager, 7–8
Oracle Enterprise Manager Application Server
Control tool, 8
Oracle Enterprise Manager Database
Control tool, 8
Oracle Enterprise Manager Grid Control tool, 8
Oracle Home software installation, 25
Oracle Net protocol, 4, 548
Oracle Spatial, 453
ORACLE_BASE root directory, 27, 548
orcl_sys connection, 34
ORDER BY clause
ascending sorting, 137–138
composite sorting, 139–142
compound queries, 385
descending sorting, 137–138
order of rows, 393
overview, 136
positional sorting, 139
Order Entry (OE) demonstration schema, 38–42
ORDER option, 503
Organisation Internationale de Normalisation
(ISO), 23, 544
original parameter, 247, 249
orphaned rows, 310
OS (operating systems), defined, 548
outer joins
defined, 310, 548
full, 338–342
versus inner, 335–336
left, 336–337
overview, 314, 334–335
right, 337–338
outer queries, 358

Index

P
paradigms, defined, 2
parent-child relationship, 335
parentheses (), 384
parsing, defined, 548
partitioned tables, 462
PATH variable, 27–28
pattern comparison, 113, 120–125, 182
percentage symbol (%), 120–121
PGAs (Program Global Areas), 548
physical storage, 13
PL/SQL (Procedural Language/Structured Query
Language), 10, 170, 548
PMON (Process Monitor), 549
positional notation, 413
positional sorting, 139
primary key constraints, 471
primary keys, 15, 18, 471, 549
private synonyms, 498–501
procedural languages, 25, 170
Procedural Language/Structured Query Language
(PL/SQL), 10, 170, 548
Process Monitor (PMON), 549
Program Global Areas (PGAs), 548
projection, 57, 104, 549
pseudo columns, 504
public synonyms, 498–501
punctuation marks, 238
pure natural joins, 312, 314

Q
q (quote) operator, 77–81
queries. See also subqueries
ad hoc, 147
blind, 60, 63
combining multiple into single
complex examples, 390–392
INTERSECT operator, 388–389
MINUS operator, 389

overview, 386–387
UNION ALL operator, 387
UNION operator, 387–388
compound, 382
generic form, 142
inner, 358
limiting rows retrieved by
Boolean operators, 126–132
comparison operators, 113–126
precedence rules, 132–135
WHERE clause, 104–113
outer, 358
repeated, 147
sorting rows retrieved by, 136–142
quote (q) operator, 77–81

R
RACs (Real Application Clusters), 5, 549
radio buttons, Connection Type, 36
RAID (Redundant Array of Inexpensive
Disks), 549
Random Access Memory (RAM), 549
range comparison, 113, 116–119, 331. See
BETWEEN operator
RAW data type, 458
raw devices, 549
RDBMS (Relational Database Management
System), 2, 549
readability, SQL statements and, 66–69
Real Application Clusters (RACs), 5, 549
records. See rows
Recovery Manager (RMAN), 550
Redundant Array of Inexpensive Disks
(RAID), 549
referential integrity, 549
REGIONS table, 41, 55–56
Registry variables, 30
relational algebra, 57, 382
Relational Database Management System
(RDBMS), 2, 549

567

568

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

relational structures. See also tables
data normalization, 14–22
overview, 10–11
rows, 11–14
tables, 11–14
relational theory, 57
relations, defined, 549. See also tables
repeated queries, 147
REPLACE function, 172, 193–196, 549
replacement term parameter, 193
reporting aggregated data
AVG function, 281–282
COUNT function, 279–280
defined, 274–275
GROUP BY clause, 287–294
HAVING clause, 294–299
MAX function, 282–285
MIN function, 282–285
nesting, 285–287
SUM function, 280–281
syntax of, 275–278
types of, 275–278
restricting data
Boolean operators, 126–132
comparison operators, 113–126
precedence rules, 132–135
WHERE clause, 104–113
right outer joins, 336, 337–338
RMAN (Recovery Manager), 550
Role drop-down box, 36
ROLLBACK command, 409, 427, 431, 433–434
rollbacks, 412
ROUND function
date, 214–216
numeric, 172, 196–198
ROWID data type, 459
rowid identifier, 512, 550
row-level results, 294
rows
controlling order of returned, 393–395
defined, 456

