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OracCertPrs5.5 / OCA Oracle Database 12c / Roopesh Ramklass / 182028-0 / Front Matter
OCA Oracle Database 12c:
SQL Fundamentals I
Exam Guide
(Exam 1Z0-061)
®
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OracCertPrs5.5 / OCA Oracle Database 12c / Roopesh Ramklass / 182028-0 / Front Matter
ABOUT THE AUTHOR
Roopesh Ramklass (Canada) is an Oracle Certified Master with expertise in
infrastructure, middleware, and database architecture. He has worked for Oracle
Global Support, Advanced Customer Services, and Oracle University. He has run
an IT consultancy and is experienced with infrastructure systems provisioning,
software development, and systems integration. He has spoken at numerous Oracle
User Group conferences and is the author of several technology books.
About the Technical Editor
Cecil Strydom has been working exclusively as an Oracle DBA for the last 13 years,
covering everything from performance tuning to disaster recovery. He has worked
with Oracle database versions 8 through 11g and has experience with RAC, Data
Guard, eBusiness, Portal, Apex, and various other Oracle applications. Cecil is an
OCP-certified Oracle DBA and is currently a systems consultant for a multinational
company based in Johannesburg, South Africa.
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OracCertPrs5.5 / OCA Oracle Database 12c / Roopesh Ramklass / 182028-0 / Front Matter
OCA Oracle Database 12c:
SQL Fundamentals I
Exam Guide
(Exam 1Z0-061)
Roopesh Ramklass
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OracCertPrs5.5 / OCA Oracle Database 12c / Roopesh Ramklass / 182028-0 / Front Matter
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With love to Ameetha for sharing your journey with me.
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OracCertPrs5.5 / OCA Oracle Database 12c / Roopesh Ramklass / 182028-0 / Front Matter
CONTENTS AT A GLANCE
1 Relational Database Design Using Oracle ....................... 1
2 Data Retrieval Using the SQL SELECT Statement ............... 55
3 Restricting and Sorting Data ................................. 109
4 Single-Row Functions ...................................... 175
5 Using Conversion Functions and
Conditional Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
6 Reporting Aggregated Data Using the Group Functions ........... 277
7 Displaying Data from Multiple Tables .......................... 315
8 Using Subqueries to Solve Problems ........................... 365
9 Using the Set Operators ..................................... 395
10 Manipulating Data ......................................... 423
11 Using DDL Statements to Create and Manage Tables ............. 475
A About the CD-ROM ....................................... 513
Glossary ................................................. 517
Index .................................................... 533
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CONTENTS
Acknowledgments .................................... xix
Preface ........................................... xxi
Introduction ........................................ xxv
1 Relational Database Design Using Oracle .......... 1
Position the Server Technologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
The Oracle Server Architecture ....................... 3
The Oracle WebLogic Server .......................... 5
Oracle Enterprise Manager ............................ 7
Cloud Computing ................................... 8
Exercise 1-1: Investigate Your Database
and Application Environment ....................... 9
Development Tools and Languages ..................... 10
Understand Relational Structures ............................. 11
Real-World Scenarios ................................ 11
Data Modeling ..................................... 12
Entities and Relations ............................... 12
Exercise 1-2: Design an Entity-Relationship Diagram
for the Geological Cores Scenario .................... 20
Rows and Tables .................................... 22
Summarize the SQL Language ................................ 26
SQL Standards ..................................... 27
SQL Commands .................................... 27
A Set-Oriented Language ............................ 28
Use the Client Tools ....................................... 29
SQL*Plus .......................................... 30
SQL Developer ..................................... 37
Create the Demonstration Schemas ........................... 42
Users and Schemas .................................. 42
The HR and OE Schemas ............................ 43
Demonstration Schema Creation ...................... 46
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✓ Two-Minute Drill ................................... 49
Q&A Self Test .......................................... 50
Lab Question ...................................... 52
Self Test Answers ................................... 53
Lab Answer ........................................ 54
2 Data Retrieval Using the SQL
SELECT Statement ............................ 55
List the Capabilities of SQL SELECT Statements ................ 56
Introducing the SQL SELECT Statement ................ 56
The DESCRIBE Table Command ...................... 57
Exercise 2-1: Describing the Human
Resources Schema ................................ 60
Capabilities of the SELECT Statement .................. 62
Execute a Basic SELECT Statement ........................... 64
The Primitive SELECT Statement ..................... 64
Syntax Rules ....................................... 69
Exercise 2-2: Answering Our First
Questions with SQL ............................... 72
SQL Expressions and Operators ........................ 75
The NULL Concept ................................. 87
Exercise 2-3: Experimenting with Expressions
and the DUAL Table .............................. 92
✓ Two-Minute Drill ................................... 96
Q&A Self Test .......................................... 98
Lab Question ...................................... 100
Self Test Answers ................................... 102
Lab Answer ........................................ 104
3 Restricting and Sorting Data ..................... 109
Limit the Rows Retrieved by a Query .......................... 110
The WHERE clause ................................. 110
Comparison Operators ............................... 119
Exercise 3-1: Using the LIKE Operator ............... 131
Boolean Operators .................................. 133
Precedence Rules ................................... 139
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Sort the Rows Retrieved by a Query ........................... 143
The ORDER BY Clause .............................. 143
Exercise 3-2: Sorting Data Using
the ORDER BY Clause ............................ 148
Ampersand Substitution .................................... 150
Substitution Variables ............................... 150
Define and Verify ................................... 157
Exercise 3-3: Using Ampersand Substitution ........... 163
✓ Two-Minute Drill ................................... 166
Q&A Self Test .......................................... 168
Lab Question ...................................... 170
Self Test Answers ................................... 171
Lab Answer ........................................ 173
4 Single-Row Functions ........................... 175
Describe Various Types of Functions
Available in SQL .......................................... 176
Defining a Function ................................. 176
Types of Functions .................................. 180
Use Character, Number, and Date Functions
in SELECT Statements ..................................... 183
Using Character Case Conversion Functions ............. 183
Exercise 4-1: Using the Case Conversion Functions ..... 188
Using Character Manipulation Functions ................ 189
Exercise 4-2: Using the Character
Manipulation Functions ........................... 202
Using Numeric Functions ............................ 203
Working with Dates ................................. 209
Using Date Functions ................................ 213
Exercise 4-3: Using the Date Functions ............... 216
✓ Two-Minute Drill ................................... 225
Q&A Self Test .......................................... 227
Lab Question ...................................... 229
Self Test Answers ................................... 230
Lab Answer ........................................ 231
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5 Using Conversion Functions and
Conditional Expressions ......................... 233
Describe Various Types of Conversion
Functions Available in SQL .................................. 234
Conversion Functions ............................... 234
Use the TO_CHAR, TO_NUMBER,
and TO_DATE Conversion Functions ......................... 237
Using the Conversion Functions ....................... 238
Exercise 5-1: Converting Dates into
Characters Using the TO_CHAR Function ............ 245
Apply Conditional Expressions in a SELECT Statement ........... 250
Nested Functions ................................... 250
General Functions .................................. 252
Exercise 5-2: Using NULLIF and NVL2
for Simple Conditional Logic ....................... 257
Conditional Functions ............................... 260
Exercise 5-3: Using the DECODE Function ........... 266
✓ Two-Minute Drill ................................... 269
Q&A Self Test .......................................... 271
Lab Question ...................................... 273
Self Test Answers ................................... 274
Lab Answer ........................................ 275
6 Reporting Aggregated Data
Using the Group Functions ...................... 277
Describe the Group Functions ................................ 278
Definition of Group Functions ......................... 278
Types and Syntax of Group Functions ................... 280
Identify the Available Group Functions ........................ 283
Using the Group Functions ........................... 283
Exercise 6-1: Using the Group Functions ............. 289
Nested Group Functions ............................. 290
Group Data Using the GROUP BY Clause ...................... 292
Creating Groups of Data ............................. 292
The GROUP BY Clause .............................. 294
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Grouping by Multiple Columns ........................ 296
Exercise 6-2: Grouping Data Based on
Multiple Columns ................................ 297
Include or Exclude Grouped Rows Using the HAVING Clause ...... 299
Restricting Group Results ............................ 300
The HAVING Clause ................................ 301
Exercise 6-3: Using the HAVING Clause ............. 304
✓ Two-Minute Drill ................................... 307
Q&A Self Test .......................................... 309
Lab Question ...................................... 311
Self Test Answers ................................... 312
Lab Answer ........................................ 314
7 Displaying Data from Multiple Tables .............. 315
Write SELECT Statements to Access Data from More
Than One Table Using Equijoins and Nonequijoins .............. 316
Types of Joins ...................................... 317
Joining Tables Using ANSI SQL Syntax ................. 322
Qualifying Ambiguous Column Names .................. 323
The NATURAL JOIN Clause ......................... 325
Exercise 7-1: Using the NATURAL JOIN ............ 327
The JOIN USING Clause ............................ 328
The JOIN ON Clause ................................ 330
Exercise 7-2: Using the NATURAL
JOIN…ON Clause ................................ 332
N-Way Joins and Additional Join Conditions ............. 333
Nonequijoins ...................................... 336
Join a Table to Itself Using a Self-Join .......................... 338
Joining a Table to Itself Using the JOIN…ON Clause ...... 338
Exercise 7-3: Performing a Self-Join .................. 340
View Data That Does Not Meet a Join
Condition by Using Outer Joins .............................. 341
Inner versus Outer Joins .............................. 342
Left Outer Joins .................................... 343
Right Outer Joins ................................... 345
Full Outer Joins .................................... 346
Exercise 7-4: Performing an Outer-Join ............... 347
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Generate a Cartesian Product of Two or More Tables ............. 349
Creating Cartesian Products Using Cross Joins ............ 350
Exercise 7-5: Performing a Cross-Join ................ 351
✓ Two-Minute Drill ................................... 356
Q&A Self Test .......................................... 358
Lab Question ...................................... 361
Self Test Answers ................................... 362
Lab Answer ........................................ 363
8 Using Subqueries to Solve Problems .............. 365
Define Subqueries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366
Exercise 8-1: Types of Subqueries .................... 367
Describe the Types of Problems That the Subqueries Can Solve ..... 369
Use of a Subquery Result Set for Comparison Purposes ..... 369
Star Transformation ................................. 371
Generate a Table from Which to SELECT ............... 372
Generate Values for Projection ........................ 373
Generate Rows to be Passed to a DML Statement ......... 373
Exercise 8-2: More Complex Subqueries .............. 374
List the Types of Subqueries .................................. 376
Single- and Multiple-Row Subqueries ................... 376
Correlated Subqueries ............................... 377
Exercise 8-3: Investigate the Different
Types of Subqueries ............................... 379
Write Single-Row and Multiple-Row Subqueries ................. 382
Exercise 8-4: Write a Query That Is Reliable
and User Friendly ................................. 383
✓ Two-Minute Drill ................................... 387
Q&A Self Test .......................................... 388
Lab Question ...................................... 392
Self Test Answers ................................... 393
Lab Answer ........................................ 394
9 Using the Set Operators ........................ 395
Describe the Set Operators .................................. 396
Sets and Venn Diagrams .............................. 396
Set Operator General Principles ....................... 398
Exercise 9-1: Describe the Set Operators .............. 399
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Use a Set Operator to Combine Multiple
Queries into a Single Query .................................. 400
The UNION ALL Operator ........................... 401
The UNION Operator ............................... 401
The INTERSECT Operator ........................... 403
The MINUS Operator ............................... 403
More Complex Examples ............................. 405
Exercise 9-2: Using the Set Operators ................ 407
Control the Order of Rows Returned .......................... 410
Exercise 9-3: Control the Order
of Rows Returned ................................ 411
✓ Two-Minute Drill ................................... 416
Q&A Self Test .......................................... 417
Lab Question ...................................... 419
Self Test Answers ................................... 420
Lab Answer ........................................ 421
10 Manipulating Data ............................. 423
Describe Each Data Manipulation Language (DML) Statement ..... 424
INSERT .......................................... 425
UPDATE ......................................... 426
DELETE .......................................... 427
MERGE .......................................... 428
TRUNCATE ...................................... 429
DML Statement Failures ............................. 429
Insert Rows into a Table ..................................... 433
Exercise 10-1: Use the INSERT Command ............ 439
Update Rows in a Table ..................................... 441
Exercise 10-2: Use the UPDATE Command ........... 444
Delete Rows from a Table ................................... 446
Removing Rows with DELETE ........................ 446
Exercise 10-3: Use the DELETE Command ........... 447
Removing Rows with TRUNCATE ..................... 449
MERGE .......................................... 450
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Control Transactions ....................................... 451
Database Transactions ............................... 451
The Transaction Control Statements ................... 455
Exercise 10-4: Use the COMMIT
and ROLLBACK Commands ....................... 456
✓ Two-Minute Drill ................................... 463
Q&A Self Test .......................................... 465
Lab Question ...................................... 469
Self Test Answers ................................... 471
Lab Answer ........................................ 473
11 Using DDL Statements to Create
and Manage Tables ............................. 475
Categorize the Main Database Objects ......................... 476
Object Types ....................................... 476
Users and Schemas .................................. 478
Naming Schema Objects ............................. 479
Object Namespaces ................................. 481
Exercise 11-1: Determine What Objects Are
Accessible to Your Session .......................... 482
Review the Table Structure .................................. 482
Exercise 11-2: Investigate Table Structures ............ 483
List the Data Types That Are Available for Columns .............. 484
Exercise 11-3: Investigate the Data Types
in the HR schema ................................ 488
Create a Simple Table ...................................... 488
Creating Tables with Column Specifications ............. 489
Creating Tables from Subqueries ....................... 490
Altering Table Definitions After Creation ............... 492
Dropping and Truncating Tables ....................... 493
Exercise 11-4: Create Tables ....................... 494
Explain How Constraints Are Created
at the Time of Table Creation ................................ 496
The Types of Constraints ............................. 497
Defining Constraints ................................ 500
Exercise 11-5: Work with Constraints ................ 504
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✓ Two-Minute Drill ................................... 506
Q&A Self Test .......................................... 507
Lab Question ...................................... 509
Self Test Answers ................................... 510
Lab Answer ........................................ 511
A About the CD-ROM ............................ 513
System Requirements ....................................... 514
Total Tester Premium Practice Exam Software ................... 514
Installing and Running Total Tester
Premium Practice Exam Software ............................. 514
Electronic Book ........................................... 515
Technical Support ......................................... 515
Total Seminars Technical Support ...................... 515
McGraw-Hill Content Support ........................ 515
Glossary ...................................... 517
Index ........................................ 533
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ACKNOWLEDGMENTS
This book was written amidst emigrating from South Africa to Canada and completed
during one of the most brutal winters in Canadian history. Concluding the task to a
standard I am satisfied with during this tumultuous time would not have been possible
without the love, support, encouragement, and friendship of my wife, Dr. Ameetha Garbharran.
■ I am grateful too for the meticulous attention to detail and wisdom of the
technical editor, my esteemed colleague and friend Cecil Strydom, who kept
me focused on the quality of this endeavor.
■ Finally, thank you to the team at McGraw-Hill Education, specifically Mary
Demery and Stephanie Evans, who kept me on track with my deadlines and
showed empathy and understanding during our intercontinental transition.
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PREFACE
The objective of this study guide is to prepare you for the 1Z0-061 exam by familiarizing
you with the knowledge tested in the exam. Because the primary focus of the book is
to help you pass the test, we don’t always cover every aspect of the related technology.
Some aspects of the technology are only covered to the extent necessary to help you understand
what you need to know to pass the exam, but we hope this book will serve you as a valuable
professional resource after your exam.
In This Book
This book is organized in such a way as to serve as an in-depth review for the
Oracle Database 12c: SQL Fundamentals exam for both novice and experienced
Oracle professionals. Each chapter covers a major aspect of the exam, with an
emphasis on the “why” as well as the “how to” of working with and supporting
Oracle SQL.
On the CD-ROM
For more information on the CD-ROM, please see the “About the CD-ROM”
appendix at the back of the book.
Exam Objective Map
At the end of the Introduction you will find an Exam Objective Map. This table
has been constructed to allow you to cross-reference the official exam objectives
with the objectives as they are presented and covered in this book. The objectives
are listed as presented by the certifying body with a chapter and page reference.
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xxii OCA Oracle Database 12c: SQL Fundamentals I Exam Guide
OracCertPrs5.5 / OCA Oracle Database 12c / Roopesh Ramklass / 182028-0 / Front Matter
In Every Chapter
We’ve created a set of chapter components that call your attention to important
items, reinforce key points, and provide helpful exam-taking hints. Take a look at
what you’ll find in each chapter:
■ Every chapter begins with Certification Objectives—what you need to know
in order to pass the section in the exam dealing with the chapter topic. The
Certification Objective headings identify the objectives within the chapter,
so you’ll always know an objective when you see it!
■ Exam Watch notes call attention to information about, and potential pitfalls
in, the exam. These helpful hints are written by authors who have taken the
exam and received their certification—who better to tell you what to worry
about? They know what you’re about to go through!
■ Exercises are interspersed throughout the chapters. These are designed to give
you a feel for the real-world experience you need in order to pass the exam.
They help you master skills that are likely to be an area of focus in 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 sidebars highlight some of the most common and confusing
problems that students encounter when taking a live exam. Designed to
anticipate what the exam will emphasize, getting inside the exam will help
ensure you know what you need to know to pass. You can get a leg up on
how to respond to those difficult-to-understand questions by focusing extra
attention on these sidebars.
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Preface xxiii
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■ Scenario & Solution sections lay out potential problems and solutions in a
quick-to-read format.
SCENARIO & SOLUTION
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.
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
multi-worded alias will raise an Oracle error. Failure to
double quote a single-word alias will result in the alias
being returned in uppercase.
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.
■ 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. It can be used for last-minute review.
■ The Self Test offers questions similar to those in the 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 Question at the end of the Self Test section offers a unique and
challenging question format that requires you to understand multiple chapter
concepts in order to answer correctly. These questions are more complex
and more comprehensive than the other questions, as they test your ability
to take all the knowledge you’ve gained from reading the chapter and apply
it to complicated, real-world situations. These questions are aimed to be
more difficult than what you will find in the exam. If you can answer these
questions, you have proven that you know the subject!
✓
Q&A
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xxiv OCA Oracle Database 12c: SQL Fundamentals I Exam Guide
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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:
1. 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. You may want to make
flashcards out of 3 × 5 index cards with the Two-Minute Drill material.
2. Reread all the Exam Watch notes and Inside the Exam elements. Remember that
these notes are written by authors who have taken the exam and passed. They
know what you should expect—and what you should be on the lookout for.
3. Review all the Scenario & Solution sections for quick problem solving.
4. Retake the Self Tests. Taking the tests right after you’ve read the chapter is
a good idea, because the questions help reinforce what you’ve just learned.
However, it’s an even better idea to go back later and answer all the questions
in the book in a single sitting. Pretend that you’re taking the live exam.
When you go through the questions the first time, you should mark your
answers on a separate piece of paper. That way, you can run through the
questions as many times as you need to until you feel comfortable with
the material.
5. Complete the exercises. Did you do the exercises when you read through 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 clear on, reread that section in the chapter.
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INTRODUCTION
There 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 now require technical staff to have certifications. 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 12c 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 and may
be taken online. The OCP qualification requires passing a third examination.
These examinations can be taken at many Oracle Testing Centers or Pearson VUE
Authorized Test Centers 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.
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 APEX development
environment.
SQL is an amazing language and the exam and this book are the first steps in
a journey that is likely to last the length of your career. Study the material and
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experiment and as you work through the exercises and sample questions and
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 exciting world of Oracle database administration is about
to begin.
Exam 1Z0-061
Official Objective Ch No. Pg No.
1.0 Retrieving Data Using the SQL SELECT Statement  
1.1 List the capabilities of SQL SELECT statements 2 56
1.2 Execute a basic SELECT statement 2 64
2.0 Restricting and Sorting Data
2.1 Limit the rows that are retrieved by a query 3 110
2.2 Sort the rows that are retrieved by a query 3 143
2.3 Use ampersand substitution to restrict and sort output at runtime 3 150
3.0 Using Single-Row Functions to Customize Output
3.1 Describe various types of functions available in SQL 4 176
3.2 Use character, number, and date functions in SELECT statements 4 183
4.0 Using Conversion Functions and Conditional Expressions
4.1 Describe various types of conversion functions that are available in SQL 5 234
4.2 Use the TO_CHAR, TO_NUMBER, and TO_DATE conversion functions 5 237
4.3 Apply conditional expressions in a SELECT statement 5 250
5.0 Reporting Aggregated Data Using the Group Functions
5.1 Identify the available group functions 6 283
5.2 Describe the use of group functions 6 278
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Official Objective Ch No. Pg No.
5.3 Group data by using the GROUP BY clause 6 292
5.4 Include or exclude grouped rows by using the HAVING clause 6 299
6.0 Displaying Data from Multiple Tables Using Joins
6.1 Write SELECT statements to access data from more than one table using
equijoins and nonequijoins 7 316
6.2 Join a table to itself by using a self-join 7 338
6.3 View data that generally does not meet a join condition by using outer joins 7 341
6.4 Generate a Cartesian product of all rows from two or more tables 7 349
7.0 Using Subqueries to Solve Queries
7.1 Define subqueries 8 366
7.2 Describe the types of problems that the subqueries can solve 8 369
7.3 Describe the types of subqueries 8 376
7.4 Write single-row and multiple-row subqueries 8 382
8.0 Using the SET Operators
8.1 Describe set operators 9 396
8.2 Use a set operator to combine multiple queries into a single query 9 400
8.3 Control the order of rows returned 9 410
9.0 Managing Tables Using DML Statements
9.1 Truncate data 10 429
9.2 Insert rows into a table 10 433
9.3 Update rows in a table 10 441
9.4 Delete rows from a table 10 446
9.5 Control transactions 10 451
10.0 Introduction to Data Definition Language
10.1 Categorize the main database objects 11 476
10.2 Explain the table structure 11 482
10.3 Describe the data types that are available for columns 11 484
10.4 Create a simple table 11 488
10.5 Explain how constraints are created at the time of table creation 11 496
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1
Relational Database Design
Using Oracle
CERTIFICATION OBJECTIVES
1.01 Position the Server Technologies
1.02 Understand Relational Structures
1.03 Summarize the SQL Language
1.04 Use the Client Tools
1.05 Create the Demonstration Schemas
✓ Two-Minute Drill
Q&A Self Test
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The content of this chapter is not directly tested by the Oracle certification examination,
but it is vital in understanding the purpose of SQL. This is considered prerequisite
knowledge that every student should have, beginning with an appreciation of how the
Oracle server technologies fit together and the positioning of each product.
The Oracle server technologies product set is more than a database. There are also
the Oracle WebLogic Server and the Oracle Enterprise Manager. Taken together,
these are the server technologies that make up the Oracle Cloud. Cloud computing
is an emerging environment for managing the complete IT environment and
providing services to users on demand.
Databases fundamentally provide the infrastructure for the organization, storage,
and retrieval of data in an efficient manner. Oracle Database 12c has evolved
from a Relational Database Management System (RDBMS) to an Object RDBMS
supporting the organization of virtually any type of information with no practical
limit on the volume of data that may be stored. The vast data volumes generated
by Amazon.com, the Large Hadron Collider (LHC) at CERN in Europe, and
many financial institutions and governments are organized and managed in Oracle
databases. Although Oracle databases have features addressing demands for scalability,
high availability, and superb performance, this guide will focus on the data
organization problem. Structured Query Language (SQL, pronounced “sequel”) is
an international standard for managing data stored in relational databases. Oracle
Database 12c offers an implementation of SQL that is generally compliant with the
current standard, which is Core SQL:2011. Full details of the compliancy are in
Appendix C 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 SQL language, the client tools, and the demonstration schemas. Several real-
world data organization scenarios are considered while discussing the concepts
behind the relational paradigm and normalizing of data into relational structures.
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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 WebLogic 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
WebLogic Server runs software that generates the web user interfaces that submit
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 cloud. The
cloud 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 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.
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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 startup and shutdown 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. A
widely used alternative is TOAD (the Tool for Application Developers) from Quest
Software, though this is licensed software. End-user applications will need to be
written with something more sophisticated than these tools, something capable of
FIGURE 1-1
The indirect
connection
between a user
and a database
Client-side components Server-side components
Instance
Server process
User process
Session components
User
Database
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managing windows, menus, proper onscreen dialogs, and so on. Such an application
could be written with the Oracle Development Tools, 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 10) 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
client-server 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 for development or training purposes), the
client-server split is still enforced. 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 concurrently. 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 dynamically add further instances running on
supplementary nodes to a database is a major part of the database’s contribution to
the cloud.
The Oracle WebLogic 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
end-user 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.
The Oracle WebLogic 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
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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, among which
Java is predominant. Applications written in Java should conform to the Java EE
(Java Enterprise Edition) standard, which defines how such applications should
be packaged and deployed. JEE and related standards are controlled by Oracle
and accepted by virtually all software developers. Oracle WebLogic Server is a
JEE-compliant application server. Oracle’s implementation of the standards allows
for automatic load balancing and fault tolerance across multiple application servers
on multiple machines through JEE 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 JEE servers from a cluster dynamically is a major part of the Oracle
WebLogic Server’s contribution to the cloud.