deleting from tables
DELETE command, 423–425
MERGE command, 426–427
TRUNCATE command, 425–426
including or excluding grouped, 294–299
inserting into tables, 406, 413–418
limiting retrieved
Boolean operators, 126–132
comparison operators, 113–126
precedence rules, 132–135
WHERE clause, 104–113
overview, 11–14
sorting retrieved, 136–142
subqueries for generating, 363–364
updating in tables, 419–422
RPAD function, 172, 186–187
Run button, 35
runtime binding. See ampersand substitution
runtime substitution. See ampersand substitution

S
SAL column, 463
SAVEPOINT command, 427–428, 431, 434–435
scalar subqueries, 358
schema objects
indexes
creating, 515–517
defined, 452
dropping, 517–520
modifying, 517–520
overview, 509
purpose of, 510–511
types of, 511–515
using, 515–517
naming, 453–455
sequences
creating, 502–504
defined, 452, 550
overview, 488, 501–502
using, 504–509

Index

synonyms, 452, 498–501, 551
views
ALTER VIEW, 493–496
complex, 492–493
CREATE VIEW, 493–496
data dictionary, 540
defined, 452
DROP VIEW, 493–496
to enforce security, 489–490
inline, 362
to make data comprehensible, 491
overview, 488–489
for performance, 491–492
to prevent errors, 490–491
retrieving data from, 497–498
simple, 492–493
to simplify user SQL, 490
schemas
defined, 38, 452–453, 550
demonstration
creating, 42–43
HR, 38–42
OE, 38–42
overview, 37
SCN (System Change Number), 550
SEARCH variable, 145
searched CASE expression, 257
second normal form, 17
Secure Sockets Layer (SSL), 551
segments, defined, 550
SELECT FOR UPDATE command, 436–438
SELECT statement
applying conditional expressions in
conditional functions, 254–261
general functions, 247–254
nesting functions, 245–246
capabilities of, 57–58
DESCRIBE Table command, 53–57
displaying data from multiple tables with
JOIN…ON clause, 322–326
JOIN...USING clause, 321–322
NATURAL JOIN clause, 319–321

569

nonequijoins, 329–331
N-way joins, 327–329
overview, 310–311
qualifying ambiguous column names,
317–318
SQL:1999 syntax, 316–317
types of joins, 311–316
executing
expressions, 74–75, 77
null values, 81–88
operators, 70–73, 75–76, 77–81
rules, 64–69
syntax, 59–64
general form using ANSI SQL:1999 syntax, 316
overview, 52–53
set operators, 382
single-row functions in
character case conversion, 177–182
character manipulations, 183–196
date built-in, 202–206
date manipulation, 206–217
numeric, 196–202
subqueries, 362
substituting expressions and text, 147
selection, 58, 104, 550
self-joins, 310, 331–334, 346, 550
self-referencing foreign key constraints, 472
self-referencing foreign keys, 39
semicolons (;), 65
sequences
creating, 502–504
defined, 452, 550
overview, 488, 501–502
using, 504–509
server parameter file (spfile), 551
server technologies
architecture, 3–5
Grid computing, 8–9
languages, 9–10
Oracle Application server, 5–7
Oracle Enterprise Manager, 7–8
overview, 2