It is important to note that Oracle’s commitment to international standards is very
strong. Applications running in the Oracle WebLogic 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 WebLogic Server toolkits can
be deployed to any third-party JEE-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 Firefox or Windows
Internet Explorer) that handles local window management, controls the keyboard,
and tracks mouse movements. The middle tier will be an Oracle WebLogic 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 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.
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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 application 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 administration staff more productive.
FIGURE 1-2
The connection
pooling model
Stateful persistent sessions
over Oracle Net
Application
server
Database
server
Stateless nonpersistent
sessions over HTTP
Browser
Browser
Browser
Browser
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Oracle Enterprise Manager comes in three forms:
■ Database Express
■ Fusion Middleware Control
■ Cloud Control
Oracle Enterprise Manager Database Express 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 Express
from a browser, and Database Express then connects to the database server. Database
Express has facilities for real-time management, performance monitoring, and running
scheduled jobs. Oracle Enterprise Manager Fusion Middleware Control is a graphical
tool for managing a Fusion Middleware deployment. These deployments typically
include Oracle WebLogic Server, an industry-leading application server which
provides the container Java virtual machines (JVMs) that host Oracle or custom
Java applications.
Oracle Enterprise Manager Cloud 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. Cloud 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).
Cloud Control provides 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 effectively manage hundreds of targets.
Cloud Computing
Critical to the concept of cloud computing is service 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 cloud work out
which clustered JEE application server can best provide it. Application servers ask
for a database service and let the cloud work out from which RAC node the data
can best be served. Within the cloud there is a mapping of possible services to
available service providers, and there are algorithms for assigning the workload and
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resources appropriately. 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 cloud is not exclusive to Oracle. At the physical level, some operating system
and hardware vendors are providing cloud-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 under-
performance at others. It should also be possible to design a cloud 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 cloud has been
implemented. The SQL will be invoked from an application server and executed by
an instance against a database: the cloud 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.
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Development Tools and Languages
The Oracle server technologies include various facilities for developing applications,
some existing within the database, others external to it.
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, although Java may also be used
outside the database. 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 such as Oracle
Application Express, JDeveloper, ADF, and many other 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.
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CERTIFICATION OBJECTIVE 1.02
Understand Relational Structures
Several real-world data organization scenarios are introduced to discuss the
relational paradigm and introduce some practical modeling techniques. 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
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.
Real-World Scenarios
This guide uses several hypothetical scenarios, including the two canned scenarios
called HR and OE provided by Oracle and frequently used as the context for exam
questions to illustrate various SQL concepts. The following scenarios evolve further
as new concepts are discussed.
Car Dealership
Sid runs a car dealership and needs a system to keep track of the cars that she buys and
sells. She has noticed business taking a dive and wants to move into the twenty-first
century and create a website that advertises available stock. She needs a system to
keep records of the cars she has bought and sold and the details of these transactions.
Geological Cores
Core samples of the earth have been collected by your local geological survey agency.
To ensure scientific rigor, the developers at GeoCore have determined that the system
must track the exact geographical location, the elemental content of the core samples,
and the dates of collection.
Order Entry
The Order Entry (OE) scenario provided as a sample by Oracle contains information
for a fictitious commercial system that tracks products, customers, and the sales orders
that have been placed.
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Human Resources
The Human Resources (HR) scenario provided as a sample by Oracle records
employees, departments, office locations, and job-related information for a typical
HR department.
Although the hypothetical scenarios described above vary in complexity, they
share several characteristics, including a potential data growth that may eventually
overwhelm a paper-based or spreadsheet-based data organization solution, as well
as a requirement for data to be manipulated (inserted, updated, and deleted) and
retrieved in an efficient manner. The challenge of producing an efficient data
organization design (also known as a data model) may be overcome with both an
understanding of how the data being organized is likely to be utilized and a few
basic data modeling techniques. The goal is to achieve an optimal balance between
data storage and data access which will provide long-term downstream cost-saving
benefits.
Data Modeling
Various formalized data modeling approaches are available, such as the Zachman
framework and the Rational Unified Process, that ultimately seek to provide a
systematic, standards-based approach to representing objects in an enterprise. There
are a multitude of notations available to model entities and their relationships. A
popular notation adopted by Oracle in its CASE tools (Computer Aided Software
Engineering) and more recently in SQL Developer is the crow’s foot notation, which
will be discussed below. Other notations, such as Relational Schema notation and UML
(Universal Markup Language), are also popular, but you must choose a notation that
is comfortable and sensible for you.
Logical modeling is based on conceptualizing objects of interest as entities and
their interactions with each other as relationships. There are many approaches to
entity-relationship diagrams, each with their benefits and limitations. A brief
discussion of entity-relationship diagrams and their notation follows.
Entities and Relations
Many Oracle professionals have adopted a framework that consists of three modeling
stages for relational database modeling. A logical model is conceived when high-level
constructs called entities, comprising various attributes and their relationships, are typically
represented together in a diagram. Entities in logical models are usually depicted as
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rectangles with rounded corners, which comprise attributes or identifiers sometimes
denoted by an “o” symbol. Attributes that uniquely identify an instance of an entity
are designated as primary keys and are sometimes denoted by “#*”. Data typing the
attributes may be done at this stage, but it is generally not reflected in the design.
The logical model is then turned into a relational model by translating the
entities into relations, commonly referred to as tables. The idea here is that sets of
instances of the entities are collectively modeled as a table. The attributes are turned
into table columns. Each instance of an entity is reflected as a tuple or row of data,
each having values for its different attributes or columns. The number of “rows in
the table” is the “cardinality of the tuples.” Usually the attributes that are unique
for each row are called unique keys, and typically a unique key is chosen to be the
primary key (which is discussed later). The relationships between the entities are
often modeled as foreign keys, which will also be explored below.
Relations in relational models are usually depicted as rectangles. At this stage
there is typically more detail in terms of data typing for the attributes, and primary
and foreign key attributes are also reflected with a “P” and an “F,” respectively, in the
relational model. Finally, the relational model is engineered into a physical model by
implementing the design in a relational database.
Crow’s foot notation is often used to depict relationships in logical and relational
data models. The relationships between the entities can be one of the following and
will be explored in the context of the Car Dealership scenario.
■ 1:N One-to-many
■ N:1 Many-to-one
■ 1:1 One-to-one
■ M:N Many-to-many
Consider the scenario of Sid’s car dealership introduced earlier. You could model
the likely data as an entity consisting of the following car-related attributes: Make,
Model, Engine Capacity, and Color. Information regarding the buying and selling
of cars is also required, so you could add Purchase Date, Sold Date, Sellers Name,
Sellers SSN (Social Security number), Sellers Company, the same details for the
buyer, and finally the Purchase Price and Selling Price, as in Figure 1-3.
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Sample transactional data stored in a table based on this entity may look like
Figure 1-4, which shows three rows of data in a table comprising fourteen columns
called CAR_DEALERSHIP. The commands to create tables and populate them with
data will be discussed later in this book. For now, there are several more fundamental
important things to notice. Tables store data in rows, also called records. Each data
element is found at the intersection of a row and a column, also called a cell. It is
fairly intuitive and much like a spreadsheet.
The first two records in the CAR_DEALERSHIP table include the following
information:
■ A silver Mercedes A160 with a 1600cc engine capacity that belonged to
Coda, a private seller with SSN 12345, was bought by Sid with SSN 12346
from Sid’s Cars, for $10,000 on 1 June 2013.
■ A silver Mercedes A160 with a 1600cc engine capacity that belonged to Sid,
with SSN 12346, from Sid’s Cars, was bought by Wags, with SSN 12347,
from Wags Auto, for $12,000 on 1 August 2013.
FIGURE 1-3
A single car
dealership entity
Car Dealership
Make
Model
Engine Capacity
Color
Purchase Date
Sold Date
Sellers Name
Sellers SSN
Sellers Company
Buyers Name
Buyers SSN
Buyers Company
Selling Price
Purchase Price
FIGURE 1-4
Sample data in the
CAR_
DEALERSHIP
table
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Notice the repetition of data. Each record contains duplicate information for
the cars being bought or sold and for the customer doing the buying or selling.
Unnecessary duplication of data usually indicates poor design since it is wasteful and
often requires needless maintenance. If this maintenance is not carefully done, this
design allows errors (sometimes referred to as insert update and deletion anomalies) to
creep in and reduces the overall integrity of the data.
Database normalization refers to modeling data using multiple entities with
relationships between them, which may reduce or entirely eliminate data redundancy.
There are many types of normal forms that have been defined theoretically, but
relational database design primarily focuses on the following three:
■ First normal form (1NF) deals with the issue of eliminating unnecessary
repeating groups of data. An example of a repeating group in Figure 1-4
would be the first four columns on the first two rows where descriptive
information about the car is repeated. You could define a new Cars entity
that uniquely identifies a specific car using the Car ID primary key attribute
as well as the Make, Model, Engine Capacity, and Color attributes. The Car
ID identifier is then used in the related Transactions entity to avoid repeating
groups of data.
■ Second normal form removes attributes from the entity (1NF) that are not
dependent on the primary key. In the proposed Cars entity described above,
the Color attribute is not dependent on a specific car. You could define a
new Colors entity that uniquely identifies a specific color using the Color
ID primary key attribute. The Color ID can then be referenced by the Cars
entity.
■ Third normal form removes all interdependent attributes from a 2NF entity.
The buyers and sellers of cars each have a uniquely identifying Social Security
number (SSN). Their names, however, are interdependent on the SSN
attribute. You could define a new Customers entity that uniquely identified a
customer using the Customer ID primary key attribute, where interdependent
information like the customer’s name and company are stored.
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.
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Note: In the context of performance tuning, it is intentional and acceptable
to duplicate data in entities. When data is normalized across multiple entities
instantiated as multiple tables that must be joined together, the Oracle server
processes need to physically fetch data from multiple tables and join them in
memory buffers to produce the required results set. The extra IO required to
query or manipulate normalized data sometimes justifies denormalizing data
models to reduce disk IO operations and hence increase performance. This is
common in Data Warehouse (DWH) and Decision Support Systems (DSS) but
is an exception rather than the rule in Online Transaction Processing (OLTP)
systems.
Consider the logical data model in Figure 1-5. The car-related data has been
modeled as the Cars entity. The customer’s (buyers and sellers) information is
essentially the same, so they have been modeled as the Customers entity with the
FIGURE 1-5
The car
dealership
entity-relationship
diagram
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Customer Type attribute to differentiate between Purchasers and Sellers. The sales
and purchases are recorded in the Transactions entity, while a lookup entity called
Colors keeps track of different colors.
There are several advantages to conceptualizing this design as four interrelated
entities. Firstly, the data has been normalized and there is no duplication of data.
A practical benefit of multiple entities, each tracking a single construct like Cars,
Customers, Colors, and even Transactions, is the ease of data maintenance. New
colors can be added, each with a unique code, and as new cars are purchased, these
colors, defined and maintained in one place, can be used to describe multiple cars
with the same color. You could improve the sophistication of this model by defining
entities for Tires, Security Systems, Tracking Devices, or Audio Visual add-ons. You
could equally enhance the details collected for each car, such as VIN and engine
numbers; or for each customer, such as address and bank details; but this hypothetical
scenario serves to illustrate several concepts and obviously cannot be used in a
production application scenario without further enhancements.
Primary Keys
Each entity in Figure 1-5 has a primary key attribute that uniquely identifies a tuple
or row of data denoted by a #* adjacent to the attribute name. Each value of the Car
ID primary key is unique in the entity. Multiple rows cannot share the same primary
key value. Similarly, Color ID uniquely identifies each row in the Colors entity, as
do Customer ID and Transaction ID in the Customers and Transactions entities,
respectively.
Relationships
The lines in Figure 1-5 linking the various entities are known as relations. The
crow’s foot notation expresses the cardinality of the relationships between the
entities—one-to-one, one-to-many, many-to-one, and many-to-many. The crow’s
foot notation explicitly illustrates the entity with the many side of the relationship
with multiple “feet,” while the entity on the one side has one foot. Attributes in a
one-to-one relationship are identical, while many-to-many relationships indicate
that multiple tuples in entity A have the same attribute values as many tuples in
entity B. Both one-to-one and many-to-many relationships are not very common
and sometimes point to flaws in the relational model. One-to-many and many-to-
one relationships occur frequently when modeling relational entities. They relate
attributes in two entities in a master-detail relationship. From the point of view of
the relationship between the Cars and Colors entities (the order is significant), for
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example, many records in the Cars entity will be one Color. Many cars could have
the same single Color ID attribute indicating they are the same color. The Colors
entity is the master or lookup entity, while the Cars entity is the detail entity in this
relationship. From the point of view of the relationship between Colors and Cars,
one Color can be associated with many Cars. So it is just a matter of perspective
whether a relationship is one-to-many or many-to-one; it all depends on which
direction of the relationship you consider. The other relationships indicated by the
crow’s feet show that a single car can be bought and sold multiple times, hence the
one-to-many relationship between the Cars and Transactions entities and lastly that
one customer can perform many transactions (like buying and selling many cars).
Referential Integrity and Foreign Keys
These relationships introduce the concept of referential integrity, which ensures
data consistency and integrity by guaranteeing that an attribute (say attribute A)
belonging to the entity on the one side of the relationship must be unique, while
the attribute (say attribute B) on the entity on the many side must have a value
that is in the set of unique values described by attribute A. Attribute B is called
a foreign key since it has a referential dependency on attribute A. Consider the
Colors-Cars relationship based on the Color ID attribute. Referential integrity
ensures that the Color ID attribute in each tuple in the Cars entity must have a
value that is identical to exactly one instance of the Color ID attribute in the Colors
entity. This guarantee is central to relational modeling since the joining of the Cars
and Colors entities on the Color ID attribute allows the Colors.Color (this is dot
notation) attribute to be matched with a related tuple in the Cars entity. The Color
ID attribute in the Cars entity is the foreign key that is related to the unique key,
which is the Color ID attribute in the Colors entity, where it also happens to be the
primary key. It is very common that foreign keys in an entity are based on primary
keys in a related entity, but this is not the rule.
Note: Foreign keys in an entity are based on unique keys in a related entity,
but those unique keys do not have to be the primary key; they just have to
be unique.
The logical model in Figure 1-5 would typically evolve into a relational model
typically with more data-typing details and clearer primary and foreign keys, as in
Figure 1-6.
The relational model could be engineered into a physical model where actual
tables and other database constructs (discussed later in this chapter) are created.
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FIGURE 1-6
Relational
model of the car
dealership
The sample data in Figure 1-4 transferred into the physical model built from the
relational model above would produce four datasets, as in Figure 1-7.
The first two rows of data in the Transactions dataset can be interpreted as
follows:
■ A transaction with TX ID 100 describes the purchase of a car with Car ID 1
by Sid’s dealership from a customer with Customer ID 2 for $10,000 on 1 June
2013. You lookup Customer ID 2 and see it was a sale from Coda, a private
seller with SSN 12345. You lookup Car ID 1 and determine that it was a 2001-
A160 Mercedes with Color ID 1, which you further resolve to be Silver.
■ A transaction with TX ID 101 describes the sale of Car ID 1 to Customer ID
4, which you resolve to be a dealer called Wags from Wags Auto with SSN
12347, for $12,000 on 1 August 2013.
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Based on the earlier descriptions offered in a single-entity design, nothing has
been lost by organizing the sample data across the four-entity design. However, much
has been gained. There is no duplication of data. There is a clarity and elegance that
will facilitate ease of maintenance of this data as new cars are bought and sold and as
new customers transact with Sid’s dealership.
EXERCISE 1-2
Design an Entity-Relationship Diagram for the Geological Cores Scenario
In this exercise, you will gain familiarity with basic relational modeling by
completing an entity-relationship diagram for the Geological Core scenario
introduced earlier. To recap: Core samples of the earth have been collected by
FIGURE 1-7
Sample data using
the car dealership
relational model
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your local geological agency. To ensure scientific rigor, the developers at GeoCore
have determined that the system must track the exact geographical location, the
elemental contents of the core samples, and the date of collection.
1. The GeoCore developers have proposed four entities with their respective
attributes as per the following table. Draw a diagram with these entities
reflecting the primary keys.
Entity Attribute
Elements Atomic Number (Primary Key)
Element Description
Symbol
Atomic Mass
Contents Content ID (Primary Key)
Atomic Number
Quantity
Depth Depth ID (Primary Key)
Start Depth
End Depth
Cores Core ID (Primary Key)
Depth ID
Collection Date
Longitude
Latitude
Content ID
2. The Elements and Contents entities share a many-to-one relationship
through the Atomic Number attribute. The Cores entity shares a many-to-one
relationship with the Depth entity through the Depth ID attribute and a
many-to-one relationship with the Contents entity through the Content
ID attribute. Your task is to update the entity diagram to reflect these
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relationships. Your completed Entity-Relationship diagram should closely
resemble this illustration:
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.
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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. Tables 1-1 through 1-4 are
examples of the tabulated data definitions (a subset of data and structures from the
sample schema known as SCOTT provided by Oracle):
TABLE 1-1
The DEPT Table
Column Name Description Data Type Length
DEPTNO Department number Numeric 2
DNAME Department name Character 14
TABLE 1-2
The EMP Table
Column Name Description Data Type Length
EMPNO Employee number Numeric 4
ENAME Employee name Character 10
DEPTNO Department number Numeric 2
TABLE 1-3
Row Data from
the DEPT Table
DEPTNO DNAME
10 ACCOUNTING
20 RESEARCH
30 SALES
40 OPERATIONS
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Looking at the layout of the DEPT and EMP Tables 1-1 and 1-2, 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. Table 1-3 shows the rows in the DEPT table stored in DEPTNO
order, but this is a matter of chance, not design: relational tables do not impose any
particular ordering on their rows. Table 1-4 shows that department number 10 has
one employee, and department 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
TABLE 1-4
Row Data from
the EMP Table
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
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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.
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 OLTP and DSS
systems. 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,
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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.
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 of 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.
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,
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 non-tabular data.
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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 7), 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 primary SQL commands, separated into commonly used groups:
The Data Manipulation Language (DML) commands:
■ SELECT
■ INSERT
■ UPDATE
■ DELETE
■ MERGE
The Data Definition Language (DDL) commands:
■ CREATE
■ ALTER
■ DROP
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■ 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 Chapter 11. DCL, which has to do with security, is
only briefly mentioned: it falls more into the domain of the database administrator
than the developer.
According to all the documentation, 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.
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 set-
oriented 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 all the rows 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. The
result: 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
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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 non-unique, 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.
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.
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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 author 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
Unix server (or the other way round) provided that Oracle Net has been configured
to make the connection.
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/12.1.0/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
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executable code and some of the configuration files. The PATH must include the
bin directory contained within the Oracle Home. The LD_LIBRARY_PATH
should include the lib directory also contained within the Oracle Home, but in
practice you may get away without setting this. Figure 1-8 shows a Linux terminal
window and some tests to see if the environment is correct.
In Figure 1-8, 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 last element of the PATH and the first element
of the LD_LIBRARY_PATH variables, respectively. Then which confirms that
the SQL*Plus executable file really is available, 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 logon 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 admin123, and the database is identified
by coda.
FIGURE 1-8
Checking the
Linux session
setup
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Following the logon, the next lines of text display the version of SQL*Plus being
used, which is 12.1.0.0.2, 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 logon 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 and 11g, 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 a 12c database. There are no problems with mixing versions:
a 12c SQL*Plus client can connect to a 10g database, and a 10g SQL*Plus client
can connect to a 12c 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 location of the executable file launched
by the shortcut will, typically, be something like the following:
D:\app\oracle\product\12.1.0\dbhome_1\BIN\sqlplus.exe
However, the exact path will be installation specific. Figure 1-9 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 12.1.0.0.2 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/admin123@coda
A change some people like to make to the shortcut that launches SQL*Plus is to
prevent it from immediately presenting a logon prompt. To do this, add the NOLOG
switch to the end of the command:
sqlplus /nolog
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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.
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 1-10 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_OraDb12c_home1
The final element of this navigation path will have a different name if there have
been several 12c installations on the machine.
FIGURE 1-9
A database logon
with SQL*Plus for
Windows
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Note the values of the Registry variables ORACLE_HOME and ORACLE_
HOME_NAME. These will relate to the 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
facilities, but the most common means of identifying oneself to the database is by
presenting a username and a case-sensitive password. There are two typically used
forms of connect identifier for identifying the database: either by providing 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@ocp12c.oracle.com:1521/orcl.oracle.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
FIGURE 1-10
The Oracle
Registry variables
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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 (typically
contained within the network/admin subdirectory of the ORACLE_HOME) or to
contact an LDAP directory such as Microsoft’s Active Directory or Oracle’s Oracle
Internet Directory (OID).
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 1-11 illustrates some of the more common problems.
First, the user launches SQL*Plus from a Windows operating system prompt, using
the NOLOG switch to prevent the immediate logon prompt. No problem so far.
FIGURE 1-11
Some common
logon problems
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Second, from the SQL> prompt, the user issues a connection request, which fails
with a well-known error:
ORA-12154: 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.
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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 12c 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 SQL
Developer as well.
To install SQL Developer, unzip the ZIP file. That’s all. It does require a JDK,
the Java Runtime Environment, to be available; this comes from Oracle. But if
an appropriate JDK 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 JDK must be preinstalled. Download it from Oracle’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 "1.7.0_21"
Java(TM) SE Runtime Environment (build 1.7.0_21-b11)
Java HotSpot(TM) 64-Bit Server VM (build 23.21-b01, mixed mode)
If it does not, using which java may help identify the problem: the search path
could be locating an incorrect version.
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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
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 1-12 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 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
FIGURE 1-12
The SQL
Developer User
Interface
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branches beneath list all the possible object types that can be managed. Expanding
the branches would list the objects themselves. The right pane has an upper part
prompting the user to enter a SQL statement and a lower part that will display the
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:
■ File A normal Windows-like file menu, from which one can save work and
exit from the tool.
■ Edit A normal Windows-like edit menu, from which one can undo, redo,
copy, paste, find, and so on.
■ View The options for customizing the SQL Developer user interface.
■ Navigate Facilities for moving between panes and for moving around code
that is being edited.
■ Run Enables execution and debugging of the SQL statements, SQL script,
or PL/SQL block that is being worked on. Debugging facilitates stepping
through code line by line rather than running the entire block of code.
■ Versioning Supports the creation of a code versioning repository to track
different versions of your SQL programs.
■ Tools Links to external programs, including Data Modeler and SQL
Worksheet
■ Help It’s pretty good.
SQL Developer can be a very useful tool, as it is highly 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 1-13 shows the
window where connections can be defined. To reach this window, click the “+”
symbol visible on the Connections tab shown previously in Figure 1-12.
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
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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.
Assuming that you are using SQL Developer to connect to an Oracle database
rather than to a 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 among five options:
■ Basic 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.
■ TNS If a name resolution method has been configured, then an alias for
the database can be selected (from the local tnsnames.ora file) or entered,
rather than the full details needed by the Basic option.
■ LDAP Database service definitions and their connection details that are
stored in a LDAP directory service may be queried by specifying the LDAP
Server details.
FIGURE 1-13
How to define
a new database
connection
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■ Advanced 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.
■ Local/Bequeath If the database is running on the same machine as your
SQL Developer client, this option allows you to connect directly to a server
process bypassing the network listener.
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.
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.
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 My Oracle Support. 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 My Oracle Support 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.
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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 (DBCA). 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.
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.
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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. Unless synonyms are available,
only a user connected as HR could get to the table by referring to EMPLOYEES
without a schema name qualifier. A synonym is a construct that makes an object
accessible to other users without requiring its schema name as a prefix.
The HR and OE Schemas
The HR demonstration schema consists of seven tables, linked by primary key to
foreign key relationships. Figure 1-14 illustrates the relationships between the tables,
as an entity relationship diagram.
FIGURE 1-14
The HR entity
relationship
diagram
REGIONS
COUNTRIES
LOCATIONS
DEPARTMENTS
EMPLOYEESJOB_HISTORY
JOBS
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Two of the relationships shown in Figure 1-14 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 1-5 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 (*).
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.
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TABLE 1-5
The Tables and
Columns on the
HR Schema
Table Columns
REGIONS #* region_id
o region_name
COUNTRIES #* country_id
o country_name
\o region_id
LOCATIONS #* location_id
o street_address
o postal_code
* city
o state_province
\o country_id
DEPARTMENTS #* department_id
* department_name
\o manager_id
\o location_id
EMPLOYEES #* 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
JOBS #* job_id
* job_title
o min_salary
o max_salary
JOB_HISTORY #* employee_id
#* start_date
* end_date
\* job_id
\o department_id
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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.
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 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
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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 execution. Typical values for the default tablespace and temporary
tablespace are USERS and TEMP, but these must already exist. After completion,
you will be connected to the database as the new HR user. To verify this, run
these statements.
SQL> show user;
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.
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 or example 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.
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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 Cloud computing.
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.
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Two-Minute Drill 49
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TWO-MINUTE DRILL
Position the Server Technologies
❑The Oracle database stores and manages access to user data.
❑The Oracle WebLogic 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.
✓
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SELF TEST
Position the Server Technologies
1. What components of the IT environment can Oracle Enterprise Manager Cloud 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
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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 WHILE..
B. Iteration, based on FOR..DO
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
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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.
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SELF TEST ANSWERS
Position the Server Technologies
1. þ E. Cloud Control can manage the complete environment (according to Oracle Corporation).
ý A, B, C, and D are incorrect. All of these can be managed by Cloud Control.
2. þ A, C, and D. SQL, PL/SQL, and Java can all run in the database.
ý B and E are incorrect. 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
3. þ B. Third normal form is the usual form aimed for by systems analysts when they normalize
data into relational structures.
ý A, C, and D are incorrect. A is incorrect 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.
4. þ A. The relational model uses two-dimensional tables.
ý B, C, and D are incorrect. B is incorrect because two dimensions is the limit for relational
structures. C and D are incorrect because they refer to nonrelational structures (though there
are facilities within the Oracle database for simulating them).
Summarize the SQL Language
5. þ D. In a set-oriented language, one command can affect many rows (a set), whereas a
procedural language processes rows one by one.
ý A, B, and C are incorrect. A is incorrect because while rows should have a unique
identifier in a well designed application, this is not actually a requirement. B is incorrect
because users cannot be grouped in the Oracle environment. C is incorrect because (even
though the statement is correct) it is not relevant to the question.
6. þ A, B, and C. These are all procedural constructions, which are not part of a set-oriented
language. They are all used in PL/SQL.
ý D and E are incorrect. These are SQL’s transaction control statements.
Use the Client Tools
7. þ B and C. B is correct because SQL Developer can be installed in its own directory. C is
correct because passwords can be saved as part of a connection definition (though this may not
be a good idea).
ý A and D are incorrect. A is incorrect because the Oracle Net protocol lets SQL Developer
connect to a number of versions of the database. D is incorrect because LDAP is only one of
several techniques for name resolution.
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8. þ A and E. A is correct because SQL Developer is written in Java and therefore requires a
Java Runtime Environment. E is correct because SQL Developer needs a graphics terminal to
display windows.
ý B, C, and D are incorrect. B is incorrect because SQL Developer uses JDBC to connect
to databases, not OCI. C is incorrect because, while SQL Developer can use LDAP or a
TNSNAMES.ORA file, it can also use and store the basic connection details. D is incorrect
because SQL Developer is a completely independent product.
Create the Demonstration Schemas
9. þ C. The demonstration schemas can be created in any database, either at database creation
time or by running scripts later.
ý A, B, and D are incorrect. A and B are incorrect because, while they may be good practice,
they are not a technical requirement. D is incorrect because it fails to understand that a schema
can only be (and always is) created with a user.
10. þ B. A schema and a user are inseparable.
ý A, C, and D are incorrect. A is incorrect because there is no such command. C and D
are incorrect because they assume the impossible: that you can separate a user from his or her
schema.
LAB ANSWER
Figure 1-15 shows a solution.
FIGURE 1-15
An entity-
relationship
diagram
describing the
OE schema
CUSTOMERS ORDERS ORDER_ITEMS
WAREHOUSES INVENTORIES PRODUCT_
INFORMATION
PRODUCT_
DESCRIPTIONS
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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
✓ Two-Minute Drill
Q&A Self Test
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This 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.
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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.
SQL*Plus has an extensive runtime command environment which can be
explored using the online documentation or the HELP INDEX command. This
lists available SQL*Plus commands such as the SHOW command, which shows
the value of a SQL*Plus variable. For example, SHOW USER shows the name
of the currently connected user.
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 such as 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 intuitive:
DESC[RIBE] <SCHEMA>.tablename
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This command will 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 are connected, the “<SCHEMA>.”
portion of the command may be omitted. Figure 2-1 illustrates the use of the SHOW
user command to verify that the current connected user is HR. While connected
to the database as the HR user, the EMPLOYEES table is described from SQL*Plus
with the DESCRIBE employees command and the DEPARTMENTS table
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
FIGURE 2-1
Describing the
EMPLOYEES,
DEPARTMENTS,
and DUAL tables
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data types are discussed in detail in Chapter 11. For the current discussion, the
following explanation of the different column data types is sufficient.
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 may be to the
right of the decimal point and up to 6 may be to the left. If more than two digits are
to the right of the decimal point, the number will be rounded to 2 decimal places as
long as there are at most 8 digits. 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,
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 that is restricted by the NOT
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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
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, click the Connect button.
2. Navigate to the SQL Worksheet, which is the tab titled Worksheet.
3. Type in the command DESCRIBE jobs. Terminating this command with
a semicolon is optional.
4. Execute the DESCRIBE command, either by pressing the 5 key or by
clicking the solid green triangular arrow icon located on the toolbar above
the SQL Editor.
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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.
SQL Developer provides an alternative to the DESCRIBE command when it
comes to obtaining the structural information of tables.
7. Navigate to the LOCATIONS table using the Tree navigator located on the
left frame underneath the connection name.
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8. 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
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
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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 Chapter 1, 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 Chapter 7.
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.
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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:
■ The primitive SELECT statement
■ Syntax rules
■ SQL expressions and operators
■ The NULL concept
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;
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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 as a shorthand, time-saving symbol instead of typing in SELECT column1,
column2,…,columnX, to select all the columns. The FROM clause specifies which
table to query to fetch (project) 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 that belong 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
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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 statement shown in Figure 2-3, executed in SQL*Plus.
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
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.
FIGURE 2-3
Projecting specific
columns from the
JOB_HISTORY
table
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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,
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 been
eliminated. This is, of course, the row that contains a JOB_ID value of ST_CLERK
and a DEPARTMENT_ID value of 50.
FIGURE 2-4
Selecting unique
JOB_IDs from the
JOB_HISTORY
table
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FIGURE 2-5
Selecting unique
DEPARTMENT_
IDs from the
JOB_HISTORY
table
FIGURE 2-6
Unique JOB_ID,
DEPARTMENT_
ID combinations
from JOB_
HISTORY
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The ability to project specific columns from a table is very useful. Coupled
with the ability to remove duplicate values or combinations of values this
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.
Syntax Rules
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 mixed case,
uppercase for reserved words and lowercase for the remainder of the statement for
illustrative purposes. Many developers, including the author of this book, prefer to
write their SQL statements in lowercase. There is also a common misconception
that SQL reserved words need 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 of ST_CLERK in the JOB_ID column.
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Metadata about different database objects is stored by default in uppercase in
the data dictionary. If you query a database dictionary table to return a list of tables
owned by the HR schema, it is likely that the table names returned are stored in
uppercase. This does not mean that a table cannot be created with a lowercase
name; it can be. It is just more common and is the default behavior of the Oracle
server to create and store tables, columns, and 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 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, region_id FROM countries;
SELECT city, location_id,
state_province, country_id
FROM locations
/
SQL statements may be
submitted to the database in lowercase,
uppercase, or mixed case. You must pay
careful attention to case when interacting
with character literal data and aliases.
Asking for a column called JOB_ID or
job_id returns the same column, but
asking for rows where the JOB_ID value
is PRESIDENT is different from asking for
rows where the JOB_ID value is President.
Character literal data should always be
treated in a case-sensitive manner.
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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
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
the development process 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.
A common technique
employed by some exam question designers
measures 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.
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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, APEX, and JDeveloper) and other
third-party 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.
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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:
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;
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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.
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
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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.
Operators with the same level of precedence are evaluated from left to right.
Round brackets may therefore be used to enforce non-default 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
expose 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.
Seven terms 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 and hours that an employee filled a particular position.
Consider employee number 176 on the ninth row of output. This employee started as
a Sales Manager on January 1, 2007, and ended employment on December 31, 2007.
Therefore, this employee worked for exactly one year, which, in 2007, consisted of
365 days or 2,920 hours.
Precedence Level Operator Symbol Operation
Highest ( ) Brackets or parentheses
Medium / Division
Medium * Multiplication
Lowest −Subtraction
Lowest + Addition
TABLE 2-1
Precedence
of Arithmetic
Operators
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The number of days for which an employee was employed can be calculated by
using the fifth term in the SELECT clause, which is an expression. This expression
demonstrates that arithmetic performed on columns containing date information
returns numeric values that represent a certain number of days. The sixth term
closely resembles the fifth but calculates the number of hours worked by additionally
multiplying by 8 (assuming an 8-hour workday).
To enforce operator precedence of the subtraction and addition operations in the
sixth term, the subexpression end_date-start_date+1 is enclosed in round brackets and
then multiplied by 8 to obtain the correct number of hours worked. The seventh
term is evaluated by first multiplying 1 by 8, which is added to end_date-start_date,
which returns the incorrect results.
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
FIGURE 2-7
Arithmetic
expression to
calculate number
of hours worked
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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):
3*100/5 + 20/5
FIGURE 2-8
Use of the
concatenation
and arithmetic
operators
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Step Expression 1 Expression 2
4. The next subexpression to be evaluated
is: 300/5:
60 + 20/10 - 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
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
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 generally 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 non-performant query.
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 Worked. 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
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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 Worked” 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 keyword not found where expected” error
is returned when a multi-worded alias is not double quoted.
FIGURE 2-9
Use of column
and expression
aliases
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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, column 43 at the point
of the problem: the word Worked. 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 Worked 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 Worked” 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 Days_
Worked, with only the title letters of the alias being capitalized, the expression
heading was returned as DAYS_WORKED: 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 formal 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 Worked” 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 author believes it is a good
SQL coding habit to form.
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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.
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
FIGURE 2-10
Use of the AS
keyword to
specify column
aliases
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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.
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;
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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' FROM dual;
What about character literals that contain single quotation marks? For example,
plural possessives pose a particular problem for character literal processing (the
possessive case of most English plural nouns are formed by adding an apostrophe to
the end if it ends with an “s”; an apostrophe and an “s” are added to singular nouns).
Consider the following statement:
SELECT 'Plural's have one quote too many' FROM dual;
FIGURE 2-11
Using the DUAL
table
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As the example in Figure 2-12 shows, executing this statement causes an ORA-
01756 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
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
or multibyte character or the four brackets: (round brackets), {curly braces}, [square
FIGURE 2-12
Error while
dealing with
literals with
implicit quotes
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brackets], or <angle brackets>. 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'delimitercharacter 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.
FIGURE 2-13
Use of two
single quotes
with literals with
implicit quotes
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FIGURE 2-14
The alternate
quote (q)
operator
The NULL Concept
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.
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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 non-mandatory column until
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.
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.
INSIDE THE EXAM
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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 contain
null values. One of the nullable columns is the COMMISSION_PCT column.
Figure 2-15 shows the first five rows of data from the EMPLOYEES table. This
is sufficient to illustrate that all 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 displayed in the Query Result tab when a null
value is encountered, as with the COMMISSION_PCT column. SQL Developer
supports customizing this default description of null column data.
FIGURE 2-15
Null values in the
COMMISSION_
PCT column
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FIGURE 2-16
Null arithmetic
always returns a
null value.
The column aliased as “Null Arithmetic” is an expression made up of
COMMISSION_PCT + SALARY + 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.
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, while the arithmetic operations
involving null values always result in null.
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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 data 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 multi-worded 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 by making use of outer joins.
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 in which
staff worked in a job could be calculated. For how many years did staff work 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 Worked. 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 Worked from
the JOB_HISTORY table.
3. The expression can be calculated by dividing 1 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 Worked"
FROM job_history;
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4. Executing the preceding SELECT statement yields the results displayed in
the following illustration:
Question 2: Query the JOBS table and return a single expression of the form:
The Job Id for the <job_title’s> job is: <job_id>.
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.
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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;
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
6,000 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
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Certication Summary 95
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 113,142,857
square units.
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, SQL*Plus, or any number of Oracle or other
third-party 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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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 the whole number portion can have at
most (x-y) digits.
❑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.
✓
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❑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, uppercase, or
mixed case. Be careful when interacting with character literals since these are
case-sensitive.
❑Arithmetic operators and the string concatenation operator acting on
column 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 OCA 12c 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.
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.
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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;
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;
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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
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?
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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_HISTORY
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 salesperson.
Extract product information in the format <PRODUCT_NAME> with code: <PRODUCT_
ID> has status of: <PRODUCT_STATUS>. 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
1. þ B. A projection is an intentional restriction of the columns returned from a table.
ý A, C, and D are incorrect. 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.
2. þ A. Columns with NUMBER(8,2) data type can store, at most, eight digits, of which, at
most, six digits are to the left of the decimal point. Although A is the correct answer, 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.
ý B, C, D, and E are incorrect, as they can legitimately be stored in this data type. C is
allowed since the fractional portion is rounded to two decimal places. 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.
3. þ B and E. The result of arithmetic between two date values represents a certain number of
days.
ý 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.
4. þ A and D. The scale of the VARCHAR2 data type, specified in brackets, determines its
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.
ý B and C are incorrect. 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
5. þ D. Unique JOB_ID values are projected from the EMPLOYEES table by applying the
DISTINCT keyword to just the JOB_ID column.
ý A, B, and C are incorrect, 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
distinct combination of JOB_ID and EMPLOYEE_ID values. This has the effect of returning
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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.
6. þ B and D. B and D represent the two illegal statements that will return syntax errors if they
are 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.
ý A and C are incorrect, as they are the legal statements. 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.
7. þ B and D. B and D do not return null values since character expressions are not affected
in the same 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.
ý A and C are incorrect. They return null values because any arithmetic expression that
involves a null will return a null.
8. þ D. An asterisk is the SQL operator that implies that all columns must be selected from a
table.
ý 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.
9. þ B. The key to identifying the correct result lies in understanding the role of the single
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.
ý 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.
10. þ D. The literal expression '6 * 6' is selected once for each row of data in the REGIONS
table.
ý 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:
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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 at least 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_HISTORY 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.
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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:'||product_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 %"
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These three expressions, along with the two regular columns, form the SELECT clause executed
against the PRODUCT_INFORMATION table as shown next:
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 197,016,573 square miles.
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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
3.03 Ampersand Substitution
✓ Two-Minute Drill
Q&A Self Test
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Limiting 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 with the ORDER BY clause, which provides data sorting capabilities.
Ampersand substitution provides a way to reuse the same statement to execute different queries
by substituting query elements at runtime. This chapter concludes with an exploration of this
technique of runtime binding in SQL statements.
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
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the SELECT and FROM clauses results in every row of data stored in the table
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 that 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 that 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 that 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
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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
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, department_id
FROM employees
WHERE salary = department_id;
FIGURE 3-1
Two ways to
select numeric
values in a
WHERE clause
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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 24000, and the range of DEPARTMENT_ID values is 10 to 270.
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.
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.
FIGURE 3-2
Using the
WHERE clause
with numeric
expressions
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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';
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 that have JOB_ID column
values that are either Sa_Rep or sa_rep, as shown in Figure 3-3.
FIGURE 3-3
Using the WHERE
clause with
character data
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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, last_name, first_name
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
single quote characters are necessary to delimit them. This condition stipulates that
only rows that 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
FIGURE 3-4
Character
column-based
WHERE clause
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values with one or more character columns. The following four clauses demonstrate
some of the options for character-based conditions:
Clause 1: WHERE 'A '||last_name||first_name = 'A King'
Clause 2: WHERE first_name||' '||last_name = last_name||' '||first_name
Clause 3: WHERE 'SA_REP'||'King' = job_id||last_name
Clause 4: WHERE job_id||last_name ='SA_REP'||'King'
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.
FIGURE 3-5
Equivalence
of conditional
expressions
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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
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-MON-
RR. 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.
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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.
The century component for the literal '01-JAN-99' becomes the previous
(twentieth) century, 19, and 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 + 1 returns a DATE value that is
1 day later than START_DATE. So the following expression is legitimate, as shown
in Figure 3-6:
SELECT start_date, employee_id
FROM job_history
WHERE start_date + 1 = '25-MAR-06';
FIGURE 3-6
Using the
WHERE clause
with numeric
expressions
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This query returns rows from the JOB_HISTORY table containing a START_
DATE value equal to 1 day before 25-MAR-2006. Therefore, only rows with a value
of 24-MAR-2006 in the START_DATE column will be retrieved.
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
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.
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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';
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,
FIGURE 3-7
Conditions based
on the equality
operator
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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. For
example, the following query may be issued to obtain a list of LAST_NAME and
SALARY values for employees who earn more than $5,000:
SELECT last_name, salary
FROM employees
WHERE salary > 5000;
Similarly, to obtain a list of employees who earn less than $3,000, 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 1: WHERE salary <= 3000;
Clause 2: WHERE salary >= 5000;
Clause 3: WHERE salary <> department_id;
Clause 4: WHERE salary != 4000+department_id;
Clause 1 returns those rows that 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, while Clauses 3 and 4 demonstrate the two forms of the
Operator Description
< Less than
> Greater than
<= Less than or equal to
>= Greater than or equal to
<> Not equal to
!= Not equal to
TABLE 3-1
Inequality
Operators
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“not equal to” operators. Clause 3 returns the rows that 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 that
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 compared
in lexicographical order. Based on the database character set and NLS (National
Language Support) settings, each character string is either evaluated using a binary
or linguistic comparison method. In English type character sets, the comparison
semantics operate as discussed in this chapter. With the default binary comparison
method, characters are assigned a numeric value with blanks having a lesser value
than other characters. 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 its ordinal character values: K (75), i (105), n (110), and
g (103). For each row in the EMPLOYEES table, the LAST_NAME column data is
similarly converted into numeric values for each character which are then compared
in turn with the numeric values of the characters in the literal 'King'. For example,
the row with LAST_NAME='Kaufling' is compared as follows: The first character
in both strings is 'K' with an equal ordinal value of 75. So the second character
(i=105) is compared with (a=97). Since (97 < 105) or (a < i), 'Kaufling' <
'King' and 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 always occurs
earlier than the third of June of the same year. Therefore, the numeric value of the
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date 02-JUN-2008 is less than the numeric value of the date 03-JUN-2008. Consider
the following query:
SELECT last_name, hire_date
FROM employees
WHERE hire_date < '01-JAN-2003';
This query retrieves the last name and hire date of each employee record
containing a HIRE_DATE value that is earlier than '01-JAN-2003'. Rows with
employee HIRE_DATE on or before 31-DEC-2002 will be returned, while rows
with employee HIRE_DATE values later than the first of January 2003 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.
FIGURE 3-8
Conditions based
on the inequality
operators
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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.
Suppose you want the last names and salaries of employees who earn a salary in
the range of $3,400 and $4,000. A possible solution using the BETWEEN operator is
as follows:
SELECT last_name, salary
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 $3,400 to $4,000 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
“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
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 on or after 24-JUL-
1994 but on or before 07-JUN-1996:
SELECT first_name, hire_date
FROM employees
WHERE hire_date BETWEEN '24-JUL-1994' AND '07-JUN-1996';
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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
FIGURE 3-9
The BETWEEN
operator
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single quotes. You may include as many literals in the set as you wish. Consider the
following example:
SELECT last_name, salary
FROM employees
WHERE salary IN (3000,4000,6000);
The SALARY value in each row is compared for equality to the literals specified
in the set. If the SALARY value equals 3000, 4000, or 6000, the LAST_NAME and
SALARY values for that row are 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, salary
FROM employees
WHERE salary = 3000
OR salary = 4000
OR salary = 6000;
This statement will return an employee’s LAST_NAME and SALARY 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.
FIGURE 3-10
The IN operator
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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. Numeric values
compared using the LIKE operator are automatically typecast as character data.
LIKE is accompanied by two wildcard characters: the percentage symbol (%)
and the underscore character (_). The percentage symbol is used to specify zero or
more characters, while the underscore character specifies one wildcard character.
A wildcard may represent any character.
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 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.
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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 “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
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, followed by a lowercase “a” which in turn is
followed by zero or more characters.
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
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could be any characters). The fifth letter must be an “i”, and the percentage symbol
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_%';
FIGURE 3-11
The wildcard
symbols of the
LIKE operator
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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 *
FROM jobs
WHERE job_id LIKE 'SA\_%' ESCAPE '\';
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 '$';
FIGURE 3-12
The ESCAPE
identifier and the
LIKE operator
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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
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”.
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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:
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.
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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 and COMMISSION_
PCT columns from the EMPLOYEES table for those rows that have NULL values
stored in the COMMISSION_PCT column:
SELECT last_name, commission_pct
FROM employees
WHERE commission_pct IS NULL;
This WHERE clause reads naturally and retrieves only the records that 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.
FIGURE 3-13
Using the IS
NULL operator
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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
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.
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
TABLE 3-2
AND Operator
Truth Table
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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”.
FIGURE 3-14
Using the AND
operator
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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.
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;
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 begins with are also returned.
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
TABLE 3-3
OR Operator
Truth Table
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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-MAR-2008 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.
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.
FIGURE 3-15
Using the OR
operator
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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.
Condition X NOT Condition X
FALSE TRUE
TRUE FALSE
NULL NULL
TABLE 3-4
The NOT
Operator Truth
Table
TABLE 3-5
Conditions
Negated by the
NOT Operator
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
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 that 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.
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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
both conditions D and E are true.
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
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;
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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.
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';
Precedence
Level 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
TABLE 3-6
Operator
Precedence
Hierarchy
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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.
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.
FIGURE 3-16
Operator
precedence in the
WHERE clause
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Boolean operators OR
and AND allow multiple WHERE clause
conditions to be specied. 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.
FIGURE 3-17
Effect of
condition clause
ordering due to
precedence rules
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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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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,
SALARY, and COMMISSION_PCT 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, salary, commission_pct
FROM employees
WHERE job_id IN ('SA_MAN','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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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.
FIGURE 3-18
Sorting data using
the ORDER BY
clause
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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.
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 2. 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. Tw o
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 fulfills 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;
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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.
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.
FIGURE 3-19
Composite
sorting using
the ORDER BY
clause
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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 speciers, 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.
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%'
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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.
8. Executing the statement returns a set of results matching this pattern as
shown in the following illustration.
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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.
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
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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.
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  key to
submit the value. Oracle then prompts you to enter a value for the EMPNO variable.
FIGURE 3-20
Single ampersand
substitution
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After typing in a value, for example, 0, and hitting , 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,
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 appears to
retrieve the FIRST_NAME and LAST_NAME columns from the EMPLOYEES
table for those rows that contain the same set of characters in both 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
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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 that contained the identical string of characters.
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
FIGURE 3-21
Double
ampersand
substitution
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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.
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
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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;
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;
FIGURE 3-22
Substituting
column names
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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
FIGURE 3-23
Substituting
expressions and
text
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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.
Dene 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.
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.
INSIDE THE EXAM
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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.
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, and SALARY is input. This value is substituted
and the statement executes. A subsequent execution of this statement within the
same session does not prompt for any COLNAME value 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 and FIRST_NAME is substituted, changing the query.
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
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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
FIGURE 3-24
The UNDEFINE
command
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executed, and the explicitly defined substitution variable EMPNAME is referenced
by both queries. Finally, the variable is UNDEFINED.
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
FIGURE 3-25
The DEFINE
command
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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.
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.
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;
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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, while 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.
FIGURE 3-26
The VERIFY
command
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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.
3. The SELECT clause is
SELECT &&EMPLOYEE_ID EMPLOYEE_ID, FIRST_NAME, SALARY, SALARY
* 12 AS "ANNUAL SALARY", &&TAX_RATE "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
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8. Executing this statement returns the set of results shown in the following
illustration.
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.
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Certication Summary 165
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 (and including) 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 parenthesized
(single-quoted if non-numeric), comma-delimited (if more than one value) 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.
✓
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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.
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 that 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 $1,000 to $4,000. 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;
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%';
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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 two 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 > 2000 AND SALARY < 5000
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,
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;
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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.
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
1. þ C. The SELECT clause facilitates projection by specifying the list of columns to be
projected from a table, while the WHERE clause facilitates selection by limiting the rows
retrieved based on its conditions.
ý A, B, and D are incorrect. The FROM clause specifies the source of the rows being
projected and the ORDER BY clause is used for sorting the selected rows.
2. þ B. The IN operator efficiently tests whether the JOB_ID for a particular row is either
SA_REP or MK_MAN, while the BETWEEN operator efficiently measures whether an
employee’s SALARY value falls within the required range.
ý A, C, and D are incorrect. A and C exclude employees who earn a salary of $1,000 or
$4,000, 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 $1,000 and $4,000. However, the second part (the OR clause),
correctly tests for JOB_ID equal to MK_MAN but ignores the SALARY condition.
3. þ C. The character literals being compared to the JOB_ID column by the IN operator must
be enclosed by single quotation marks.
ý 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.
4. þ B. The LIKE operator tests the DEPARTMENT_NAME column of each row for values
that contain the characters "er". The percentage symbols before and after the character literal
indicate that any characters enclosing the "er" literal are permissible.
ý 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.
5. þ A and D. The IS NULL operator correctly evaluates the COMMISSION_PCT column
for 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.
ý B and C are incorrect. NULL values cannot be tested by the equality operator or the
IN operator.
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6. þ A and C. Each of these conditions tests for SALARY values in the range of $2,000 to
$5,000.
ý B, D, and E are incorrect. B excludes values like $2,500 from its set, D excludes the
boundary values of $2,000 and $5,000, and E is illegal since it is missing the SALARY column
name reference after the AND operator.
Sort the Rows Retrieved by a Query
7. þ C. The terms specified in an ORDER BY clause can include column names, positional
sorting, numeric values, and expressions.
ý A, B, and D are true.
8. þ C. Positional sorting is performed, and the third term in the SELECT list, COMMISSION_
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.