570

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

server tier, 4
service names, 550
Session Identifier (SID), 550
sessions, defined, 4, 550
SET command, 152, 154
set comparison, 113, 119–120, 365, 369.
See IN operator
SET DEFINE ON|OFF command, 152
set operators
combining multiple queries into single
complex examples, 390–392
INTERSECT operator, 388–389
MINUS operator, 389
overview, 386–387
UNION ALL operator, 387
UNION operator, 387–388
controlling order of rows returned, 393–395
general principles of, 384–386
overview, 382–383
Venn diagrams and, 383–384
SET UNUSED command, 466
SGA (System Global Area), 550
SID (Session Identifier), 550
SID (System Identifier), 550
simple CASE expression, 257
simple views, 492–493
single ampersand substitution, 143–145
single quotes ('), 77–81
single values, 285
single-row functions
character case conversion
INITCAP, 179–182, 543
LOWER, 177–179
overview, 171
UPPER, 179
character manipulations
CONCAT, 183–184
INSTR, 189–190
LENGTH, 184–185
LPAD, 186–187
overview, 171
REPLACE, 193–196

RPAD, 186–187
SUBSTR, 190–193
TRIM, 187–189
date built-in
date arithmetic, 205–206
date storage, 202–204
SYSDATE, 204–205
date manipulation
ADD_MONTHS, 208–211
LAST_DAY, 213–214
MONTHS_BETWEEN, 207–208
NEXT_DAY, 211–213
overview, 173, 206
ROUND, 214–216
TRUNC, 216–217, 460
numeric
MOD, 199–202, 546
overview, 172–173
ROUND, 196–198
TRUNC, 198–199
overview, 170–171
single-row subqueries, 365–366, 369–372
SMON (System Monitor), 551
sort merge joins, 511
sorting data, with ORDER BY clause, 136–142
Source button, 35
source number parameter, 197, 199
source tables, 311, 408
spfile (server parameter file), 551
SQL. See Structured Query Language
SQL Developer
AUTOCOMMIT command in, 435–436
database connections, 35–37
installing and launching, 33–34
User Interface, 34–35
SQL*Loader utility, 405
SQL*Plus
AUTOCOMMIT command in, 435–436
database connections, 30–32
on Linux, 27–28
overview, 26
on Windows, 28–30

Index

SQL:1999 syntax, 316–317
SSL (Secure Sockets Layer), 551
star transformation, 361–362
STAR_TRANSFORMATION_ENABLED
initialization parameter, 362
START WITH option, 503
statement terminators, 65–66
STDDEV function, 278
Structured Query Language (SQL). See also
SELECT statement
commands, 23–24
defined, 551
set-oriented language, 25
standards for, 23
subqueries
for comparison purposes, 360–361
correlated, 366–369
creating tables from, 464–465
defined, 358–360
for generating rows passed to DML statements,
363–364
for generating tables, 362
for generating values for projection, 362
multiple-row, 365–366, 369–372
single-row, 365–366, 369–372
for star transformation, 361–362
substitution. See ampersand substitution
SUBSTR function (substring), 172, 190–193, 551
SUM function, 280–281, 551
synonyms, 452, 498–501, 551
syntax errors, 410
SYS user, 453
SYSDATE function, 173, 204–205
sysdba privilege, 36, 551
sysoper privilege, 551
System Change Number (SCN), 550
System Global Area (SGA), 550
System Identifier (SID), 550
System Monitor (SMON), 551
SYSTEM user, 453
systems, defined, 551

571

T
table aliases, 317
tables
DDL statements
altering definitions, 465–466
constraints, 469–477
creating from subqueries, 464–465
creating with column specifications, 462–464
data types available for columns, 457–461
dropping, 466–469
main database objects, 450–456
table structure, 456–457
truncating, 466–469
defined, 551
deleting rows from
DELETE command, 423–425
MERGE command, 426–427
TRUNCATE command, 425–426
displaying data from multiple
Cartesian products, 342–346
JOIN…ON clause, 322–326
JOIN...USING clause, 321–322
NATURAL JOIN clause, 319–321
nonequijoins, 329–331
N-way joins, 327–329
outer joins, 334–342
overview, 310–311
qualifying ambiguous column names,
317–318
self-joins, 331–334
SQL:1999 syntax, 316–317
types of joins, 311–316
on HR schema, 40–41
inserting rows into, 413–418
overview, 11–14
parent-child relationship, 335
subqueries for generating, 362
updating rows in, 419–422
tablespaces, defined, 551
target tables, 311, 408