ý 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
9. þ B. A session-persistent substitution variable may be referenced using an ampersand symbol
from within any SQL statement executed in that session.
ý 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.
10. þ D. The first time this statement is executed, two single ampersand substitution variables
are 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.
ý 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 '1024x768' or '1280x1024'. 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 conditions are:
PRODUCT_NAME LIKE 'GP%1024x768%' or
PRODUCT_NAME LIKE 'GP%1280x1024%'
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%1024x768%') OR (PRODUCT_NAME LIKE 'GP%1280x1024%')
8. The ORDER BY clause is
ORDER BY LIST_PRICE DESC
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9. Executing this statement returns the set of results matching this pattern as shown in the
illustration:
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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
✓ Two-Minute Drill
Q&A Self Test
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Functions 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
Dening a Function
A function is a program written to optionally accept input parameters, perform
an operation, and 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 may be mandatory while others may be optional. The second
component is the data type of its resultant value. Upon execution, only one value of
a predefined data type is returned by the function. The third encapsulates the details
of the processing performed by the function and contains the program code that
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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
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
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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.
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.
rpad('#PASSWORD#',11,'#') = #PASSWORD##
lpad('#PASSWORD#',11,'#') = ##PASSWORD#
The TRIM function literally trims off leading or trailing (or both) character
strings from a given source string:
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, statistical, and financial calculations
have been exposed as built-in functions by Oracle.
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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
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(date1, date2) function returns the number of months
between two dates, while the ADD_MONTHS(date, 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) function returns the last day of the month that the
specified date falls into, while the NEXT_DAY(date, 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
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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.
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 Core SQL:2011, a standard endorsed by
both ISO (International Organization for Standardization) and ANSI (American
National Standards Institute). SQL functions have been standardized, and Oracle has
documented those that are fully or partially compliant to the SQL:2011 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, usually 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 user-specified
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.
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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 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 (to obtain the position of the last character) producing
the following ORDER BY clause:
ORDER BY substr(region_name, length(region_name),1)
FIGURE 4-1
A single-row
function
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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 6.
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.
FIGURE 4-2
Functions in
SELECT, WHERE,
and ORDER BY
clauses
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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),
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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;
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.
FIGURE 4-3
The LOWER
function
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Consider writing an alternative query to return the same results without using the
LOWER function in the WHERE clause. It could be done as follows:
SELECT first_name, last_name, lower(last_name)
FROM employees
WHERE last_name LIKE '%ur%'
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 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%';
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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
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;
FIGURE 4-4
The UPPER
function
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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.
FIGURE 4-5
The INITCAP
function
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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
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%'
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7. Executing this statement returns employees’ names containing the characters
“li” as shown in this illustration:
Using Character Manipulation 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.
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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.
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.
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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 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),
FIGURE 4-6
The CONCAT
function
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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 automatically 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 10 characters and the remaining columns from the
COUNTRIES table.
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
FIGURE 4-7
The LENGTH
function
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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(string, length, [padstring]) and
RPAD(string, length, [padstring]),
where string represents the source string, length is an integer value that represents
the final length of the string returned, and padstring specifies the character string to
be used as padding. If LPAD (or RPAD) is used, the padstring is added to the left (or
right) of the source string until it reaches the target length. Note that if the length
parameter is smaller than or equal to the length of the source string, then no padding
occurs and a substring of the source string having length number of characters is
returned. If padstring is omitted, the character string used for padding defaults to a
single blank space.
The following queries illustrate the usage of this function:
Query 1: SELECT lpad(1000+200.55,14,'*') FROM dual;
Query 2: SELECT rpad(1000+200.55,14,'*') FROM dual;
Query 3: SELECT lpad(SYSDATE,14,'$#') FROM dual;
Query 4: SELECT rpad(SYSDATE,4,'$#') FROM dual;
Query 1 returns a 14-character string: “*******1200.55”. The numeric expression
is evaluated to return the number 1200.55. This number is cast as the string
“1200.55” of length seven (including the decimal point). To achieve the final length
of 14 characters, 7 asterisks are left padded to the string. Query 2 returns the string
“1200.55*******”.
The LPAD function in query 3 has a target string length of 14 characters. Assume
that SYSDATE returns a nine-character date value: “17-DEC-07”. This date is
converted into a string, and the padding string is systematically applied to reach
the target length. It returns: “$#$#$17-DEC-07”. Note that although the padding
string consists of two characters ($#), 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 nine-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”.
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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
to those in Figure 4-6 but have been rendered more readable 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),
FIGURE 4-8
The LPAD and
RPAD functions
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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. When no
direction of trimming is specified, then spaces are trimmed from both sides
of the string.
■ TRIM(trailing trimstring from s) removes all occurrences of trimstring from
the end of the string s if it is present.
■ 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 is 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
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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).
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;
FIGURE 4-9
The TRIM
function
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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 is 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 begins 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 string “on” is found beginning at the second position of the
DEPARTMENT_NAME column.
FIGURE 4-10
The INSTR
function
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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.
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 cast 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”.
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;
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Query 3 extracts the substring beginning at position 5. Since the third parameter
is not specified, the default extraction length is equal to the number of characters
from and including the start position (position 5 in this case) to the end of the source
string, which is six characters long. 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 three characters from the end of the string. Therefore, the 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.
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.
FIGURE 4-11
The SUBSTR
function
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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 3 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 ten-character
string based on the SALARY column in the
EMPLOYEES table. If the SALARY value is less
than ten characters long, zeros must be added to
the left of the value to yield a ten-character string.
Is this possible?
Yes. The LPAD function may be used as follows:
SELECT LPAD(SALARY,10,0)
FROM EMPLOYEES;
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;
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
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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”.
FIGURE 4-12
The REPLACE
function
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EXERCISE 4-2
Using the Character 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.
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, among 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
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”
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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 zero, then the source number is rounded
to the nearest whole number. If it is 1, then the source number is rounded to the
nearest tenth. If it is 2, then the source is rounded to the nearest hundredth, and so
on. The following queries illustrate the usage of this function:
Query 1: SELECT round(1601.916718,1) FROM dual;
Query 2: SELECT round(1601.916718,2) FROM dual;
Query 3: SELECT round(1601.916718,-3) FROM dual;
Query 4: SELECT 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 (6) is found three places
TABLE 4-1
Decimal Precision
Descriptions
Decimal Precision Significant Rounding Digit Decimal Position
–3 1 Thousands (n ×1000)
–2 2 Hundreds (n ×100)
–1 3 Tens (n ×10)
0 4 Units (n ×1)
1 5 Tenths (n ÷10)
2 6 Hundredths (n ÷100)
3 7 Thousandths (n ÷1000)
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to the left of the decimal point and the source number is rounded up to the nearest
thousand to yield 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. The TRUNC function takes two parameters. Its syntax is:
TRUNC (source number, [decimal precision]),
FIGURE 4-13
The numeric
ROUND function
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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 truncation 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 1: SELECT trunc(1601.916718,1) FROM dual;
Query 2: SELECT trunc(1601.916718,2) FROM dual;
Query 3: SELECT trunc(1601.916718,-3) FROM dual;
Query 4: SELECT 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
round(1601.916718,2) was performed since the digit in position (n + 1 = 3) after the
decimal point is 6 (greater than 5) and 1601.92 would have been returned. 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 to the nearest thousand, as
shown earlier in Table 4-1. Therefore, the source number is truncated at its thousands
position 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 up to 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 the dividend
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 dividend, leaving
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the remainder equal to the dividend. 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 1: SELECT mod(6,2) FROM dual;
Query 2: SELECT mod(5,3) FROM dual;
Query 3: SELECT mod(7,35) FROM dual;
Query 4: SELECT 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 2 as the remainder. 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 2.2 as the remainder.
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 to distinguish between even and odd numbers.
FIGURE 4-14
The numeric
TRUNC function
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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 4. As Figure 4-15 shows, the MOD function allocates the numbers 0 to 3
to each row in a round-robin manner.
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 rst 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.
FIGURE 4-15
The MOD
function
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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 examine 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 SQL Developer. Note that previous versions
of SQL Developer used forward slashes as the default date separator when displaying
dates. Figure 4-16 shows the contents of the START_DATE column after querying the
JOB_HISTORY table using SQL*Plus.
FIGURE 4-16
Default date
separators in
SQL*Plus
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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.
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. The RR date format mask differs from the YY format mask since it is used
to specify different centuries based on the current and specified years. The century
component assigned to a date value when it is inserted or updated when a
two-digit year value is submitted depends on the NLS_DATE_FORMAT setting
in that session. If the NLS_DATE_FORMAT is set to the default DD-MON-RR
format, the century component is automatically assigned based on the following rules:
■ If the last two digits of the current year and specified year lie between 0 and
49, the current century is returned. Suppose the current year is 2014. The
date assigned for 24-JUL-04 is 24-JUL-2004.
■ If the last 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 year is 2014. The date assigned for 24-JUL-94 is 24-JUL-1994.
■ If the last two digits of the current and specified years lie between 50 and 99,
the current century is returned by default. If the current year is 2055, the date
assigned for 24-JUL-94 is 24-JUL-2094.
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  and 
HH24 Twenty-four-hour time
MI Minutes
SS Seconds
TABLE 4-2
Date Format
Masks
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■ If the last 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. If the
current year is 1975, the date assigned for 24-JUL-17 is 24-JUL-2017.
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
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
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.
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 |
Date Format. Change the default display mask (DD-MON-RR) to (DD-MON-
RR HH24:MI:SS).
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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.
The HIREDATE column for employees with DEPARTMENT_ID values of 30 is
subtracted from the date item 02-JUN-2006 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-2006, a negative
number is returned.
FIGURE 4-17
The SYSDATE
function and date
arithmetic
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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.
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-MON-
RR 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(date1, date2),
The function computes the difference in months between date1 and date2. If the
date1 occurs before the date2, 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 months_between(SYSDATE+31, SYSDATE),
months_between(SYSDATE+61, SYSDATE),
months_between(SYSDATE+92, 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;
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
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the MONTHS_BETWEEN (29-MAR-2008, 29-DEC-2007) function returns exactly
three months.
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 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.
Query 3 is identical to Query 2 except it multiplies the result of Query 2 by 31.
Expectedly it returns 1.03225806 × 31 = (1 x 31) + (0.03225806 × 31) = 31 + 1 = 32.
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.
FIGURE 4-18
The MONTHS_
BETWEEN
function
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A common mistake is
to assume that the return data type of
single-row 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 other data types. 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.
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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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-2012.
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
FIGURE 4-19
The ADD_
MONTHS
function
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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-2012', 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 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
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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 day of the week parameter may be longer than three characters, but any
characters following a valid abbreviation are ignored; for example, Sunday may be
referenced as “sun”, “sun dance”, “sun hat”, 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. 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.
FIGURE 4-20
The NEXT_DAY
function
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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 590.423.4567. 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 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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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.
The date ROUND function takes one mandatory and one optional parameter. Its
syntax is:
ROUND(source date, [date precision format]),
FIGURE 4-21
The LAST_DAY
function
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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 1 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
1 and 6, 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.
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-JUN-2009
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.
FIGURE 4-22
The date
ROUND function
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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
single-row 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 frequent mistake 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.
INSIDE THE EXAM
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 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
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Certication Summary 223
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.
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 high-level
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.
FIGURE 4-23
The date TRUNC
function
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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
provide a starting point for your discovery of their usefulness.
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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 the 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. Any fractional component returned 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.
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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.
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? (Choose the best answer.)
SELECT LENGTH('How_long_is_a_piece_of_string?') FROM DUAL;
A. 29
B. 30
C . 24
D. None of the above
4. What value is returned after executing the following statement? (Choose the best answer.)
SELECT SUBSTR('How_long_is_a_piece_of_string?', 5,4) FROM DUAL;
A. long
B. _long
C . ring?
D. None of the above
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5. What value is returned after executing the following statement? (Choose the best answer.)
SELECT INSTR('How_long_is_a_piece_of_string?','_',5,3) FROM DUAL;
A. 4
B. 14
C . 12
D. None of the above
6. What value is returned after executing the following statement? (Choose the best answer.)
SELECT REPLACE('How_long_is_a_piece_of_string?','_','') FROM DUAL;
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? (Choose the best answer.)
SELECT MOD(14,3) FROM DUAL;
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? (Choose the best answer.)
SELECT ADD_MONTHS(SYSDATE,-1) FROM DUAL;
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
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10. Assuming SYSDATE=30-DEC-2007, what value is returned after executing the following
statement? (Choose the best answer.)
SELECT TRUNC(SYSDATE,'YEAR') FROM DUAL;
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 that 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
1. þ B and C. Single-row functions execute once for every record selected in a dataset and may
either take no input parameters, like SYSDATE, or many input parameters.
ý A and D are incorrect. A function by definition returns only one result and there are many
functions with no parameters.
2. þ A and D. The LOWER function converts the case of the input string parameter to its
lowercase equivalent, while INITCAP converts the given input parameter to title case.
ý B and C are incorrect. They are not valid function names.
Use Character, Number, and Date Functions in SELECT Statements
3. þ B. The LENGTH function computes the number of characters in a given input string
including spaces, tabs, punctuation marks, and other nonprintable special characters.
ý A, C, and D are incorrect. The string literal has 30 characters including underscore
characters and the question mark.
4. þ A. The SUBSTR function extracts a four-character substring from the given input string
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”.
ý B, C, and D are incorrect. 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.
5. þ B. The INSTR function returns the position that the nth occurrence of the search string
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.
ý A, C, and D are incorrect. Position 4 is the first occurrence of the search string and
position 12 is the third occurrence if the search began at position 1.
6. þ C. All occurrences of the underscore character are replaced by an empty string, which
removes them from the string.
ý 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.
7. þ C. When 14 is divided by 3, the answer is 4 with remainder 2.
ý A, B, and D are incorrect.
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8. þ A. The –1 parameter indicates to the ADD_MONTHS function that the date to be returned
must be one month prior to the given date.
ý 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.
9. þ A. Since the first of January 2009 falls on a Thursday, the date of the following Wednesday
is six days later.
ý 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.
10. þ C. The date TRUNC function does not perform rounding and since the degree of 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.
ý 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 and 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, 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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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:
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5
Using Conversion Functions
and Conditional Expressions
CERTIFICATION OBJECTIVES
5.01 Describe Various Types of Conversion
Functions Available in SQL
5.02 Use the TO_CHAR, TO_NUMBER, and
TO_DATE Conversion Functions
5.03 Apply Conditional Expressions in a
SELECT Statement
✓ Two-Minute Drill
Q&A Self Test
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Functions 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
the data it can store. A NUMBER column cannot store character information.
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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 W H E R E 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, 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.
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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 3: SELECT mod('11',2) FROM dual;
Query 4: SELECT mod('11.123',2) FROM dual;
Query 5: SELECT mod('11.123.456',2) FROM dual;
Query 6: SELECT mod('$11',2) FROM dual;
Queries 3 and 4 implicitly convert the character strings “11” and “11.123” into the
numbers 11 and 11.123, respectively, before the MOD function evaluates them and
returns the results 1 and 1.123. Query 5 returns the error “ORA-1722: invalid number”
when Oracle tries to perform an implicit character to number conversion. It fails because
the string “11.123.456” is not a valid number. Query 6 also fails with the invalid
number error, since the dollar symbol cannot be implicitly converted into a number.
Implicit character to date conversions are possible when the character string
conforms to the following date patterns: [D|DD] separator1 [MON|MONTH]
separator2 [R|RR|YYYY]. D and DD represent a single and two-digit day of the
month. MON is a three-character abbreviation, while MONTH is the full name
for a month. R and RR represent a single and two-digit year. YYYY represents a
four-digit year. The separator1 and separator2 elements may be most punctuation
marks, spaces, and tabs. Table 5-1 demonstrates implicit character to date conversion,
listing several function calls and the results SQL Developer returns.
TABLE 5-1 Examples of Implicit Character to Date Conversion
Function Call Format Results
add_months('24-JAN-09',1) DD-MON-RR 24/FEB/09
add_months('1\january/8',1) D\MONTH/R 01/FEB/08
months_between('13*jan*8', '13/feb/2008') DD*MON*R, DD/MON/
YYYY
–1
add_months('01$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
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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-
M O N - R R'),'D a y').
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
three-character-long string USD, the following command may be issued:
ALTER SESSION SET nls_currency='USD';
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 such as width, currency symbol, the position
of a decimal point, and group (or thousands) separators, and must be enclosed
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
TABLE 5-2
Syntax of
Explicit Data
Type Conversion
Functions
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in single 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
six 9s in the format mask indicate to the TO_CHAR function that leading zeros must
FIGURE 5-1
National
language support
(NLS) session
parameters
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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.
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, regardless of the case of the characters used to store the job title. 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.
Format
Element Description 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
(comma is default)
09999G999 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
TABLE 5-3 Numeric Format Masks
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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]),
FIGURE 5-2
CHAR function
with numbers
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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
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
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
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component of the current system date is extracted for conversion to a character type.
Second, 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 zero or more spaces to yield a nine-character-long
string. Since January is eight characters long, one space is added, but if the month
was September, no spaces would be automatically added. 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 “Januaryis a special
time”. Note there is no space between January and the literal “is a special time” as
a result of the fm modifier. 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 sign
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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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 day such as the 17th or 31st. The 'Mo nt h' 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 four-digit year component.
TABLE 5-6
Date Format
Masks for Time
Components
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
TABLE 5-7
Miscellaneous
Date Format
Masks
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
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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 13th of January, Two Thousand One.
1. Start SQL Developer and connect to the HR schema.
2. The WHERE clause is
WHERE TO_CHAR(HIRE_DATE,'fmDay') = 'Saturday'
The fm modifier is necessary to remove trailing blanks since a comparison
with a character literal is performed and an exact match is required.
FIGURE 5-3
TO_CHAR
function with
dates
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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:
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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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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 2008. The format mask
matches 01 to MM, 12 to DD, and 2008 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]),
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 it returns the error “ORA-1722: invalid number.” Query
2 matches the dollar symbol, comma, and period from the string to the format mask,
FIGURE 5-4
The TO_DATE
function
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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
function was then used to convert these eight characters, including a decimal
point, into a number that 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.
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')
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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
FIGURE 5-5
The TO_
NUMBER
function
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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.
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. T O _ D A T E ('2 8 / 1 0 /0 9 ',' D D /
M M / R R') 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',
'f m M o n t h') 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 L E N G T H('O c t o b e r') function is evaluated and the query returns
the number 7.
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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;
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
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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 three-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
FIGURE 5-6
The NVL function
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COMMISSION_PCT columns are also selected. The difference between the queries is
in the calculated expression aliased as MONTHLY_COMMISSION. Due to the NVL
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. 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 N V L 2(' b c', N o t
bc','No Substring') function is evaluated. The ifnotnull parameter, the string
'Not bc', is returned.
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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),
FIGURE 5-7
The NVL2
function
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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(1234,1233+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 implicitly evaluated, and the NULLIF function finds
1234 equivalent to 1233+1 (1234), so it returns a null value. 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
FIGURE 5-8
The NULLIF
function
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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 four-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.
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 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 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
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6. Executing this statement returns employees’ names and the NAME_LENGTHS
expression as shown in the following illustration:
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))
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The data type COALESCE returns if a not null value is found that 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.
FIGURE 5-9
The COALESCE
function
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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.
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. Following is an example of
if-then-else logic: 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.
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;
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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 m atch' is returned. Query 3 searches through the comparison
parameters for a match. The character terms 'c o m p 1' and 'c o m p 2' are not equal
to search, so the results true1 and true2 are not returned. A match is found in the
third comparison term 'c o m p 3' (parameter 6), which contains the string search.
Therefore, the third result term iftrue3 (parameter 7) containing the string 't r u e3'
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
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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.
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.
FIGURE 5-10
The DECODE
function
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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
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.
INSIDE THE EXAM
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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 C O N C A T('1','2 3') 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 not equal to 50, 80, 90, 100, or 110 are retrieved along
with two numeric expressions and one CASE expression, as shown in Figure 5-11.
The numeric expression aliased as MONTHS returns a truncated value obtained
by calculating the months of service between the 1st of January 2013 and an
employee’s hire date using the MONTHS_BETWEEN function.
Five categories of loyalty classification based on every two years of service are
defined by truncating the quotient obtained by dividing the total months of service
by 24. 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, the remaining WHEN…THEN statements catch 13 rows
and 3 rows are caught by the ELSE statement. The set is then sorted by months.
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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
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
FIGURE 5-11
The CASE
expression
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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('01-JAN-2013',hire_date)) months,
trunc(months_between('01-JAN-2013',hire_date)/24) "Months divided by 24",
CASE
WHEN trunc(months_between('01-JAN-2013',hire_date)/24) < 2 then 'Intern'
WHEN trunc(months_between('01-JAN-2013',hire_date)/24) < 3 then 'Junior'
WHEN trunc(months_between('01-JAN-2013',hire_date)/24) < 4 then 'Intermediate'
WHEN trunc(months_between('01-JAN-2013',hire_date)/24) < 5 then 'Senior'
ELSE 'Furniture'
END loyalty
FROM employees
where department_id NOT IN (50,80,90,100,110)
ORDER BY months;
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.
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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:
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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.
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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, such as 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.
❑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.
✓
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❑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.
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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? (Choose the best answer.)
SELECT LENGTH(3.14285) FROM DUAL;
A. Explicit conversion
B. Implicit conversion
C . 7
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? (Choose the best answer.)
SELECT TO_NUMBER(1234.49, 999999.9) FROM DUAL;
A. 1234.49
B. 001234.5
C . 1234.5
D. None of the above
4. What value is returned after executing the following statement? (Choose the best answer.)
SELECT TO_CHAR(1234.49, '999999.9') FROM DUAL;
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? (Choose the
best answer.)
SELECT TO_CHAR(SYSDATE, 'fmMONTH, YEAR') FROM DUAL;
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? (Choose the
best answer.)
SELECT TO_CHAR(SYSDATE, 'fmDDth MONTH') FROM DUAL;
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? (Choose the
best answer.)
SELECT TO_CHAR(TO_DATE(TO_CHAR(SYSDATE,'DD'),'DD'),'YEAR') FROM DUAL;
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? (Choose the best answer.)
SELECT NVL2(NULLIF('CODA','SID'),'SPANIEL','TERRIER') FROM DUAL;
A. SPANIEL
B. TERRIER
C . NULL
D. None of the above
9. What value is returned after executing the following statement? (Choose the best answer.)
SELECT NVL(SUBSTR('AM I NULL',10),'YES I AM') FROM DUAL;
A. NO
B. NULL
C . YES I AM
D. None of the above
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10. If SYSDATE returns 12-JUL-2009, what is returned by the following statement? (Choose the
best answer.)
SELECT DECODE(TO_CHAR(SYSDATE,'MM'),'02','TAX DUE','PARTY') FROM DUAL;
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 any requested reference date in the current year.
The list should retrieve rows from the CUSTOMERS table that 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.
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SELF TEST ANSWERS
Describe Various Types of Conversion Functions Available in SQL
1. þ B. The number 3.14285 is given as a parameter to the LENGTH function. There is a data
type mismatch, but Oracle implicitly converts the parameter to the character string “3.14285”
allowing the function to operate correctly.
ý 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.
2. þ D. Dates are only converted into character strings using TO_CHAR and not the
TO_DATE function.
ý A, B, and C are correct statements.
Use the TO_CHAR, TO_NUMBER, and TO_DATE Conversion Functions
3. þ D. An “ORA-1722: invalid number” error is returned because the statement is trying to
convert a number with two digits after the decimal point using a format mask that caters for just
one digit after the decimal point. If the expression was TO _ NUMBER(1234.49, '999999.99'),
the number 1234.49 would be returned.
ý A, B, and C are incorrect.
4. þ C. For the number 1234.49 to match the character format mask with one decimal place,
the number is first rounded to 1234.5 before TO_CHAR converts it into the string “1234.5”.
ý 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'.
5. þ B. The MONTH and YEAR components of the format mask separated by a comma and a
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.
ý A, C, and D are incorrect. If the format mask was 'MON, YYYY' or 'M O N-Y Y', A and C,
respectively, would be returned.
6. þ A. The DD component returns the day of the month in uppercase. Since it is a number, it
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.
ý 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'.
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7. þ B. The innermost nested function is TO _ CHAR(SYSDATE,'DD'), which extracts the
day component of SYSDATE and returns the character 12. The next function executed is
T O _ D A T E('1 2','D D ') 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.
ý A, C, and D are incorrect.
8. þ A. The NULLIF function compares its two parameters and, since they are different, the
first parameter is returned. The N V L 2('C O D A', 'S P A N I E L',' T E R R I E R') function call returns
SPANIEL since its first parameter is not null.
ý B, C, and D are incorrect.
9. þ C. The character literal “AM I NULL” is nine characters long. Therefore, trying to obtain
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.
ý A, B, and D are incorrect.
10. þ B. The innermost function TO _ CHAR(SYSDATE, 'MM') results in the character string
“07” being returned. The outer function is D E C O D E('0 7','0 2',' T A X D U E','P A R T Y '). Since
“07” is not equal to “02” the else component “PARTY” is returned.
ý 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 the day component
of the DATE_OF_BIRTH is between two days ago and seven days from a provided reference
date in the current year. The reference date may be handled using an ampersand substitution at
runtime. This solution utilizes the DDD date format mask described in Table 5-4 which provides
the day of the year from a date.
3. The condition of the WHERE clause is
TO_CHAR(DATE_OF_BIRTH,'DDD') - TO_CHAR(&REFDATE,'DDD') BETWEEN -2 AND 7
4. 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.
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5. The case tested is based on the expression in the WHERE clause discussed above. The day of
the year of the DATE_OF_BIRTH minus the day of the year of the reference date is evaluated.
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'.
6. The SELECT clause is therefore
SELECT CUST_FIRST_NAME, CUST_LAST_NAME, CUST_EMAIL, DATE_OF_BIRTH,
CASE TO_CHAR(DATE_OF_BIRTH,'DDD')) - TO_CHAR(&REFDATE,'DDD')
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
7. The FROM clause is
FROM CUSTOMERS
8. The ORDER BY clause is
ORDER BY TO_CHAR(DATE_OF_BIRTH,'MM-DD')
9. Executing this statement returns the set of results matching this pattern as shown in the
following illustration for REFDATE=to _ date('08-JAN-2014'):
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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 Include or Exclude Grouped Rows Using
the HAVING Clause
✓ Two-Minute Drill
Q&A Self Test
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Single-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 are
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
Denition 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, result3,…, resultn;
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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
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
FIGURE 6-1
Group functions
operating on
12 groups
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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
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) ;
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This syntax may be deconstructed into the following forms:
1. AVG(DISTINCT expr)
2. AVG(ALL expr)
3. AVG(expr)
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);
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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)
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.
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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)).
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
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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;
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.
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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 1: SELECT sum(2) FROM employees;
Query 2: SELECT sum(salary) FROM employees;
Query 3: SELECT sum(DISTINCT salary) FROM employees;
Query 4: SELECT sum(commission_pct) FROM employees;
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 691416. Query 3 returns a total of 409908 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 end of 2015 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.
FIGURE 6-2
The COUNT
function
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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 1: SELECT avg(2) FROM employees;
Query 2: SELECT avg(salary) FROM employees;
Query 3: SELECT avg(DISTINCT salary) FROM employees;
Query 4: SELECT avg(commission_pct) FROM employees;
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
FIGURE 6-3
The SUM
function
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of 691400. This is divided by the 107 rows with nonnull SALARY values to return
the average 6461.83178. You may expect Query 3 to return a smaller result than
Query 2, but it does not. There are 58 unique salary values, which when added yield
a total of 409908. Dividing 409908 by 58 returns 7067.37931 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 by the end of 2015. The second uses the AVG
function to calculate the average number of years for which current programmers
have been employed by the end of 2015.
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.
Character strings are converted to numeric representations of their constituent
characters based on the NLS settings in the database. When the MIN function is
FIGURE 6-4
The AVG function
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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-1995 and the latest END_DATE is 31-DEC-2007. 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
FIGURE 6-5
The MIN and
MAX functions
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the sales representatives working for the shortest and longest durations were hired on
21-APR-2008 and 30-JAN-2004, 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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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 following 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)
and 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.
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
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.
FIGURE 6-6
Nested group
functions
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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.
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.
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Grouping data and 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.
FIGURE 6-7
Unique
DEPARTMENT_
ID values in the
EMPLOYEES table
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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;
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.
It is common to see the grouping attribute in the SELECT list alongside grouping
functions. If an item, that is not a group function, appears in the SELECT list with
other group functions 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
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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 four-
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 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.
FIGURE 6-8
The GROUP
BY clause
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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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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"
FIGURE 6-9
The GROUP
BY clause with
multiple columns
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5. The FROM clause is
FROM JOB_HISTORY
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
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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 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 10 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
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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.
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. Group-level 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]];
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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:
SELECT department_id, count(*)
FROM job_history
WHERE department_id IN (50,60,80,110)
GROUP BY department_id
HAVING count(*) > 1
AND department_id > 50;
Two rows with DEPARTMENT_ID values of 80 and 110, each with a COUNT(*)
value of 2 are returned. Notice the grouping attribute DEPARTMENT_ID may be
present in the HAVING clause.
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 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 specied when a GROUP BY
clause is present. A GROUP BY clause can
be specied 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 based on the grouping attributes.
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FIGURE 6-10
The HAVING
clause
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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 18 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 18.
7. The HAVING clause is
HAVING COUNT(*) >= 18
8. The FROM clause is
FROM EMPLOYEES
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9. Executing this statement returns the days of the week on which 18 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
or grouping attributes (or both) must exist in
the SELECT clause if there is a GROUP BY
clause. The HAVING clause may occur before
the GROUP BY clause but usually follows 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 or a grouping attribute or both.
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.
INSIDE THE EXAM
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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.
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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.
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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.
Describe the Group Functions
1. What result is returned by the following statement? (Choose the best answer.)
SELECT COUNT(*) FROM DUAL;
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? (Choose the best answer.)
SELECT SUM(SALARY) FROM EMPLOYEES;
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.
A. 900
B. 1000
C . NULL
D. None of the above
4. Which values are returned after executing the following statement? (Choose all that apply.)
SELECT COUNT(*), COUNT(SALARY) FROM EMPLOYEES;
Assume there are ten employee records and each contains a SALARY value of 100, except for
one, which has a null value in their SALARY field.
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? (Choose the best answer.)
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.
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? (Choose the best answer.)
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.
A. An error is returned
B. 3
C . 4
D. None of the above
7. How many records are returned by the following query? (Choose the best answer.)
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.
A. 12
B. 11
C . NULL
D. None of the above
8. What values are returned after executing the following statement? (Choose the best answer.)
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.
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
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Include or Exclude Grouped Rows Using the HAVING Clause
9. How many rows of data are returned after executing the following statement? (Choose the best
answer.)
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.
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? (Choose the best
answer.)
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.
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
1. þ C. The DUAL table has one row and one column. The COUNT(*) function returns the
number of rows in a table or group.
ý A, B, and D are incorrect.
2. þ B. By definition, group functions can operate on multiple rows at a time, unlike single-row
functions.
ý 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
3. þ A. The SUM aggregate function ignores null values and adds nonnull values. Since nine
rows contain the SALARY value 100, 900 is returned.
ý 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.
4. þ C. COUNT(*) considers all rows including those with NULL values. COUNT(SALARY)
only considers the nonnull rows.
ý A, B, and D are incorrect.
5. þ C. The NVL function converts the one NULL value into 100. Thereafter, the average
function adds the SALARY values and obtains 1000. Dividing this by the number of records
returns 100.
ý 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
6. þ C. The dataset is segmented based on the SALARY column. This creates two groups: one
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
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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.
ý 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.
7. þ A. There are 12 distinct DEPARTMENT_ID values. Since this is the grouping attribute,
12 groups are created, including 1 with a null DEPARTMENT_ID value. Therefore, 12 rows
are returned.
ý B, C, and D are incorrect.
Include or Exclude Grouped Rows Using the HAVING Clause
8. þ C. For a GROUP BY clause to be used, a group function must appear in the SELECT list.
ý A, B, and D are incorrect. The statement is syntactically inaccurate and is disallowed by
Oracle. Do not mistake the column named MAX_SALARY for the MAX(SALARY) function.
9. þ B. Two groups are created based on their common DEPT_ID values. The group with
no null SALARY values 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 the null SALARY record 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.
ý 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 any way.
10. þ A. Two groups are created based on their common DEPT_ID values. The group with
no null SALARY values 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(NVL(SALARY,100)) > 400 clause. The group with the null SALARY
record 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.
ý 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.
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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 Generate a Cartesian Product of Two or
More Tables
✓ Two-Minute Drill
Q&A Self Test
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The 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,
collectively known as inner joins, 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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Write SELECT Statements to Access Data 317
of nonequijoins and additional join conditions. Joining is described by focusing on
the following eight areas:
■ Types of joins
■ Joining tables using ANSI SQL syntax
■ Qualifying ambiguous column names
■ The NATURAL JOIN clause
■ The JOIN USING clause
■ The 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 probably 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
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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.
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 Query 2 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.
Inner Joins
The inner 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.
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
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matching the join condition. This set is then restricted by the WHERE clause. In this
case, because the COUNTRY_NAME value must be “Canada”, a 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.
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
FIGURE 7-1
Natural joins and
other inner joins
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two discrete tables must necessarily have any 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 inner
joins to be formed on specific columns instead of on all commonly named columns.
The fourth query demonstrates the JOIN…ON format of the inner 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 inner join format.
Be wary when using 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 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. Query 3 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.
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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.
The traditional Oracle join syntax supports inner joins, 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;
FIGURE 7-2
Cross join
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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 an inner 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
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 ANSI SQL 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 with the
latest ANSI standards. It offers no performance benefits over the traditional syntax.
Inner, outer, and cross joins may be written using both ANSI SQL and traditional
Oracle SQL.
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The general form of the SELECT statement using ANSI SQL 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) (+)] ;
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 an inner 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 as 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.
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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 “ORA-25154:column part of USING clause cannot have qualifier” error would
be raised.
FIGURE 7-3
Dot notation
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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.
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 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 1: SELECT *
FROM locations
NATURAL JOIN countries;
Query 2: SELECT *
FROM locations, countries
WHERE locations.country_id = countries.country_id;
Query 3: SELECT *
FROM jobs
NATURAL JOIN countries;
Query 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
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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,
FIGURE 7-4
The natural join
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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 tables
works perfectly.
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 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.
3. The FROM clause is
FROM JOB_HISTORY JH
4. The JOIN clause is
NATURAL JOIN EMPLOYEES EMP
5. The SELECT clause is
SELECT EMPLOYEE_ID, JOB_ID, DEPARTMENT_ID, EMP.LAST_NAME,
EMP.HIRE_DATE, JH.END_DATE
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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 JOIN USING Clause
The format of the syntax for the JOIN USING clause is as follows:
SELECT table1.column, table2.column
FROM table1
JOIN table2 USING (join_column1, join_column2…);
While the 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
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or more equijoin columns to be explicitly specified in brackets after the USING
keyword. This avoids the shortcomings associated with the natural join. Many
situations demand that tables be joined only on certain columns and this format
caters for this requirement. Consider the following queries:
Query 1: SELECT *
FROM locations
JOIN countries
USING (country_id);
Query 2: SELECT *
FROM locations, countries
WHERE locations.country_id = countries.country_id;
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. The join
columns specified with the JOIN…USING syntax cannot be qualified using table
names or aliases when they are referenced in the SELECT and JOIN clauses.
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
FIGURE 7-5
Natural join
using the JOIN…
USING clause
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three identically named columns. In this example, only two of these are specified
as join columns. Notice that although the third identically named column is
DEPARTMENT_ID, it is qualified with a table alias to avoid ambiguity while the
join columns specified in the SELECT clause cannot be qualified with table aliases.
The JOIN ON Clause
The format of the syntax for the JOIN ON clause is as follows:
SELECT table1.column, table2.column
FROM table1
JOIN table2 ON (table1.column_name = table2.column_name);
The 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;
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.
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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 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.
FIGURE 7-6
Inner join using
the JOIN…ON
clause
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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
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"
4. The FROM clause is
FROM EMPLOYEES E
5. The JOIN…ON clause is
JOIN DEPARTMENTS D
ON (E. EMPLOYEE_ID=D.MANAGER_ID)
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6. Executing this statement returns 11 rows describing the managers of each
department as shown in the following illustration:
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 joined. 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. However, there are instances when primary- and foreign-key relationships
are not defined between tables. These tables may also be joined, but the results do not
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benefit from referential integrity being enforced by the database. When multiple joins
exist in a statement, they are evaluated from left to right. Consider the following
query using a mixture of natural joins and Oracle joins:
SELECT r.region_name, c.country_name, l.city, d.department_name
FROM departments d
NATURAL JOIN locations l,
countries c, regions r;
The natural join between DEPARTMENTS and LOCATIONS creates an
interim result set consisting of 27 rows since they are implicitly joined on the
LOCATION_ID column. This set is then Cartesian-joined to the COUNTRIES
table since a join condition is not implicitly or explicitly specified. 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 then joined to the REGIONS table.
Once again, a Cartesian join occurs because the REGION_ID column is absent
from any join condition. The final result set contains 2700 (675 × 4) rows and
four columns. Using natural joins mixed with Oracle joins is error prone and not
recommended since join conditions may sometimes be erroneously omitted.
The JOIN…USING and JOIN…ON syntaxes are better suited for joining multiple
tables. The following query joins four tables using the 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 equijoin
or inner join formats. The 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. 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 that the USING,
ON, and NATURAL keywords are mutually
exclusive in the context of the same
join clause.
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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.
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)]|
FIGURE 7-7
N-way joins and
additional join
conditions
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[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 16 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
doubled and compared with all MAX_SALARY values in the JOBS table. If the join
condition evaluates to true, the row is returned.
Nonequijoins are not as commonly used as equijoins. 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.
FIGURE 7-8
Nonequijoins
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CERTIFICATION OBJECTIVE 7.02
Join a Table to Itself Using a Self-Join
Storing hierarchical data in a single relational table may be 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
duplicate the table and this description is merely provided to explain the concept of
self-joining. Consider the following three 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.
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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.
FIGURE 7-9
Self-join
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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_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
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
required, however, an outer join is used to fetch them. Several variations of outer joins
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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 an inner join between the two tables. The
second and third scenarios cause many problems. Joining the EMPLOYEES and
DEPARTMENTS tables on the DEPARTMENT_ID column may result in rows
with null DEPARTMENT_ID values 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 follows:
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.
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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
left of the JOIN keyword. None of the rows from the DEPARTMENTS table contain
any of these 16 LOCATION_ID values.
FIGURE 7-10
Left outer join
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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);
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 the letter “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. These have a null value
in the “Jobs in JOB_HISTORY” column.
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. The full outer join
performs an inner join as well as a left and
right outer join.
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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);
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
FIGURE 7-11
Right outer join
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NULL DEPARTMENT_ID values shows the orphan rows in both tables. There is
one record in the EMPLOYEES table that has no DEPARTMENT_ID values, and
there are 16 departments to which no employees belong.
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
FIGURE 7-12
Full outer join
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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
6. Executing this statement returns 16 rows describing the departments to
which no employees are currently assigned as shown in the following
illustration:
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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 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)
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 cross join syntax. This technique is
described in the next section.
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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 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.
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When using the cross join
syntax, a Cartesian product is intentionally
generated. Inadvertent Cartesian products
are created when there are insufcient join
conditions in a statement. Joins that specify
fewer than N-1 join conditions when
joining N tables or that specify invalid
join conditions may inadvertently create
Cartesian products. A natural join between
two tables sharing no identically named
columns results in a Cartesian join since two
tables are joined but less than one condition
is available.
Figure 7-13 shows a cross join between the REGIONS and COUNTRIES tables.
There are 4 rows in REGIONS and 25 rows in COUNTRIES. Since the WHERE
clause limits the REGIONS table to 2 of 4 rows, the Cartesian product produces 50
(25 × 2) records. The results are sorted alphabetically, first on the REGION_NAME
and then on the COUNTRY_NAME. The first record has the pair of values, Asia
and Argentina. When the REGION_NAME changes, the first record has the pair of
values, Middle East and Africa and Argentina. Notice that the COUNTRY_NAME
values are repeated for every REGION_NAME.
EXERCISE 7-5
Performing a Cross Join
You are required to obtain the number of rows in the EMPLOYEES and
DEPARTMENTS tables as well as the number of records that would be created
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
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FIGURE 7-13
The cross join
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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 2,889
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 inner 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 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.
INSIDE THE EXAM
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.
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Certication Summary 355
The ANSI SQL join syntax is discussed in detail, and three forms of the inner
join are explored. Each form has a purpose, and the advantages and risks associated
with them are considered.
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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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 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 an inner 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.
✓
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❑An outer join is performed when rows, which are not retrieved by an inner join,
are included for retrieval in addition to the rows retrieved by the inner join.
❑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
number 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
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3. Which join is performed by the following query? (Choose the best answer.)
SELECT E.JOB_ID,J.JOB_ID FROM EMPLOYEES E
JOIN JOBS J ON (E.SALARY < J.MAX_SALARY);
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);
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
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Join a Table to Itself Using a Self-Join
7. How many rows are returned after executing the following statement? (Choose the best answer.)
SELECT * FROM REGIONS R1 JOIN REGIONS R2 ON (R1.REGION_ID=LENGTH(R2.REGION_
NAME)/2);
The REGIONS table contains the following row data:
REGION_ID REGION_NAME
1 Europe
2 Americas
3 Asia
4 Middle East and Africa
A. 2
B. 3
C . 4
D. None of the above
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
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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 $1,000. The report must contain customer first and last names, as well as 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 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
1. þ D. The queries in B and C incorrectly contain the NATURAL keyword. If this is removed,
they will join the DEPARTMENTS and EMPLOYEES tables based on the DEPARTMENT_ID
column.
ý A, B, and C are incorrect. A performs a natural join that implicitly joins the two tables on all
columns with identical names which, in this case, are DEPARTMENT_ID and MANAGER_ID.
2. þ A, B, and C. These clauses demonstrate different techniques to join the tables on both the
DEPARTMENT_ID and MANAGER_ID columns.
ý D is incorrect.
3. þ B. The join condition is an expression based on the less than inequality operator. Therefore,
this join is a nonequijoin.
ý 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 clauses was
used instead of the JOIN…ON clause.
4. þ A. This statement demonstrates the correct usage of the JOIN…USING clause.
ý 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.
5. þ B demonstrates the correct usage of the JOIN…ON clause.
ý 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.
6. þ C. The JOIN…ON clause and the other join clauses may all be used for joins between
multiple tables. The JOIN…ON and JOIN…USING clauses are better suited for N-way
table joins.
ý 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
7. þ B. Three rows are returned. For the row with a REGION_ID value of 2, the REGION_NAME
is Asia and half the length of the REGION_NAME is also 2. Therefore, this row is returned. 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.
ý A, C, and D are incorrect.
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View Data That Does Not Meet a Join Condition Using Outer Joins
8. þ A. The right outer join fetches the COUNTRIES rows that the inner join between the
LOCATIONS and COUNTRIES tables have excluded in addition to the inner join results.
The WHERE clause then restricts the results by eliminating these inner join results. This leaves
the rows from the COUNTRIES table with which no records from the LOCATIONS table
records are associated.
ý B, C, and D are incorrect.
9. þ A. This statement demonstrates the correct use of the RIGHT OUTER JOIN…ON clause.
ý 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
10. þ A. The cross join associates every four rows from the REGIONS table 25 times with the
rows from the COUNTRIES table yielding a result set that contains 100 rows.
ý 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.
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)
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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:
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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 Write Single-Row and Multiple-Row
Subqueries
✓ Two-Minute Drill
Q&A Self Test
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The 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
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 subquery
nesting permitted is unlimited in the FROM
clause and up to 255 levels in the WHERE 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.
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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);
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, country_id
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FROM employees
NATURAL JOIN departments
NATURAL JOIN locations)
GROUP BY country_id;
See Figure 8-1.
FIGURE 8-1
Different types of
subqueries
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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);
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In the preceding example, the subquery is used as an alternative to a join. The same
result could have been achieved with the following as shown in Figure 8-2.
SELECT department_name
FROM departments
JOIN employees
ON employees.department_id = departments.department_id
GROUP BY department_name;
If a subquery may return more than one row, the comparison operator must be
able to accept multiple values. These operators are IN, NOT IN, ANY, and ALL.
FIGURE 8-2
Subquery as
an alternative
to a join
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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 the large
SALES table in the demo SH schema used for recording sales transactions. Each
record captures a particular product sold to a particular customer through a particular
channel. These attributes are identified by lookup codes used as foreign keys
to dimension tables with rows that describe each product, customer, and channel. To
identify all sales of an item called “Comic Book Heroes” to customers in the city of
Oxford through Internet orders, one could run the following query:
SELECT count(quantity_sold)
FROM sales s, products p, customers c, channels ch
WHERE s.prod_id=p.prod_id
AND s.cust_id=c.cust_id
AND s.channel_id=ch.channel_id
AND p.prod_name='Comic Book Heroes'
AND c.cust_city='Oxford'
AND ch.channel_desc='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 as shown
in Figure 8-3.
SELECT count(quantity_sold)
FROM sales
WHERE prod_id IN
(SELECT prod_id
FROM products
WHERE prod_name='Comic Book Heroes')
AND cust_id IN
(SELECT cust_id
FROM customers
WHERE cust_city='Oxford')
AND channel_id IN
(SELECT channel_id
FROM channels
WHERE channel_desc='Internet');
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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 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 ('Comic Book Heroes', 'Oxford', 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
rewrite 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 the following problem based on the HR schema: Employees
are assigned to a department, and departments have a location. Each location is in
FIGURE 8-3
Subqueries
used in a star
transformation
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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, country_id
FROM employees
NATURAL JOIN departments
NATURAL JOIN locations)
GROUP BY country_id;
The subquery conceptually constructs a table with every employee’s salary and the
country in which their 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 may be used 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.
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
WHERE department_id is not null);
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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
calculate the average salary of all employees and
passes this value (a scalar quantity) to an update
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 the use of the additional WHERE clause in the subquery to
ensure that no NULL values are returned by the subquery. If this clause was absent,
no rows would be deleted.
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
FROM dual);
EXERCISE 8-2
More Complex Subqueries
In this exercise, you will write more complex 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
A subquery can be used
to select rows for insertion but not in a
VALUES clause of an INSERT statement.
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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 departments 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);
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CERTIFICATION OBJECTIVE 8.03
List the Types of Subqueries
There are three broad divisions of subquery:
■ Single-row subqueries
■ Multiple-row subqueries
■ Correlated subqueries
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 corresponds with any 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
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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 and must be used with a
comparison operator
ALL Returns rows that match all the values in a list and must be used with
a comparison operator
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
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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.
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.
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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;
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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:
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.
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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;
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 the ideal solution is to always 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.
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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.
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.
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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:
SELECT last_name
FROM employees
WHERE department_id =
(SELECT department_id
FROM departments
WHERE department_name = '&Department_name');
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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.
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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.
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 bulletproof and user friendly:
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INSIDE THE EXAM
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
single-column subquery. For the first SQL
OCA 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.
INSIDE THE EXAM
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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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.
✓
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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.
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'.
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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.
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);
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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;
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.
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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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SELF TEST ANSWERS
Define Subqueries
1. þ A, B, C, and E. Subqueries can be used at all these points.
ý D and F are incorrect. A subquery cannot be used in the GROUP BY and ORDER BY
clauses of a query.
2. þ D. Subquery nesting can be done to many levels.
ý A, B, and C are incorrect. A and C are incorrect because subqueries can be nested. B is
incorrect because the number of rows returned is not relevant to nesting subqueries, only to the
operators being used.
3. þ A. The result set of the inner query is needed before the outer query can run.
ý B, C, and D are incorrect. 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.
4. þ D. This is a correlated subquery which must be run for every row in the table.
ý A, B, and C are incorrect. 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
5. þ A and D. The two statements will deliver the same result, and neither will fail if the name
is duplicated.
ý B and C are incorrect. B is incorrect because the statements are functionally identical,
though syntactically different. C is incorrect because the comparison operator used, IN, can
handle a multiple-row subquery.
List the Types of Subqueries
6. þ A and B. A scalar subquery can be defined as a query that returns a single value.
ý C and D are incorrect. C is incorrect because a scalar subquery is the only subquery that
can be used in the SELECT LIST. D is incorrect because scalar subqueries can be correlated.
7. þ E. ALL, ANY, IN, and NOT IN are the multiple-row comparison operators.
ý A, B, C, and D are incorrect. All of these can be used.
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Write Single-Row and Multiple-Row Subqueries
8. þ C. The equality operator requires a single-row subquery, and the second subquery could
return several rows.
ý A, B, and D are incorrect. A is incorrect because the statement will fail in all circumstances
except the unlikely case where the number of employees is zero or one. B is incorrect because this is
not a problem; there need be no relationship between the source of data for the inner and outer
queries. D is incorrect because the subquery will only run once; it is not a correlated subquery.
9. þ A and B are identical.
ý C and D are incorrect. 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.
10. þ C. If a subquery returns NULL the comparison will also return NULL, meaning that no
rows will be retrieved.
ý A, B, and D are incorrect. A is incorrect because this would not cause an error. B is
incorrect because a comparison with NULL will return nothing, not everything. D is incorrect
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 > (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 subquery
should be based on EMPLOYEE_ID, not LAST_NAME.
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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 Control the Order of Rows Returned
✓ Two-Minute Drill
Q&A Self Test
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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.
CERTIFICATION OBJECTIVE 9.01
Describe the Set Operators
The set operators used in compound queries are as follows:
■ UNION Returns the combined rows from two queries, sorting them and
removing duplicates.
■ UNION ALL Returns the combined rows from two queries without sorting
or removing duplicates.
■ INTERSECT Returns only the rows that occur in both queries’ result sets,
sorting them and removing duplicates.
■ MINUS Returns only the rows in the first result set that do not appear in
the second result set, sorting them and removing duplicates.
These commands are equivalent to the standard operators used in mathematics
set theory, often depicted graphically as Venn diagrams.