572

OCA Oracle Database 11g: SQL Fundamentals I Exam Guide (Exam 1Z0-051)

TCL (Transaction Control Language) commands, 24
TCP (Transmission Control Protocol), 552
TCPS (Transmission Control Protocol with SSL), 552
tempfile, 552
Test button, 37
text, ampersand substitution for, 147–149
third normal form, 17–18, 327
third-generation languages (3GLs), 10
TIMESTAMP data type, 55, 387, 458
TIMESTAMP WITH LOCAL TIMEZONE data
type, 459
TIMESTAMP WITH TIMEZONE data type, 459
TNS (Transparent Network Substrate), 552
TNS option, 36
TO_CHAR conversion function
converting dates to characters using, 235–240
converting numbers to characters using,
232–235
defined, 228, 552
overview, 231–232
TO_DATE conversion function
converting characters to dates using,
205–206, 241–242
defined, 228
overview, 231–232
TO_NUMBER conversion function
converting characters to numbers using,
243–244
defined, 228, 552
overview, 231–232
Tool for Application Developers (TOAD), 5
Tools button, 35
Transaction Control Language (TCL) commands, 24
transaction control statements
AUTOCOMMIT command, 435–436
COMMIT command, 432
ROLLBACK command, 433–434
SAVEPOINT command, 434–435
SELECT FOR UPDATE command, 436–438
transaction isolation, 437
transactions, defined, 427–428, 552
Transmission Control Protocol (TCP), 552

Transmission Control Protocol with SSL (TCPS), 552
Transparent Network Substrate (TNS), 552
transposition errors, 410
TRIM function, 172–173, 187–189
TRUNC function (truncate)
date, 216–217, 460
numeric, 172, 198–199
TRUNCATE command
deleting rows from tables with, 425–426
overview, 409
truncate function. See TRUNC function
truncating tables, 466–469
truth tables
NOT Operator, 131
OR Operator, 128
tuples, defined, 552. See also rows
two-tier model, 4
type casting, 410, 461

U
UGA (User Global Area), 552
UIs. See user interfaces
UNDEFINE COLNAME command, 151
UNDEFINE command, 149–154
underscore character (_), 120–122
Uniform Resource Locators (URLs), 552
UNION ALL set operator, 382, 385, 387, 393
UNION set operator, 339, 382, 387–388
unique constraints, 470
unique indexes, 511
un-normalized data model, 22
UPDATE command, 406–407
UPPER function, 171, 179
uppercase, writing SQL statements in, 64–65
URLs (Uniform Resource Locators), 552
User Global Area (UGA), 552
user interfaces (UIs)
defined, 552
GUIs, 543
NETBEUI, 546
SQL Developer, 34–35

Index

users
defined, 38
indirect connection with database, 4
overview, 452–453
USING keyword, 322, 346
UTC (Coordinated Universal Time), 553

V
values
literal, 65, 77, 190
null
defined, 55, 547
foreign keys and nullable columns, 85–88
NOT NULL columns, 82–84
NULLABLE columns, 82–84
overview, 81–82
single, 285
subqueries for generating, 362
VALUES clause, 416
VARCHAR2 data type, 54–55, 229, 386, 458, 460
variables
assigning, 144
bind variables, 538
environment, 542
LD_LIBRARY_PATH, 27
PATH, 27–28
Registry, 30
SEARCH, 145
VARIANCE function, 277
Venn diagrams, 383–384
Venn, John, 383
VERIFY command, 149–150, 154–156
View button, 35
views
ALTER VIEW, 493–496
complex, 492–493
CREATE VIEW, 493–496
data dictionary, 540

defined, 452
DROP VIEW, 493–496
to enforce security, 489–490
inline, 362
overview, 488–489
for performance, 491–492
to prevent errors, 490–491
retrieving data from, 497–498
simple, 492–493
to simplify user SQL, 490
using to make data comprehensible, 491
virtualization, 8