Sets and Venn Diagrams
Consider groupings of living creatures, classified as follows:
■ Creatures with two legs Humans, parrots, bats
■ Creatures that can fly Parrots, bats, bees
■ Creatures with fur Bears, bats
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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 in
the figure also includes the universal set, represented by the rectangle. The universal
set is all elements that exist, including those that are not members of the defined sets.
In this case, the universal set would be defined as all living creatures, including those
that did not develop fur, two legs, or the ability to fly (such as fish).
That’s enough school math; now proceed to the implementation within SQL.
HUMANS PARROTSBEES
BATS
BEARS FISH
FIGURE 9-1
A Venn diagram,
showing three
sets and the
universal set
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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 possible future change
in 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 data types DATE and NUMBER, and those from the second
query could be TIMESTAMP and INTEGER. The result set of the compound
query will have columns with the higher level of precision: in this case, they would
be TIMESTAMP and NUMBER. Other than
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 an error even if the string variables
could be resolved to legitimate date and
numeric values.
The columns in the
queries that make up a compound query
can have different names, but the output
result set will use the names of the
columns in the rst query.
The corresponding
columns in the queries that make up a
compound query must be of the same
data type group.
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UNION, MINUS, and INTERSECT will 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 on the order in which the columns happen to appear in the select
lists) is not the order you want, it is possible to put a single ORDER BY clause at
the end of the compound query. It is not possible to use ORDER BY in any of the
queries that make up the whole compound query, as this would disrupt the sorting
that is necessary to remove duplicates.
A compound query will
by default return rows sorted across all
the columns, from left to right. The only
exception is UNION ALL, where the rows
will not be sorted. The only place where
an ORDER BY clause is permitted is at the
end of the compound query.
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, Americas,
Asia, Middle East, and Africa.
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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.
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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
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 the order they are
stored, followed by the rows from the second table in the order they are 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 in
the table OLD_DEPTS is 20 bytes long (padded
with spaces), whereas the DNAME in NEW_
DEPT, where it is a VARCHAR2, is only as long
as the name itself. The spaces give the row from
OLD_DEPT a higher sort value, even though
the date value is less.
A UNION ALL will return
rows grouped from each query in their
natural order. This behavior can be
modied by placing a single ORDER BY
clause at the end.
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The second query in Figure 9-3 removes any leading or trailing spaces from the
DNAME columns and chops off the time elements from DATED and STARTD. Two
of the rows thus become identical, 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.
FIGURE 9-2
A UNION ALL
with data type
conversions
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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.
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 be removed. Because of the sort, the rows will be in order
irrespective of the order in which the queries appear in the compound query.
FIGURE 9-3
UNION
compound
queries
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INSIDE THE EXAM
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.
INSIDE THE EXAM
FIGURE 9-4
INTERSECT and
MINUS
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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, NULL
FROM cats
UNION ALL
SELECT name, NULL, wingspan
FROM birds;
Note the use of 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 location, 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 it 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
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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';
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
self-documenting. 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.
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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 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.
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;
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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 (Figure 9-5):
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 (Figure 9-6):
SELECT department_id, NULL,sum(salary)
FROM employees
GROUP BY department_id
UNION
SELECT NULL, manager_id, sum(salary)
FROM employees
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GROUP BY manager_id
UNION ALL
SELECT NULL, NULL,sum(salary)
FROM employees;
FIGURE 9-5
Using the set
operators
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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
FIGURE 9-6
Using NULL as
a pseudocolumn
with the UNION
ALL operator
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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 in which
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:
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.
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 (Figure 9-7):
SELECT department_id dept, NULL mgr, sum(salary)
FROM employees
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FIGURE 9-7
Aliasing NULL
pseudocolumn
in set operation
using UNION ALL
GROUP BY department_id
UNION ALL
SELECT NULL, manager_id, sum(salary)
FROM employees
GROUP BY manager_id
UNION ALL
SELECT NULL, NULL, sum(salary)
FROM employees;
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3. Attempt to sort the results of the queries that subtotal by using UNION
instead of UNION ALL (Figure 9-8):
SELECT department_id dept, NULL mgr, sum(salary)
FROM employees
GROUP BY department_id
UNION
SELECT NULL, manager_id, sum(salary)
FROM employees
GROUP BY manager_id
UNION
SELECT NULL, NULL, sum(salary)
FROM employees;
FIGURE 9-8
Implicit sorting
using the UNION
operator
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FIGURE 9-9
Using NVL and
UNION to
improve query
output
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. There are certainly
ways to improve this output, which you should experiment with further. For
example, the NVL function may be used to zero missing DEPARTMENT_ID
and MANAGER_ID values, resulting in a more logically sorted output
(Figure 9-9).
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Certication Summary 415
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 all 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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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.
✓
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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.
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;
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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;
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;
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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.
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 substitution 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
1. þ D. UNION ALL returns rows in the order that they are delivered by the two queries from
which the compound query is made up.
ý A, B, and C are incorrect. INTERSECT, MINUS, and UNION all use sorting as part of
their execution.
2. þ A, B, C. INTERSECT, MINUS, and UNION all remove duplicate rows.
ý D is incorrect. UNION ALL returns all rows, duplicates included.
Use a Set Operator to Combine Multiple Queries into a Single Query
3. þ C. All set operators have equal precedence, so the precedence is determined by the
sequence in which they occur.
ý A, B, and D are incorrect. A and B are incorrect because set operators have equal
precedence—though this may change in future releases. D is incorrect because many set
operators can be used in one compound query.
4. þ A = 4; B = 8; C = 0; D = 4
ý Note that 16 is not used; that would be the result of a Cartesian product query.
5. þ C. Every query in a compound query must return the same number of columns.
ý A, B, D, and E are incorrect. A is incorrect because the columns can have different names.
B is incorrect 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 incorrect.
Control the Order of Rows Returned
6. þ B. You cannot use ORDER BY for one query of a compound query; you may only place a
single ORDER BY clause at the end.
ý A, C, D, E, and F are incorrect. All these lines are legal.
7. þ B. The rows from each query will be together, but there will be no sorting.
ý A, C, and D are incorrect. A is not possible with any syntax. C is incorrect because that
would be the result of a UNION, not a UNION ALL. D is incorrect because UNION ALL will
return the rows from each query grouped together.
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LAB ANSWER
1. To identify all employees who have not changed jobs, 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;
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9
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10
Manipulating Data
CERTIFICATION OBJECTIVES
10.01 Describe Each Data Manipulation
Language (DML) Statement
10.02 Insert Rows into a Table
10.03 Update Rows in a Table
10.04 Delete Rows from a Table
10.05 Control Transactions
✓ Two-Minute Drill
Q&A Self Test
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This 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
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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 in
several ways, but the most common method is with the INSERT statement. SQL
is a set-oriented language, so one command can affect one row or a set of rows. It
follows that one INSERT statement can insert one or more rows into one or 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 system, and Data Pump, 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 violates a constraint. So if
you attempt to insert a row into EMPLOYEES with an EMPLOYEE_ID that already
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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
two-dimensional 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 rows inserted into the target
table contain information that is much more
immediately comprehensible to the end users
than the raw 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 Data Pump. 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
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.
An INSERT command
can insert one row, with column values
specied in the command, or a set of rows
created by a SELECT statement.
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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 be disconcerting when it happens by accident. If you select the rows
to be updated with any column other than the primary key, you may update several
rows, not just one. If you omit the WHERE clause completely, you will update the
whole table—perhaps billions 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.
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
One UPDATE statement
can change rows in only one table, but it
can change any number of rows in that
table.
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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 may 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 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.
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TRUNCATE
The TRUNCATE command is not a DML command; it is a 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
the section “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 additional work behind the scenes that the user is not
aware of. DDL commands are not user transactions (though within the database,
they are in fact implemented as transactions—but developers cannot control them),
and there is no choice about whether to make them permanent or to reverse them.
Once executed, they are done. However, in comparison to DML, they are very fast.
Transactions consisting
of INSERT, UPDATE, and DELETE (or
even MERGE) commands can be made
permanent (with a COMMIT) or reversed
(with a ROLLBACK). A TRUNCATE
command, like any other DDL command,
is immediately permanent: it can never be
reversed.
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 usually 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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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-08', 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
FIGURE 10-1
Some examples
of statement
failure
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string passed is formatted as a European-style date, but the database has been set up to
expect an NLS_DATE_FORMAT of DD-MON-RR. Once the default date format is
altered in the session the statement succeeds.
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?
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 encounters 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 “Control Transactions.”
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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).
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
VALUES (10,'Great Britain');
INSERT INTO hr.regions (region_name, region_id)
VALUES ('Australasia',11);
INSERT INTO hr.regions (region_id)
VALUES(12);
INSERT INTO hr.regions
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
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was 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
self-documenting (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.
Very often, an INSERT statement will include functions to do type casting or
other editing work. Consider this statement:
INSERT INTO emp_copy (employee_id, last_name, hire_date, email, job_id)
VALUES (1000,'WATSON','03-Nov-13', 'jwatson@hr.com', 'SA_REP');
in contrast with this:
INSERT INTO emp_copy (employee_id, last_name, hire_date, email, job_id)
VALUES (1000,
upper('Watson'),
to_date('03-Nov-13','dd-mon-yy'),
lower('JWatson@hr.com'),
upper('sa_rep'));
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 emp_copy (employee_id, last_name, hire_date, email, job_id)
VALUES (1000 + 1,user,sysdate - 7,'jwatson@hr.com','SA_REP');
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
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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.
FIGURE 10-2
Using functions
with the INSERT
command
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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.
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 hr.regions;
This assumes that the REGIONS_COPY table already exists (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
two-dimensional 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
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 hr.employees e
FULL OUTER JOIN hr.departments d
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ON e.department_id = d.department_id
GROUP BY department_name
ORDER BY department_name;
Figure 10-3 shows the execution of the previous INSERT statement, followed by
a query showing the results.
FIGURE 10-3
Using subqueries
with the INSERT
command
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Any SELECT statement,
specied as a subquery, can be used as
the source of rows passed to an INSERT.
This enables insertion of many rows.
Alternatively, using the VALUES clause will
insert one row. The values can be literals or
prompted for as substitution variables.
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 join with no COALESCE
functions, which might be all an end user would do, might be very misleading. By
doing all the complex work in the INSERT statement, users can then run much
simpler queries against the denormalized and aggregated data in the summary table.
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 OCA 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 hr.employees
WHERE hire_date > sysdate - 30;
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
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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 multi-table 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');
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');
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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:
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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 10-4 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.
FIGURE 10-4
Examples of using
the UPDATE
statement
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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
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 10-4 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 4, 5, and 6) 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 7) is transferred to department 80 (the sales department), and then
the subquery in lines 4, 5, and 6 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
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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.
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.
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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;
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;
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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 FROM employees
WHERE employee_id=206;
DELETE FROM employees
WHERE last_name LIKE 'S%';
DELETE FROM employees
WHERE department_id=&Which_department;
DELETE FROM employees
WHERE 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.
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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'))));
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.
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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;
The following illustration shows the exercise as done from SQL*Plus:
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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 database’s data files. This is known as
an extent and will be empty. Then, as rows are inserted, 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 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.
TRUNCATE completely
empties the table. There is no concept of
row selection, as there is with a DELETE.
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The syntax to truncate a table couldn’t be simpler:
TRUNCATE TABLE table;
Figure 10-5 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 in the OCA 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, email, job_id)
VALUES (n.employee_id, n.last_name, n.salary, n.email, n.job_id);
FIGURE 10-5
The TRUNCATE
command in SQL
Developer
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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 to use the SAVEPOINT 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
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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.
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
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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
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
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before terminating the first. This can be done with PL/SQL (Oracle’s proprietary
third-generation 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 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,
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
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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.
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.
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 a session has inserted rows, other sessions
that SELECT from that 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.
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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. This is in accordance with the principle of
durability.
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
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SQL Developer sessions or one of each. The following table lists steps to follow in
each session.
Step In your first session In your second session
1SELECT *
FROM regions;
SELECT *
FROM regions;
Both sessions see the same data.
2INSERT INTO regions
VALUES (100,'UK');
INSERT INTO regions
VALUES (101,'GB');
3SELECT *
FROM regions;
SELECT *
FROM regions;
Both sessions see different results: the original data, plus their own change.
4COMMIT;
5SELECT *
FROM regions;
SELECT *
FROM regions;
One transaction has been published to the world; the other is still visible to only one
session.
6ROLLBACK; ROLLBACK;
7SELECT *
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.
8DELETE FROM regions
WHERE region_id=100;
INSERT INTO regions
VALUES (101,'GB');
8COMMIT;
9DELETE FROM REGIONS
WHERE region_id=101;
10 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.
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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;
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 TAB1, with
one column.
Command
Rows Visible to
the User
Rows Visible to
Others
TRUNCATE TABLE tab1; 0 0
INSERT INTO TAB1
VALUES ('one'); 1 0
SAVEPOINT first; 1 0
INSERT INTO tab1
VALUES ('two'); 2 0
SAVEPOINT second; 2 0
INSERT INTO tab1
VALUES ('three'); 3 0
ROLLBACK TO SAVEPOINT second; 2 0
ROLLBACK TO SAVEPOINT first; 1 0
COMMIT; 1 1
DELETE FROM tab1; 0 1
ROLLBACK; 1 1
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.
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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
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 Advanced: you will see the Autocommit 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.
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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;
This is what the first user will see, from a SQL*Plus prompt:
SELECT *
FROM regions;
REGION_ID REGION_NAME
--------- -------------------------
5 UK
1 Europe
2 Americas
3 Asia
4 Middle East and Africa
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;
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
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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 on transaction structure.
INSIDE THE EXAM
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. The
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.
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(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.
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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 add 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 made up of 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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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.
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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;
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:
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.
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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
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.
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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 while keeping its structure intact?
(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.
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 the best answer.)
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 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
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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';
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:
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';
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7. This command will hang. In the first session, release the locks by issuing a commit:
COMMIT;
8. 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');
9. In the first session, attempt to truncate the CUSTOMERS table:
TRUNCATE TABLE customers;
10. 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;
11. 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
1. þ B, C, D, and F. These are the DML commands: they can all be rolled back.
ý A and E are incorrect. COMMIT terminates a transaction, which can then never be rolled
back. TRUNCATE is a DDL command and includes a built-in COMMIT.
2. þ B. Assuming no constraint violations, the primary key can be updated like any other column.
ý A, C, and D are incorrect. A is incorrect because there is no restriction on updating
primary keys (other than constraints). C is incorrect because there is no need to do it in such a
complex manner. D is incorrect because the UPDATE will apply its own lock: you do not have
to lock the row first.
3. þ C. This is the expected behavior: the statement is rolled back, and the rest of the transaction
remains uncommitted.
ý A, B, and D are incorrect. A is incorrect because, while this behavior is in fact configurable,
it is not enabled by default. B is incorrect because, while this is in fact possible in the event of
space errors, it is not enabled by default. D is incorrect because only the one statement will be
rolled back, not the whole transaction.
Insert Rows into a Table
4. þ D. A, B, and C will all succeed, even though B will force the database to do some
automatic type casting.
ý A, B, C, and E are incorrect. A, B, and C are each valid, but D groups them into the
appropriate single valid response. E is incorrect because A, B, and C will all succeed.
5. þ D. The syntax is wrong: use either the VALUES keyword or a subquery, but not both.
Remove the VALUES keyword, and it will run. C3 and C4 would be populated with NULLs.
ý A, B, C, and E are incorrect. A is incorrect because there is no need to provide values for
columns not listed. B and C are incorrect 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 incorrect because the statement is not syntactically correct.
6. þ C. The statement is syntactically correct, and the use of “MAX(REGION_ID) + 1”
guarantees generating a unique number for the primary key column.
ý A, B, and D are incorrect. A is incorrect because the function will generate a unique
value for the primary key. B is incorrect 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 incorrect because if there is a
third column, it will be populated with a NULL value.
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Update Rows in a Table
7. þ A. This is the simplest (and therefore the best) way.
ý B, C, and D are incorrect. All these will work, but they are all needlessly complicated: no
programmer should use unnecessary statements.
8. þ D. Any arithmetic operation on a NULL returns a NULL, but all other rows will be updated.
ý A, B, and C are incorrect. A and B are incorrect because the lack of a WHERE clause means
that every row will be processed. C is incorrect because trying to do arithmetic against a NULL is
not an error (though it isn’t very useful, either).
Delete Rows from a Table
9. þ C. An UPDATE, without a WHERE clause, is the only way.
ý A, B, and D are incorrect. A is incorrect because there is no such syntax: a DELETE
affects the whole row. B is incorrect because there is no such syntax: a TRUNCATE affects the
whole table. D is incorrect 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.
10. þ A and C. The TRUNCATE will be faster, but the DELETE will get there too.
ý B and D are incorrect. B is incorrect because this will remove the table as well as the
rows within it. D is incorrect because the rows will still be there—even though they are populated
with NULLs.
Control Transactions
11. þ B. The principle of isolation means that only JOHN can see his uncommitted transaction.
ý A, C, and D are incorrect. A is incorrect because transaction isolation means that no other
session will be able to see the changes. C and D are incorrect because a committed transaction
can never be rolled back.
12. þ C. Transaction isolation means that no other session will be able to see the changes until
they are committed.
ý A, B, and D are incorrect. A is incorrect because locking is not relevant; writers do not
block readers. B is incorrect 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 incorrect because savepoints are only markers in a transaction; they do not affect publishing
changes to other sessions.
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13. þ A, C, and F. COMMIT and ROLLBACK are the commands to terminate a transaction
explicitly; TRUNCATE will do it implicitly.
ý B, D, and E are incorrect. B is incorrect because DELETE is a DML command that can be
executed within a transaction. D and E are incorrect because creating savepoints and rolling
back to them leaves the transaction in progress.
LAB ANSWER
Figure 10-6 shows the first five steps of the exercise.
Figure 10-7 shows the final seven steps of the exercise, as seen from the point of view of the
first session.
FIGURE 10-6
Steps 1 through
5 of the lab
exercise
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FIGURE 10-7
The final steps of
the lab exercise,
including the
TRUNCATE
error
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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 Create a Simple Table
11.05 Explain How Constraints Are Created
at the Time of Table Creation
✓ Two-Minute Drill
Q&A Self Test
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There are several types of data objects in a database that can be accessed using SQL. The
most commonly used type of object is the table. Tables come in various forms, but SQL
is independent of their forms. 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 SQL statement will address only the table
with which the other objects are associated. This chapter details table creation.
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 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:
SELECT object_type, count(object_type)
FROM dba_objects
GROUP BY object_type
ORDER BY object_type;
OBJECT_TYPE COUNT(OBJECT_TYPE)
------------------- ------------------
CLUSTER 10
CONSUMER GROUP 12
CONTEXT 6
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DIMENSION 5
DIRECTORY 9
EDITION 1
EVALUATION CONTEXT 13
FUNCTION 286
INDEX 3023
INDEX PARTITION 342
INDEXTYPE 12
JAVA CLASS 22018
JAVA DATA 322
JAVA RESOURCE 820
JOB 11
JOB CLASS 11
LIBRARY 177
LOB 769
LOB PARTITION 7
MATERIALIZED VIEW 3
OPERATOR 60
PACKAGE 1240
PACKAGE BODY 1178
PROCEDURE 118
PROGRAM 17
QUEUE 37
RESOURCE PLAN 7
RULE 1
RULE SET 21
SCHEDULE 2
SEQUENCE 204
SYNONYM 26493
TABLE 2464
TABLE PARTITION 199
TRIGGER 413
TYPE 2630
TYPE BODY 231
UNDEFINED 6
VIEW 4669
WINDOW 9
WINDOW GROUP 4
XML SCHEMA 93
42 rows selected.
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
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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.
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 out of scope of the SQL exam. Briefly, a
view is a stored SELECT statement that can be addressed as though it were a table.
It is nothing more than a named 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.
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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
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 its 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. These users store code and data required by various database options.
For example, the user MDSYS stores the objects used by 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.
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■ 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.
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,
Object names must be no
more than 30 characters. The characters
can be letters, digits, underscore, dollar,
or hash.
FIGURE 11-1
Using double
quotes to use
nonstandard
names
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remember that object names are always case sensitive. In this example, the two
tables are completely different:
CREATE TABLE lower (c1 date);
Table created.
CREATE TABLE "lower" (col1 varchar2(2));
Table created.
SELECT table_name
FROM user_tables
WHERE lower(table_name) = 'lower';
TABLE_NAME
------------------------------
lower
LOWER
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.
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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.
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
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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).
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;
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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. In 12c
database, a new system parameter, MAX_STRING_SIZE, allows string data types to
be much larger than in previous versions when it is changed from its default value of
STANDARD to EXTENDED.
The following are the data types for alphanumeric data:
■ VARCHAR2 Variable-length character data, from 1 byte to 4000 bytes
if MAX_STRING_SIZE=STANDARD or 32767 bytes if MAX_STRING_
SIZE=EXTENDED. The data is stored in the database character set.
■ NVARCHAR2 Like VARCHAR2, but the data is stored in the alternative
national language character set, one of the permitted Unicode character sets.
■ CHAR Fixed-length character data, from 1 byte to 2000 bytes, in the
database character set. If the data is not the length of the column, then it will
be padded with spaces.
For ISO/ANSI compliance, you can specify a VARCHAR data type, but any
columns of this type will be automatically converted to VARCHAR2.
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The following is the data type for binary data:
■ R AW Variable-length binary data, from 1 byte to 4000 bytes if
MAX_STRING_SIZE=STANDARD or 32767 bytes if MAX_STRING_
SIZE=EXTENDED. 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.
The following are the data types for numeric data, all variable length:
■ NUMBER 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.
■ FLOAT 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.
■ INTEGER Equivalent to NUMBER, with scale zero.
The following are the data types for date and time data, all fixed length:
■ DATE 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.
■ TIMESTAMP 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 WITH TIMEZONE 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 LOCAL 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.
■ INTERVAL YEAR TO MONTH Used for recording a period in years and
months between two DATEs or TIMESTAMPs.
■ INTERVAL DAY TO SECOND Used for recording a period in days and
seconds between two DATEs or TIMESTAMPs.
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The following are the large object data types:
■ CLOB Character data stored in the database character set, size effectively
unlimited: (4GB –1) multiplied by the database block size.
■ NCLOB Like CLOB, but the data is stored in the alternative national
language character set, one of the permitted Unicode character sets.
■ BLOB Like CLOB, but binary data that will not undergo character set
conversion by Oracle Net.
■ BFILE A locator pointing to a file stored on the operating system of the
database server. The size of the files is limited to 4GB.
■ LONG 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 RAW Like LONG, but binary data that will not be converted by
Oracle Net. Any LONG RAW columns should be converted to BLOBs.
The following is the ROWID data type:
■ ROWID 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.
All examinees will be
expected to know about these data types:
VARCHAR2, CHAR, NUMBER, DATE,
TIMESTAMP, INTERVAL, RAW, LONG,
LONG RAW, CLOB, BLOB, BFILE, and
ROWID. Detailed knowledge will also
be needed for VARCHAR2, NUMBER,
and DATE.
The VARCHAR2 data type must be qualified with a number indicating the
maximum length of the column. If a value is inserted into the column that is less
than this, it is not a problem: the value will only take up as much space as it needs.
If the value is longer than this maximum, the INSERT will fail with an error. If the
value is updated to a longer or shorter value, the length of the column (and therefore
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the row itself) will change accordingly. If it is not entered 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 significant decimal digits, where
the most significant digit is the left-most nonzero digit, and the least significant
digit is the right-most known digit in the number. The scale is the number of digits
from the decimal point to the least significant digit. A positive scale is the number
of significant digits to the right of the decimal point to (and including) the least
significant digit. A negative scale is the number of significant digits to the left of the
decimal point to (but not including) the least significant digit.
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 day 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
FIGURE 11-2
Use of type
casting functions
and automatic
type casting
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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 '23-Nov-13',
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.
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.
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,
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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.
■ Index clusters Can denormalize tables in parent-child relationships so that
related rows from different tables are stored together.
■ Hash clusters Force a random distribution of rows, which will break down
any ordering based on the entry sequence.
■ Partitioned tables Store rows in separate physical structures, the partitions,
allocating rows according to the value of a column.
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 Specications
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.
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
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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 10 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:
INSERT INTO scott.emp (empno, ename, sal)
VALUES (1000, 'John', 1000.789);
1 row created.
SELECT *
FROM scott.emp;
EMPNO ENAME HIREDATE SAL COMM
---------- ---------- --------- ---------- ----------
1000 John 19-NOV-13 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.
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
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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 the section titled “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:
SELECT *
FROM emp_dept
WHERE rownum < 10;
ENAME DNAME SERVICE
--------------- --------------- ----------
Gietz Accounting 4203
De Haan Executive 4713
Kochhar Executive 3001
Chen Finance 2994
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Faviet Finance 4133
Popp Finance 2194
Sciarra Finance 2992
Urman Finance 2834
Austin IT 3089
9 rows selected.
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 Denitions 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;
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■ Renaming columns:
ALTER TABLE emp
RENAME COLUMN hiredate TO recruited;
■ Marking the table as read-only:
ALTER TABLE emp
READ ONLY;
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.
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
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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.
Oracle 12c includes a recycle bin option, which is enabled by default. This
allows any dropped table to be restored unless it was dropped with the
PURGE option or if the recycle bin option has been disabled.
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 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.
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8. Connect to the database as HR with SQL*Plus.
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 be 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.
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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.
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
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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
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
A primary key constraint
is a unique constraint combined with
a not null constraint.
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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 employees, where e-mail address, Social Security number, and
employee number should all be 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
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 insert 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 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 foreign key constraint in
a child table must reference the columns
of either a unique constraint or a primary
key constraint in the parent table.
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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 of the MANAGER_NUMBER column must refer back to
a valid EMPLOYEE_NUMBER. If an employee is his own manager, then the row
would refer to itself.
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.
Dening 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.
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.
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Consider these two table creation statements (to which line numbers have
been added):
1 CREATE TABLE dept(
2 deptno NUMBER(2,0) CONSTRAINT dept_deptno_pk PRIMARY KEY
3 CONSTRAINT dept_deptno_ck CHECK (deptno BETWEEN 10 AND 90),
4 dname VARCHAR2(20) CONSTRAINT dept_dname_nn NOT NULL);
5 CREATE TABLE emp(
6 empno NUMBER(4,0) CONSTRAINT emp_empno_pk PRIMARY KEY,
7 ename VARCHAR2(20) CONSTRAINT emp_ename_nn NOT NULL,
8 mgr NUMBER(4,0) CONSTRAINT emp_mgr_fk REFERENCES emp (empno),
9 dob DATE,
10 hiredate DATE,
11 deptno NUMBER(2,0) CONSTRAINT emp_deptno_fk REFERENCES
dept(deptno)
12 ON DELETE SET NULL,
13 email VARCHAR2(30) CONSTRAINT emp_email_uk UNIQUE,
14 CONSTRAINT emp_hiredate_ck CHECK(hiredate>= dob + 365*16),
15 CONSTRAINT emp_email_ck
16 CHECK ((instr(email,'@') > 0) AND (instr(email,'.') > 0)));
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 “ORA-00054: 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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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 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 birthdate, 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).
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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 birthdate. 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
does make the code less self-documenting, as the complete table definition will then
be spread over several statements rather than being in one.
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 12c documentation
set devotes 86 pages to it, with another
104 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.
INSIDE THE EXAM
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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');
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INSERT INTO emp
VALUES (9999,'New emp',99);
TRUNCATE TABLE dept;
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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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 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 in line with its column or at the table level after
the columns.
❑Table-level constraints can be more complex than those defined in line.
❑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.
✓
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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.
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.
D. Heap tables cannot be indexed.
E. Tables in a heap do not have a primary key.
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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. R AW
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.)
A. There will be an error because of the impossible condition.
B. No table will be created because the condition returns FALSE.
C . The table T1 will be created but no rows inserted because the condition returns FALSE.
D. The table T1 will be created and every row in REGIONS inserted because the condition
returns a NULL as a row filter.
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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.
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.
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SELF TEST ANSWERS
Categorize the Main Database Objects
1. þ D. The schema will default to the current user.
ý A, B, C, and E are incorrect. A is incorrect because all tables must be in a schema. B is
incorrect because the creation will succeed. C is incorrect because the SYS schema is not a
default schema. E is incorrect because while there is a notional user PUBLIC, he does not have
a schema at all.
2. þ A. Indexes have their own namespace.
ý B, C, D, and E are incorrect. Stored procedures, synonyms, tables, and views exist in the
same namespace.
3. þ D and E. D violates the rule that a table name must begin with a letter, and E violates
the rule that a table name cannot be a reserved word. Both rules can be bypassed by using
double quotes.
ý A, B, and C are incorrect. These are incorrect because all will succeed (though A and B
are not exactly sensible).
Review the Table Structure
4. þ A and C. A heap is a table of variable length rows in random order.
ý B, D, and E are incorrect. B is incorrect because a heap table can only be one table. D and
E are incorrect because a heap table can (and usually will) have indexes and a primary key.
List the Data Types That Are Available for Columns
5. þ A, C, D, E, and F. All these are variable length data types.
ý B is incorrect. CHAR columns are fixed length.
6. þ D. The number will be rounded to 1 digit, and the string will be cast as a date.
ý A, B, and C are incorrect. Automatic rounding and type casting will correct the “errors,”
though ideally they would not occur.
7. þ D. STRING is not an internal data type.
ý A, B, and C are incorrect. CHAR, FLOAT, and INTEGER are all internal data types,
though not as widely used as some others.
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Create a Simple Table
8. þ C. The condition applies only to the rows selected for insert, not to the table creation.
ý A, B, and D are incorrect. A is incorrect because the statement is syntactically correct.
B is incorrect because the condition does not apply to the DDL, only to the DML. D is incorrect
because the condition will exclude all rows from selection.
9. þ D. Check and not null constraints are not dependent on any structures other than the
table to which they apply and so can safely be copied to a new table.
ý A, B, C, and E are incorrect. A is incorrect because not null and check constraint will be
applied to the new table. B, C, and E are incorrect 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
10. þ C and D. Unique and primary key constraints are enforced with indexes.
ý A and B are incorrect. Check and not null constraints do not rely on indexes.
11. þ C. A constraint violation will force a rollback of the current statement but nothing else.
ý A, B, D, and E are incorrect. A is incorrect because all statements that have succeeded
remain intact. B and D are incorrect because there is no commit of anything until it is specifically
requested. E is incorrect 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 2000000 AND 3999999),
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))
);
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512 Chapter 11: Using DDL Statements to Create and Manage Tables
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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.
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A
About the CD-ROM
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514 Appendix: About the CD-ROM
The CD-ROM included with this book comes complete with Total Tester customizable
practice exam software with more than 150 practice exam questions and the electronic
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12-Appendix.indd 514 2/26/14 12:06 PM
Technical Support 515
OracCertPrs5.5 / OCA Oracle Database 12c / Roopesh Ramklass / 182028-0 / Appendix
select the Total Tester program. Select Remove, and Windows will completely
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12-Appendix.indd 515 2/26/14 12:06 PM
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OracCertPrs5.5 / OCA Oracle Database 12c / Roopesh Ramklass / 182028-0 / Glossary / Blind Folio 517
Glossary
ACID Atomicity, consistency, isolation, and durability. Four characteristics that
a relational database must be able to maintain for transactions.
AL16UTF16 A Unicode fixed-width, 2-byte character set, commonly specified for
the NLS character set used for NVARCHAR2, NCHAR, and NCLOB data types.
alias (1) In Oracle SQL, an alternative name for a table, view, or set of data
used in a query. (2) 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.
API Application Programming Interface. A defined method for manipulating
data, typically implemented as a set of PL/SQL procedures in a package.
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518 Glossary
ASCII American Standard Code for Information Interchange. A standard
(with many variations) for coding letters and other characters as bytes.
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.
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.
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.
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.
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Glossary 519
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 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.
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.
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 Xeon or an Oracle SPARC T6.
data blocks The units into which data files are formatted.
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520 Glossary
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 file The disk-based structure for storing data.
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.
DBA Database Administrator. The person responsible for creating and managing
Oracle databases—this could be you.
DBA role A preseeded role provided for backward compatibility that includes
all the privileges needed to manage a database, except those 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.
DDL Data Definition Language. The subset of SQL commands that change
object definitions within the data dictionary: CREATE, ALTER, DROP, and
TRUNCATE.
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Glossary 521
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.
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.
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: INSERT, UPDATE, DELETE, and MERGE.
DNS Domain Name Service. The TCP mechanism for resolving network names
into IP addresses.
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 opposed to OLTP work.
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 that can be
used by application software and by shell scripts.
equijoin A join condition using an equality operator.
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522 Glossary
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.
GMT Greenwich Mean Time. Now referred to as UTC, this is the time zone of
the meridian through Greenwich Observatory in London.
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.
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.
inconsistent backup A backup made while the database was open.
incremental backup A backup containing only blocks that have been changed
since the last backup was made.
INITCAP A function that accepts a string of characters and returns each word in
title case.
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Glossary 523
inner join When equijoins and nonequijoins are performed, rows from the source
and target tables are matched. These are referred to as inner joins.
instance recovery The automatic repair of damage caused by a disorderly
shutdown of the database.
INSTR A function that returns the positional location of the nth occurrence of
a specified string of characters in a source string.
I/O Input/output. The activity of reading from or writing to disks—often the
slowest point of a data processing operation.
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.
JEE Java 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.
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524 Glossary
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.
LAST_DAY A function used to obtain the last day in a month given any valid
date item.
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.
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.
MOD 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.
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Glossary 525
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.
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.
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.
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526 Glossary
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.
OCM Oracle Certified Master. An advanced qualification you may ultimately wish
to pursue.
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.
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.
ORACLE_BASE The root directory into which Oracle products are installed.
ORACLE_HOME The root directory of any one Oracle product, typically for
a given release. This value is located beneath ORACLE_BASE.
Oracle Net Oracle’s proprietary communications protocol, layered on top of an
industry standard protocol.
OS Operating system. Typically, in the Oracle environment, this will be a version
of Unix (perhaps Linux) or Microsoft Windows.
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Glossary 527
outer join A join performed when rows, which are not retrieved by an inner
join, are included for retrieval.
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.
PMON Process monitor. The background process responsible for monitoring the
state of user sessions against an instance.
primary key The column (or combination of columns) whose value(s) can be
used to uniquely 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.
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.
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.
raw device An unformatted disk or disk partition.
RDBMS Relational Database Management System. Often used interchangeably
with DBMS.
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528 Glossary
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.
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.
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 A database object, within a schema, that can generate consecutive
numbers.
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.
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Glossary 529
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
uniquely identify a session logged on to an Oracle instance.
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.
table A logical two-dimensional data storage structure, consisting of rows and
columns.
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530 Glossary
tablespace The logical structure that abstracts logical data storage in tables from
physical data storage in data files.
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).
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.
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Glossary 531
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 Window System 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.
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13-Gloss.indd 532 2/26/14 5:04 PM
INDEX
A.O. in date
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rmat m
asks,
243
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44
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43
2
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451-453
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243
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494
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179
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211
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76
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algebra,
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s,
79-81
purpose,
66
ca
hl
cs,
32
4
ALL
keyword
AVG,
286
COU T, 280,
28
4
IN
S
ERT
, 438
MAX
and
MIN
,
288
s
uh
quc
ri
es
, 370, 377, 382-383
SU
M,
285
ALL_
OBJECTS
view,
47
8
ALTER
T
ABLE
co
mm
an
d,
49
2-493
altering table definitions, 492-493
alte
rn
a
ti
ve
quote operator, 8
4-
87
AM
in
date
fo
rmat m
asks,
24
4
a
mbi
gu
ous
column names, 323-325
American atio
nal
Standards Institute (ANSI)
jo
in
s,
322-323
SQ
L st:
md:mJs,
27
:impers:ind
(&)
s
ul:6ticuti
on
define and
verify,
157
-
162
varinblc~.
Sec
s
uh
sri r
uri
on
va
riabl
es
AND operator
overvi
ew,
134-
13
6
precedence,
14
0
rani:,rcs,
124
an
oma
li
es,
insert update and
de
letion,
15
ANSI (American National Standards
lnscitu
te)
jo
in
s,
322-323
SQL
standa
rd
s,
27
A Y operators, 370, 377, 3 2-383
a
pp
l
ica
ti
on
ti
c~
in
web
app
li
cations, 6
arirhme
ti
c and
ar
ithmetic operators
d;ites, 2
11
-2
12
wi
th
ULL,
90
o
verYiew,
76-78
precedence, 139
row
updat
es,
44
2
AS
keyword,
81-82
ascend
in
g sort
s,
144-146
a:.
te
ri
sks
( *)
co
lumn
s,
65
COUN
T,
280
evaluation order,
76
precedence,
14
0
atomicity
in
ACID test, 451-
45
2
attention to detail,
71
nnributes
entiti
es
, 12-13
gro
uping,
29
4
S 3 4 OCA Oracle Database I 2
c:
SQL Fundamentals I
Ex
am
Guide
AUTOCOMMIT
feature,
459
aucomatic
type
casting, 487-4
88
AVG
func
li
o
n,
280-281. 286-287
B
.C.
in
date
form
at
masks, 243
Basic
Co
nn
ection T
ype
secrin
g,
40-41
BC
in
dat
e
fo
nn
at
masks, 243
BET\'VEE
operato
r
non
eq
uijoins,
337
pr
ece
den
ce
,
140
mn
ge
compa
risons, 124-1
25
BFI
LE
d
aca
type, 4
86
Binary Large
Objects
(B
LO B
s),
59,
4
83
binding,
nintim
c,
15
l
black boxes, functions as, 1
77
bl
ank
spaces vs.
NULL,
87
blind q ueries,
65
BLO Bs (Binary Large
Objcc
c
s).
59.
4
86
Boolean
operato
rs
AN
D, 134-136
NOT,
137-1
38
O
R,
13
6-
137
ove
rview, 133-134
r.inges, 124
roth
o
pti
on in
trimmin
g,
1
95
oo
und
ary
\·
alues for ranges, 124
bracket:.
delim1cers,
86
ro
und.
See
r
ou
nd
bmckcts
()
C
lan
guage, 10
C.'lpitn
lized case, 1
86-
1
87
ca
r dealership scenari
o,
11
referential inccgril)' a
nd
foreign
key~.
I 8-20
relationships, 1
3-
17
cmd
inality
of
tuples, 13
carri
age
recums, 70
Ca
rt
esian
pr
odl
1ccs, 320-321, 349-353
case
co
nv
ersion functions.
See
character
conve
rsio
ns
CASE
expressions, 262-266
CAS
E function, 181
case
sens
itivity
c
hara
c
ter
-based
co
nditi
ons
,
114
passwords,
34
SELECT Srnremenr, 69
CASE
(Computer
Aided
So
ftware Engineering)
too
ls,
12
CC
in dace for
mat
masks
<k'l>c
ript
ion, 210, 243
precisio
n,
22
1
cc
nturi
e in dace format mask:.
description,
210,
243
pr
ecisio
n,
22
1
C
HAR
d
f't
ta
rype,
59,
484
ch
arac
t
er
co
nv
ersions, 181, 183,
23
7
ch
arac
ter$ co daces, 247-248
c
hara
cters Lo
numbe
rs, 248-249
daces
co
chf'lraccers, 2
41
-2
46
exe
rcise, 188-1
89
implicit,
235
-236
I
NITCAP,
1
86-
1
87
LO
WER.
1
83
-1
85
numbers co c
ha
racters, 238-2
41
UPPER,
18
5-1
86
ch
arac
ter function
s,
177-1
78,
1
89
<...'O
'CAT,
1
90-
191
exercise, 202-203
1
"STR.
196-198
LENGTll
,
191
-192
LP
AD
a
nd
RPAD, 192-194
REPLA
CE,
1
99-201
SU
BS
TR
, 198-199
TR
IM, 194-196
ch
arac
ter sets
in
inequ
alicy rests, 1
22
ch
arac
ters
BETWEEN
ope
rat
or, 124
co
n
ca
tena
tion, 82-83
co
nversions.
See
charac
rer
conversions
COUN
T,
280
functions.
See
character
functions
in
cq
ua
li
cy,
1
22
Joining, 190-191
lil
cra
ls,
ll
4, 190-191,
244
pntcern
compar
isons,
127
- 1
28
qu
ery
co
nditi
ons,
11
4-116
check consrraincs, 500
child-parenc relacionships
fo
reign key constn'lincs, 499
oucer
jo
ins, 342
clienr-server model,
4-5
clie
nt
ciers,
4-6
client t
oo
ls,
29-30
SQ
L Developer, 37-41
SQ
L• Plus,
30-36
CLOB darn type, 486
cloud computing, 8
-9
Cloud
Co
ntrol, 8
clusters, 6, 4
89
columns
aliases,
79
-8
1
ambiguous names, 323-325
averages, 286-287
combining,
67-68
data types, 59,
484-488
dropping, 4
93
grouping
by,
296-298
joining, 190-191
l
og
ical models,
13
spec
if
ications,
489-490
substituting names, 154-155
tables, 57
totals, 285-286
combining
columns,
67-68
que
ri
es, 4
00
comma
s(,)
dat
e
fo
nna
t masks, 244
IN operato
r,
l25
number fonnat masks, 240
COMM
IT
command, 45 I, 453-457
comparison operators, 119
Boolean.
See
Bo
olean ope
rat
ors
equality and inequality,
11
9-123
ULL. 1
32-
133
pattern, 1
27-
132
range
s,
124-125
set comparisons, 1
25-
127
subquerics,
376
\XIHEREclause, 113
compmisons, subqueries
for,
369-371
complex subqueries, 374-375
composite incqual iry operators,
1.2
l
composite keys, 4
97
co
mp
osite sorting, 146-14 7
compound quer
ies
with sets, 398
Index
S3S
Co
mputer Aided Software Engineering (CASE) cools, 12
CONCA
T function, 178, 1
90-19
1
concatenation
characrers and strings, 82-83, 1
89-19
1
literals, 83
conditional exp
re
ss
ion
s,
250
certifi
ca
tion summa
ry,
268
conditional functions, 260-267
general functions.
See
genernl functions
lab answer, 275-276
lab question, 273
nested functions, 250-252
self test, 271-273
self
test
an
swers, 274-275
two-minute drill, 269-270
co
nditions, 260
CASE, 262-266
DECODE, 260-262, 266-267
WHERE
clause, 113-119
co
nn
ec
ti
on pooling mode
l,
7
Co
nn
ection Type setcing, 40
co
nn
ec
ti
ons
database, 34
-36
details, 35
SQL
Developer, 39-41
co
nsistency in
ACID
test, 452
constraints
check
,500
creating, 4
96-497
defining, 500-503
errors
fr
om, 432
cables, 425
un
ique,
497-49
8
working with, 504-505
concrol
taremenc
s
for
tran
sac
ti
on
s,
455-456
conversion functions, 181, L83, 234-235
certification summa
ry
, 268
characters to dates, 247-248
5 l 6
OCA
Or
acle Database I
2c:
SQL
Fund
amentals I
Exam
Gu
id
e
conversion functions (cont.)
characters
to
numbers, 248-249
daces co
cha
racters, 241-
246
exercise,
188-189
explicic,
237
implicit, 235-236
INlT
CAP, 186-187
ln
b ;inswer,
275-276
l
ab
qu
estion,
273
L
OW
ER,
183-185
numbers
co
characters,
238-2
41
self test,
271
-
273
self test answers,
27
4-275
rwo-minure drill, 269-270
UPPER, 185-1
86
working with,
238
correlated subqu
er
ies,
377-378
COUNT
function,
280,
284-285
CREATE
SCHEMA
comma
nd,
42
CREATE
TAB
LE command,
489,
49
1
CREA
TE
USER
com
mand,
42
cross joins,
320--321
,
350--353
crow's foot
notation,
13,
18
currency number fo
rmat
masks,
240
curr
ent
system date,
21
1
D
dnte format mnsks,
242
number
fonnat
masks. 240
D
ata
Co
nt
ro
l Language
(DCL)
commands,
28
Dara Definiti
on
Language
CDD
L) scaremencs,
44
9
certification su
mm
;iry,
505
cons
traint
s.
See
constraints
dma
dictiom1ry,
479
dropping tables, 493-494
ln
b answer,
51
1
-5
12
l
ab
question,
509
lisr, 27-28
self test,
507-509
self cest
an
swers, 51
0--5
11
table creacion, 488-492, 494-4
96
rable definicions, 492-494
TRUN
CA
TE,
429, 449,
493-494
1:wo-minuce drill,
506
darn dictionary
description,
57,
72
extents
in,
483
cable locaci
on
in,
44
9
updating, 4
79
data
in
tab
les,
57
Dara Manipul
at
ion Language (DML) scaccmencs, 27-28
ce
rtifi
cation
summary, 4
62
DELETE,
427-428
fai
lures,
429-433
inse
rtin
g rows inco cables, 425-426, 433-440
lab
nnswer,
473
-
474
lab
quescion, 469-470
merging rows inco tables,
428,
4
50-45
I
overview,
424-425
rows
for,
373-374
self test, 465-466
self cesr answers, 4 71-4
73
transactions, 451-456
TRUNCATE,
429
two-minute drill,
463-464
upd
at
ing rows
in
tabl
es,
426-427,
441-445
data
m
ode
ling,
12
Data Pump tool,
425
dnca tier
in
web applications, 6
data
types
attr
ibutes,
13
ove
rview, 484-488
sets,
398
Database
Con
fi
guration Assistant
(DBCA),
42.
46
d<1tabase
connectio
ns
SQL
Developer,
39-41
SQL
*Plus, 34-36
Database Express, 8
database
sess
i
on~
in
three
ri
er
env
ironment, 6-7
database tiers, 5
database transactions, 451-453
control
statements, 455-456
isolation, 4
61
starts
and
ends, 453-455
with
TRUNCATE,
429
date
functions,
179,
2
13
ADD
_
MO
TH
S.
215-216
exercise, 216-21 7
LAST
_D
AY.
219-220
MO
THS_BETWEEN,
213
-
215
EXT
_DAY,
217-218
R
OUND,
220-221
SYSDATE,
21
1
TRUNC, 222-223
dnce lirerals
joining, 190
rosicion in search
~c
r
i
n
gs,
l
96
st
ring r
ep
laceme
nt
, 199
ul
~
trin
g
extraction,
198
d
f!
c
es
find
DA
TE
darn rype,
59,
209,
485, 4
87
arichmecic operations,
77,
21
1-212
BETWEEN operato
r,
124-
12
5
character conversions, 241
-2
48
COUNT,
280
dntabase storage, 200-211
explicit con
ve
rsions,
237
f
un
ction~.
See
dale fw1ctions
implic
it
con
vers
ions, 235-236
inequality, 1
22
length, 191
literals.
See
date literals
padding,
192-193
query conditions, 17-19
scring rc
pl
acemcl\l,
199
DAY
in
fo
rmat masks
description,
210,
243
precision,
22
1
day
of
the
week function, 2
17
-2
18
days in date format
masks
, 210, 242-243
DBA_OBJE
CTS
vi
ew,
477-478
DBCA (Onrnbnse
Co
nf
igurmion Assisrnnt), 4
2,
46
OC
L (Data Control Language) co
mm
a
nd
s,
28
DD
in dale formnt masks
description, 210, 2
42
precision,
22
1
DD-MON-RR
fo
rm
al mask, 210
DOD
in
dace
for
mat
m11
ks, 242
DDL.
See
Dac.1
Definition Language (DDL) statements
DDL
wb,
494-4
95
decimal
poi
nt
s in number
fo
nn
ar
masks,
240
DECODE function, I I
overview,
2©-262
working with, 266-267
DEFAULT clause
column specifications, 489-490
nor null constminr ,
498
DEFINEcommand, 1
51,
1
5~
1
61
defining
cons
c
rainc~,
500-503
functions,
17
6-
1
80
deletion :md DELETE commnnd
anoma
li
es,
15
overvi
ew,
427-428
rows
from
rabies, 446-450
demonstration schemas
crearing, 46-47
HR
. 43-46, 60-62
OE, 43-46
u
ser.;,
42-43
descendi
ng
sores, 144-1
46
DESC
RI
BE
co
mmand
overvi
ew,
57-60
t
ab
le definitions,
234
wo
rking with, (i()-62
detail
anention
t
o,
71
co
nn
ect
ions,
35
relation hip, 18
Index
S37
development tools and langu;iges overview, 10
dic
c
ionari~,
daca
description, 57,
72
CXlC
nt
' in,
483
rable locmion m,
44
9
updating, 4
79
difference between dates, 2
11
dirty reads,
453
DISPLAY
env
ironment vnriable, 38
DISTINCT
keyw
ord
AV0,
286
COUN
T,
28
4
group
fu
ncrions, 283-284
MAX
and MI
N,
288
purpose, 66-67
SUM,
285
WHERE clause,
111
di\'ision operntions,
75
evaluati
on
o
rd
er,
76,
1
40
MOD, 1
79,
206-208
with lU
LL
,
90
DM
L.
See
Da
ra Manipulation Language (DML)
stacements
do
ll
ar signs ($) in number
fo
rmac
m11
sks,
240
5 l 8
OCA
Oracle Database I
2c:
SQL
Fund
amentals I
Exam
Guide
do c norn
ri
on
attributes, 18
col
umn
na
me
s, 323-325
do
ub
le ampersand
(&&)substit
ut ion,
152
-
154
do uble pipe
sy
mb
o
ls
( I I )
concate
n
ation,
82-83,
18
9
pre
ce
den
ce
,
140
do uble quo
tati
on marks (")
aliases, 80-81
dare formar
ma
sks,
244
nam
es,
480
DROP
TAB
LE
com
m
and,
4
93
dro
ppin
g
co
lumn
s,
493
tables, 493-494
dup
licate
inf
orma
ti
on, 1
5-l
7
dur
abi
lity
in
ACID
tes
t,
453
DY
in dare formar masks, 242
ec
ho
co
mmand
, 31
Edit me
nu
in
SQ
L Develope
r,
39
EE
EE in
nu
mber form
at
masks, 240
eleme
nt
s in sets,
397
E
LSE
stateme
nt
s, 264-265
END
srareme
nr
s, 263-264
end-user sessio
ns
in
th
ree
ti
er e
nvir
o
nment,
6-7
e
nd
s of
tran
sact
ions, 453-455
Eng
li
sh type cha
ract
er sets, 1
22
e
ntiti
es, 1
2-
17
e
ntity
-relations
hip
diagrams, 20-21
equal signs ( =)
wi
th
ANY
,
383
co
mparison Ope
r'c1
tors,
121
is
equa
l
co
operato r, 120
pr
ece
den
ce,
140
equality
and
EQUAL
ope
rntor
co
mp
arison Oper'
c1
tors,
119-1
23
NUL
L, 255-258
subqueries, 3 71, 3
76
WHERE
clau
se,
11
.1
ESCAPE
ident
ifier,
129
-
131
evaluarion o
rd
er
of
operarors,
76
even
numbers,
206
excl
amat
ion
po
i
nts
( ! )
date
fonnat
ma
sks. 2
44
n
ot
equa
l ro
ope
rator, l 21
pr
ecedence, 140
subque
ri
es,
376
explicit t
ype
cast
ing,
23
7,
43
2
expressions
co
nditi
onal.