W
warning prompts, 407
web applications, defined, 6
WHEN...THEN statement, 257
WHERE clause
case conversion functions, 182
character-based conditions, 107–110
date-based conditions, 110–113
versus HAVING clause, 294, 296
and joins, 328
nesting subqueries, 358
numeric-based conditions, 105–107
overview, 104–105
and rows, 274, 407, 465
subqueries, 372
TO_DATE function, 242
wildcard symbols, 120–122
Windows, SQL*Plus on, 28–30
WITH CHECK OPTION keywords, 494
WITH READ ONLY keywords, 494

X
XML (Extensible Markup Language), 553
X-Windows, 553

573

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Aerospace and Defense Manufacturing
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Communications and Media
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Consumer Sector/Consumer Packaged Goods
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4
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98 Other Business and Services

DESCRIBES YOUR PRIMARY JOB
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Corporate Management/Staff

▫ 01 Executive Management (President, Chair,
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(VP/Director/ Manager/Controller,
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03 Sales/Marketing Management
(VP/Director/Manager)
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IS/IT Staff

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06
07
08
09
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11
98

Programming Management
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▫ 01 Digital Equipment UNIX
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99

04
05
06
07
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5

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Other
None of the above

Hardware
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IN YOUR JOB, DO YOU USE OR PLAN TO
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Software

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01
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19

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CASE
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File Management
Finance
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Workstation

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28
29
30
31
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33

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Modems
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34
35
36
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40
98

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Tools

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99
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Oracle E-Business Suite

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01
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11
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15
16

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Oracle9i
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Oracle9i Application Server
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Server/Software

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24
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30
31
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98

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PURCHASE OF ANY OF THE FOLLOWING?
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▫ 01
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▫ 05
99 ▫

2 WHICH OF THE FOLLOWING BEST

▫ 02 Finance/Administrative Management

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99

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7 WHAT OTHER DATABASE PRODUCTS ARE
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99
8

▫

Access
▫ 08
Baan
▫ 09
dbase
▫ 10
Gupta
▫ 11
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▫ 12
Informix
▫ 13
Ingres
▫ 14
Other
None of the above

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Progress
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VSAM

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9

01
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98

01
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05

BEA
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Sun
Other

DURING THE NEXT 12 MONTHS, HOW
MUCH DO YOU ANTICIPATE YOUR
ORGANIZATION WILL SPEND ON
COMPUTER HARDWARE, SOFTWARE,
PERIPHERALS, AND SERVICES
F O R Y O U R L O C A T I O N ? (check only one)

▫
▫
▫
▫
▫
▫

01
02
03
04
05
06

Less than $10,000
$10,000 to $49,999
$50,000 to $99,999
$100,000 to $499,999
$500,000 to $999,999
$1,000,000 and over

10 W H A T I S Y O U R C O M P A N Y ’ S Y E A R L Y

S A L E S R E V E N U E ? (please choose one)

▫
▫
▫
▫
▫

01
02
03
04
05

$500, 000, 000 and above
$100, 000, 000 to $500, 000, 000
$50, 000, 000 to $100, 000, 000
$5, 000, 000 to $50, 000, 000
$1, 000, 000 to $5, 000, 000
100103

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Subject                         : Team DDU
Title                           : OCA Oracle Database 11g : SQL Fundamentals I Exam Guide (Exam 1Z0-051)
Author                          : John Watson , Roopesh Ramklass
ISBN                            : 0071597867
Create Date                     : 2008:07:21 01:31:13+08:00
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