See
co
nditi
onal expressions
exe
rcises, 92-95
jo
ining,
190-191
lit
era
ls
in,
83-84
substituting,
155-157
WHERE
clause,
11
3
extents,
449,
4
83
e
xt
rac
ting sub:mings, 198-199
fi
le me
nu
in
SQ
L Developer,
39
fi
ll
mo
de
(fm) operator, 242-243
first
no
rm
al
form
(I
NF)
, 15
FLOAT darn type, 485
fm
(fill
mo
de)
ope
rat
o
r,
2
42
-2
43
FOR
UPDATE
clau
se,
459
-460
for
ei
gn-key
co
ns
tra
i
nt
s,
496
dropp
ing rabies, 494
overvie
w,
499
-500
for
eig
n keys, 18-20, 4
83
jo
ins,
333
l
og
i
ca
l models,
13
NUL
LABLE
co
lu
mn
s,
91
self-referencing, 44
format masks for daces
descriptions, 2
10,
243-2
44
pr
ecision,
22
1
forma
tt
ing
dates ro
charac
ters
co
nversion,
24
1
-2
44
numbers to c
ha
rac
ters
co
nversion, 238-240
forw
ar
d slashes
(/)
date
fo
nn
at
masks, 244
eva
lu
at
ion order,
76
precedence, 140
srncemen
r.s,
70
fracrions in d:ires,
211,
213
FROM
clause, 65
uh
.
1u
e
r
i~.
366, 372-373
W1
!ERE
clause, I I I
full
oucer joins, 346-347
f
unc
tions, 1
76
characters.
Sec
character functions
conversion.
See
co
nversion functions
d,ucs.
See
date
functions
defining,
176
- 180
group.
Sec
group funclions
mulriple-row, 1
82
nested,
177.
180, 250-252
num
eric.
See
numeric
fu
nc
ri
ons
s
in
gl
e-
row.
See
s
in
gle-row functions
Fusion Middleware, 8
G in num
ber
format
ma~ks,
240
general
fu
nctions,
252
COA
L
ESCE,
25~259
l
LJLLIF,
255-25
VL, 252-254
VL2,
254-255, 257-25
geological core· scenario
Jescripuon,
11
enr
icy-relati
ons
h1
p diagrams,
20-21
gr
ea
ccr ch
an
or
eq
ual
operators(>•)
nonequijo1ns,
337
subqueri
l>s,
3
76
gr
ea
te r
than
signs (
>)
and
ope
rators
wich ANY a
nd
ALL.
383
comparisons, 121
precedence,
140
subqueries, 3 71, 3
76
GROUP
BY
cl
ause
multiple
co
lumns, 296-298
overview,
29
4-
2%
subqueries,
367
group functions, 278
AVG,
286-287
cenificauon
summary,
306
COU.
T, 2 4-2 5
definition,
278-2
0
G R
OU
P
BY
clnuse, 294-298
groups
of
data,
292-293
H
AVING
clause,
299-305
l
ab
answe
r,
314
l
ah
qucsci
on,
3
11
MAX
and Mh ,
287
-
289
ncsced, 290-291
self tesc, 309-311
se
lf
tl>:.t
an~wcrs,
312-3
14
SU
M, 285-286
rwo-minute drill, 307-308
cypes
and
sy
nrax
, 280-282
working with, 283-28
4,
289
group-l
eve
l results,
300
grouping a
uribu
te,
29
4
hash
clusrers, 4
89
HAVING
dau:.c
description,
299
overview, 300-301
u
bq
uerie.,
367,
376
working with,
.301-.305
heap
rnhle~.
4
HELP INDEX
comman
d.
57
Help
menu
in
SQ
L Developer,
39
HH
in
date
fo
rmat
masks
de~cripcion,
210, 244
precision,
22
1
l-IH
12
in
date
fo
rm
at
ma
sks, 2
44
l-IH24
in
da
re
fo
rm
at
masks, 21
0,
244
hi
gh
wacer
m11rk
s,
44
9
hours
DATE darn type, 4
87
dAte fonnnc masks,
244
precision,
22
1
I l
uman
Reso
ur
ces ( l IR) scenario,
12
I in
d1lle
format
m~ks,
243
if-then-else functions, 260-262
implicit
data
type conve!'l>ion, 235-236
Index
S39
540
OCA
Or
acle Data
ba
se
I 2c: SQL Fundamentals I
Exam
Gu
ide
IN opernror
preceden
ce,
140
scc
co
mpari on.s, 125-1
27
subqueries, 3 70, 3 77
WHERE
clause,
443
indentation,
71
index
clu
tcrs, 4
89
index organized
tab
l
es,
489
indexes with unique
cons
trnints,
497
inequnliry
ope
rarors, 119-123
IN
ITC
AP
funccio
n,
l 77, 186-187
in
l
in
e views for su
bq
ueries, 372-373
inn
er jo
in
s
vs
. o
ut
er jo
in
s, 342-343
ove
rview, 318-320
inn
er
quc
r
ic~,
366
input
paramc
l
c~
list
fo
r f
un
ctions
, L
76
I
NSERT
I
NTO
sratemenr,
436
inserting
an
d I
NS
ERT st.1tem
en
t
o,·crvicw, 4
25
-426
rows
int
o cables, 433-440
update a
nd
del
e
ti
on
a
no
m
ali~.
15
I
NSTR
funct
ion, 1
78,
196-198
INTEGER
d
am
type, 4 5
interface
in
SQ
L Developer, 38-39
l
NTERSECI
operator
de
scription,
396,
399
working with, 4
03
inter
sec
tion of sets,
397
INT
ERVAL D
AY
TO
SECO
D data type, 485
INTER
VAL
YEAR TO
MONTH
dara type, 485
is
eq
ual
to
ope
rat
or ( =
),
1
20
IS
NULL
operfltor
nu
ll
co
mp
arison
s,
1
32
-
13
3
pre
ce
den
ce
,
140
I
SO
(O
rg
ani
sa
ti
on I
nt
e
rn
ationa le
de
Normalisation)
SQ
L sran
ch
:mls,
27
isola
ti
on
AC
ID
te~l.
45
2-453
trnnsnctions, 461
IW in dace
fo
rm
at
ma
~.
243
IY
in
dace form.it
ma
sks, 243
IYY
in date
fo
nn
at
masks, 243
IYYY
in
J
nte
fo
rm
at m
ns
ks,
243
J in
dat
e format
ma
sks,
243
Java EE
(J
ava
Encerpris
e
&hcion)
s
candard
, 6
Java language, 10
J
ava
Runt
ime Environme
nt
(JRE), 37-3
JOI
N 0 1 clause, 330-331, 338-34 L
JO IN
US
I
NG
clnu~c.
328-330
joi
nin
g, 18
chnrncter l
it
era
ls,
co
lumn
s,
a
nd
expression
s,
190-191
c
ha
rncters :
md
strings, 82-83
SE
LE
CT
srncem
cn
t,
63
joins
a
mb
iguous
co
lumn
na
me • 323-325
ANS
I
SQ
L synrnx,
322
-
323
O m
ei.
i
nn
pr
od
u
ct..~
.
349-353
ce
rtification s
umm
ar
y,
354-355
cross, 320-321, 350-353
mn
er, 318-320
J
OIN
0 clause, 330-331
JO IN U I G
cl11u
se, 328-330
lah
a
n
~
w
c
r,
363-364
lab question,
36
1
n-way j
oi
ns, 333-336
ATURA
L JO I clause, 325-328
ATURA
LJ
OI
· ... O 1 claus
e,
332-333
nonequijo
in
~
.
336-337
Oracle
sy
nt
a
x,
321-322
ourer,
320
,
34
I -
348
self-joins,
338
-3
41
self
t~t.
358-361
self rest
nn
swers, 362-363
two-
minut
e drill, 356-357
types, 316-3
17
JR
E Uava
Runt
ime Envi
ro
nm
e
nt
), 37-38
Julian dates, 243
k
ey~
a
nd
key
co
n
~
tr
ai
nt
,
43
2
co
lumns, 4
97
DELETE,
446
drop
pin
g rabies,
494
e
nt
ities, 17
foreign, 18-20, 4 3, 499-500
joins,
333
l
og
ical m
ode
l
s,
13
ULL
AB
LE
co
lumns, 91
primar
y,
498-499
se
l
f-
referencing,
44
,
500
UPDAT
E.
427
keywords,
64
Lin
num
be
r
fo
rmat m
as
k
s,
240
la rgest i
tem
s in groups, 287-289
LA
ST
_DAY fun
cc
ion, 1
79,
219-220
LD_LIB
RARY
_PAT H vari
ab
le, 30-31
L
DAP
Co
nnection Type
se
tti
ng, 40
lending characters, trimming, 1
95
left ou
te
r joins, 343-3
44
LENGT H
fu
nction, 1
77,
1
80,
191- 1
92
less ch
an
or equal
to
(
<•)
opcracors
no
n
eq
ui
jo
in
s,
337
s
ubq
u e
r
i~.
3
76
less than igns
(<)
nnd operamrs
wi
th
A Y and ALL,
383
comparisons,
12
1
precedence, 1
40
suhqueries,
37
1,376
lib I ibrary,
31
LI
KE ope
r11
ror
pat
te
rn
co
mpari
so
ns, 127-
13
2
precedence, 1
40
li
ne bre;iks, 70
Li
nu
x.
SQ
L *Plus on, 30-32
literals
ADD_
M
ON
TH
S,
215
cha
ra
ctcr·
l-n
scd
co
ndition
s,
11
4
concaten
at
in
g,
83
date-based
co
nditions, 117
da
te
fo
nn
at
m,isks, 2
44
in
exp
ressions, 83-84
IN operato r, 1
25-126
joining, 1
90-
1
91
LAST
_DA
Y, 219
l
eng
th
, 1
92
MONTHS_
BET\
VE
E
N,
213-21
4
EX
T_OAY,
217
pad
din
g, 1
92
patte
rn
co
m
pa
rison
s,
1
27
-1
28
posi
ti
on
in
search strin
gs,
1
96
string re
pl
aceme
nt
, 1
99
substr
in
g excr
act
ion, 198
In
dex
541
LocaVBcqu
eat
h
Co
nn
ection
T\
•
pc
scccing,
41
l
oc
k
ing
rows,
460
l
og
i
ca
l model
fo
r
e
nti
t
ic~,
12-13
l
og
i
ca
l operarors
AND,
13
4-1
36
NO
T, 1
37
- 1
38
OR, 136-
13
7
overv
iew, 1
33-
134
rn
nges, 124
logon string, 31-32
l
ogon
to d
awbaJ>cs,
35-36
LONG dnrn type, 4
86
LONG RAW
da
ta
typ
e,
4
86
lookup e
nt
iti
c:.,
17-18
L
OWE
R fun
ct
ion, 1
77,
183-1
85
lower
c.'ll>e
co
nversions t
o,
1
83-
1
85
SE
LE
CT
Smreme
nc
,
69
LP
AD
function, 1
78,
1
92-
194
mandatory
co
lumn ,
59
master-detail rel
at
ionsh ips
ou
te
r
j
o
in
~.
34 2
overview, 17- 18
M
AX
f
un
ct ion
multiple-ro
w,
382
ove
rvi
ew,
28
1-282, 287-289
wi
th
scalars,
44
3
MER
GE
co
mmand, 4
28,
450-451
meridian indicntors in dace fo
rm
at m
as
ks, 2
44
metad
ma
rnbles,
57
uppercase,
70
mec
e<>
r shower prohahilit
y,
77
MI
in
dace
fo
rm
at
masks
descr
ipno
n, 210, 244
precision,
22
1
5 4 l
OCA
Or
acle Database I
2c:
SQL Fundamentals I
Exam
Gu
id
e
MI in number fo
rm
at
m;i
sks,
240
middle tier
in
w
eb
applications, 6
MIN function
overview,
28
1- 2
82,
287
-
289
with scalars, 443
minus
of
set
s,
397
MINU
S operator for sets
desc
ri
p
ti
on,
396,
399
working with, 403-404
minus sign
s(-
)
date format masks.
24
4
evalu
at
ion order,
76
num
ber fo
nn
al masks, 240
preceden
ce
, 1
40
minut
es
DATE
data
t)•pe, 487
d
ate
formal masks, 2
44
pr
ec
isi
on,
22
1
mixed case,
69
MM
in
dat
e format masks
descrip
ti
on, 242
pr
ec
ision,
22
1
MOD
fu
nc
ti
on,
17
9, 206-208
MON
in
date format m
as
ks,
210,
242
MONTI I
in
dat
e format m
ac;
ks, 242
mo
nth
s
DATE da
ta
type, 487
date fo
nn
at
masks, 210,
242
pr
ec
ision, 22 1
MONTH
S_
BETWEEN fun
ct
ion
description, 1
79
working with, 21
3-215
multiple
co
lumn
s,
grouping
by,
296-298
multiple lines, spmming, 71
multiple-row func
ci
ons, 1
82
multipl
e-
row subqueries, 37
6-3
77
, 382-383
multiple
tab
les.
Sec
joins
multiplication operations evaluation order, 76, 1
40
n-way join
s,
333-336
nam
es
;imbiguous, 323-325
schema o
bj
ec
ts, 4 79-481
sets,
399
substit
utin
g,
154
-
155
namcspaccs, 48
1-4
82
Na
ti
onal
lan
guage Support
(NLS
)
set
tin
gs
co
nversion funct
io
ns, 238-239
ine
qu
ality tests, 122
NATURA
LJ
OI clause
inner joins, 318
ove
rvie
w,
325-328
NA
T
URA
LJ
O IN ...
01
clmise, 332-333
Navigate me
nu
in
SQ
L D
ev
eloper, 39
NCLOB
data
ty
pe, 486
nested fu
nctio
n
s,
177
groups, 290-291
overvie
w,
250-252
single-ro
w,
180
nested subqueries, 366-367
nested rabies, 482
New Table option, 4
94
NEXT
_DA
Y fun
ct
ion, 179, 2
17-
2
18
9 in
numb
er
fo
rm
at
masks, 240
NLS (
aL
ional
La
nguage Suppo
rt
)
se
tc
in
gs
co
nversion functions, 238
-2
39
inequality tests, I
22
NLS_CURRENCY se
ttin
g, 238-239
LS_DATE_FORi\itAT se
uin
g,
210
NLV funccio n, 414
NOLOG
s
wit
ch, 32-33,
35
non
eq
ui
jo
ins, 336-337
normali
za
tio
n,
11
, 15
n
ot
eq
ual to operator
(!•)
pr
eceden
ce
,
14
0
subquerics, 3
76
NOT
IN
ope
rat
or in subqueries, 370-371, 377
NO
T NULL
co
lu
mn
s,
88-90
NO
T
NU
LL
co
nstraint, 59-60, 4
96,
498
NOT
operator
ove
rvie
w,
13 7- 1
38
pr
ece<
len
ce,
14
0
ranges, 124
NUL
L
va
lu
es
AND
operator, 1
34
co
mpa
ri
son ,
13
2
-1
33
co
ncepts,
87-9
1
de
termining,
252-25
4
equfl
li
cy
r
es
t
s,
255-258
fo
reign k
ey
co
nstraint
s,
499
OR opcr
aco
r,
136
sets,
405
s
in
gle-r
ow
fun
c
ti
on
s,
181
sorting,
144
s
ubqueri
es
, 3 7 l
unique constraints, 497-498
ULLABL
E
co
lumn
s,
88-91
NULLIFfunc
ci
on, 181, 255-258
U
LLS
FIRST
keywo
r
ds
in
so
res,
l4'1
ULLS
L
AS
T
keyw
o
rd
s
in
so
re
s,
144
numbers
and
NUMBER
data lype, 485
column
s,
59
COU T,
280
precisi
on and
sca
le,
487
qu
e
ry
co
n<liLi
ons,
111
-1 13
n
um
e
ric
conversions
characters, 238-24 l, 248-249
ex
plicit,
23
7
impli
cit,
235
-
236
numeric expr
ess
ions
length, 191
padding, 192-193
string replacement,
199
numeric
(uncLi
ons,
203
MOD,
2C6-208
ove
rvi
e
w,
17
8-
179
ROUND, 203-204
TRUNC, 205-206
numeric lite
ral
s
joining, 190
pos
ition
in
sea
rch strin
gs,
196
string re
plac
ement,
199
substring extractio
n,
198
NVARCHAR2 data type,
484
VL
fun
ct
ion, 181, 252-254
NV
L2 function, 181,
25
4-255, 257-258
o
bj
ects,
47
6
nam
e.s
pac
es,
481-482
sc
hema
nam
es,
479-481
cable
strucwre, 482-484
types,
476-477
u
sers
a
nd
sc
hem
as
, 4 78
-4
79
OC
I (Oracle Call
lncerfa
ce) librar
ies,
10
odd
num
bers,
206
ON
DELETE
CASCADE co
mm
flnd
,
499
ON
DELETE
SET
ULL
command,
499
ON
ke)
•
wo
rd
fo
r inner jo
ins
, 318
o
nl
ine documenta
ti
on,
57
OR operator
TN
opern
ro
r,
12
6-
12
7
o
verview,
136-13 7
pipe
symbo
l, 64
pr
ecede
nce,
14
0
rnn
ges,
124
Index
S4]
ORA-00904: SA_
RE
P:
inva
lid
identifier error,
114
,
15
2
O
RA
-00905>
: inval
id
number of
arg
um
e
nt
s error,
253
O
RA-
0091
8:
column a
mbi
guously de
fined
erro
r,
324
ORA-00923:
FROM
ket.vo
rd
not
fo
und where expecc
e<
I
error,
79
-8
1
ORA-00932:
ln
co
nsis
tenc
data
ty
p
es
error, 2
59
,
286
ORA-00934:
gro
up
function is not
allowed
here error,
294
ORA-00
935
: group function is n
esce<
l coo
deeply
e
rr
o
r,
29
1
ORA-00937: not a s
in
gle-gro
up
group func
ti
on erro
r,
294
ORA-009
42:
ca
ble or v
iew
cl
Ol..'S
not
ex
ist er
ro
r,
79
ORA-00979: not a GROUP
BY
e
xpr
essio
n e
rr
or, 294-295
ORA-0
1017:
inval
id
u
se
rn
ame/
pa
ss
wo
r<l;
l
og
in
den
ied
er
ror,
36
O
RA-01427:
single-row s
ubqu
ery
returns more than one
ro
w erro
r,
380-381, 385,
443
ORA
-0172
2: inva
li
d number er
ror,
236, 248, 327
ORA-01756 er
ro
r,
85
O
RA-
0184
0:
input
va
lu
e is not long enough for date
fonnat er
ro
r,
24 7
ORA-01
862:
the n
um
e
ri
c
va
lu
es
does not match the
length of the
formflt
item
error, 24 7
O
RA
-12
154:
TN
S:
co
uld
not r
eso
l
ve
the
co
nnect
id
entifier s
pe
ci
fied
error, 36
ORA-125
14
: TNS: l
iste
ner d
oes
not c
urr
ently know of
service
requcsced
in conn
ecco
r descriptor error, 36
ORA-1
25
41
: T
NS:
no
li
stener error, 36
O
RA
· I
555;
sn
aps
hot c
oo
o
ld
error,
45
2
ORA-251
54:
column part of US
IN
G clau
se
cannot have
qual
ifi
er er
ror,
324
Oracle Ca
ll
Interface
(OC
I)
li
bra
ri
es,
10
Oracle
Deve
lopme
nt
T
oo
l
s,
S
5
44
OCA
Or
acle Database I
2c:
SQL Fundamentals I
Exam
Gu
id
e
Oracle
Ent
erprise Manager, 7-8
ORAC
LE_H
OME
var
inble, 30-31, 34
ORAC
L
E_HOME_NAME
var
iable,
34
Oracle
join sy
ntax
,
32
1
-322
Oracle
ec prococol, 4
Or
acle
Se
rver
Arc
hi
tec
tur
e, 3-4
Oracle
WebL
og
ic Server,
5-
7
ORD
ER
BY
claus
e,
14
3
asce
ndin
g
and
desce
ndin
g
so
rts,
14
4-
146
co
mposire
so
r
es,
146-147
exercise, 148-149
positional sorts,
14
6
sets, 399, 401, 41 l
Order
Ent
ry
(OE)
scenario,
11
o
rd
er o f op
era
t
or
evaluation,
76
order of rows rcru
rn
ed
with
set
operators,
41
0-414
ord
ina
l te
xt
in
dat
e
fo
rmat
ma
sks,
244
Or
ga
ni
sm
i
on
l
nt
e
rn
arionale de
No
rma
li
sacion
(ISO)
SQ
L
sta
nd
ards, 27
ORGAl'\ll
ZAT
I
ON
HEAP
option, 4
89
orp
han rows, 499
o
ut
er
joi
ns, 320
exe
rcise, 347-348
full, 346-347
vs.
in
ner, 342-3
43
l
ef
t, 34
3--3
44
ove
rview, 341
-3
42
rig
ht
, 345-346
o
ut
er
que
ries,
366
o
wn
ers, 4
79
P.M
. in
date
form
at
masks, 244
pQdding
charac
ters, 192-194
parnmerers
li
scs,
176
pa
re
nt
-c
hild relationships
for
eign
key
co
nscrainrs, 499
o
ut
er
jo
in
s,
342
paremhe
ses
()
IN operato
r.
1
25
precedence,
76
, 139-140
sets,
398
parririo
ned
r
ab
ies,
489
passwo
rd
s
databases, 34
SQ
L
Dev
elope
r,
39
PATH
var
iable, 30-31
patt
e
rn
co
mparisons,
127
-
132
per
ce
nta
ge signs (%) in pa
tt
em
co
mpari
so
n
s,
127-1
28,
13
1
peri
ods
(.)
au
ributes,
18
co
l
umn
nam
es
,
323--325
date
fom1at masks, 2
44
number
fo
rmat
ma
sks,
240
pe
rmi
ssions, access, 4
32
pipe
sy
mb
ol ( I )
co
n
cate
nntion, 82-83,
189
OR,64
pr
eceden
ce,
14
0
PL
/SQ
L l
ang
uiige, JO
pla
ceh
olders
in
subst
itutio
n
var
i
nb
les,
150
plus signs (
+)
eva
luation order,
76,
l 40
PM
in
dot
e fo
rm
at
masks
, 2
44
populati
ng
tab
les, 425-4
26
position
in
strings, 196-198
positional n
otat
ion, 433-434
positional
sort
ing,
146
positional
text
in
date format masks, 2
44
pound
sign
s(#)
in dare formar masks, 2
44
PR
numb
er fo
rm
at masks, 240
preceden
ce
eva
lu
ation
order, 76
s
eis
,
398
WHERE
clause.
139
-1
42
precision
dates, 220-223
numbers,
59,
4
87
ro
un
ding, 203-204
trun
catio
n, 205-206
prima
ry
keys, 4
83
co
ns
cr
aincs, 4
32,
498-499
DE
LET
E,
446
en
tit
ies, I 7
jo
ins,
333
l
og
i
ca
l m
ode
ls,
13
UPDATE,
427
prioriry
in
ser
s,
398
pr
oc
edural
lan
guages, 28-29
projc.:c
ci
ons
SELECT,
62-63
sulxJueri
e:.
for, 3
73
propri
eta
ry Oracle
join
syntax, 321-322
puncrnacion
dace format masks, 2
44
e
rr
ors,
79
PURGE
op
ti
on,
494
Qin
dare
form11r
masks
description,
243
precis ion,
22
1
q operato
r,
85-87
qu
alifying ambiguous
co
lumn
nam
e:.,
323-325
qu
a
rr
er
in
dace fo
nnar
masks
d
e:.c
ription, 2
43
precision,
22
1
que
:.
ti
on ma
rk
s U)
in
<lat
e fonnat m
asks,
2
44
qu
ota
tion marks
('
,")
aliases, 80-81
cha
rn
ccer-ba:.ed
co
nditi
ons,
114
date·oosed
cond
i
tion~.
117
dare formar ma k , 2
42,
2
44
IN operato
r,
125
-
12
6
literals, 4-87
names, 480
number fonnauing.
239
range
co
mparison opcrnrors, 124-1
25
Rational Unified Process, 12
RAW
dat;i rype, 4
85
RDBMS (Relational Database
Man
ageme
nt
Sysccm), 2
read
-o
nl
y rabies,
493
readahility,
71
real-world
sce
narios,
11
- 12
referential integrity, 18-20
Registry Editor,
33
-3
4
relation:il nlgehra,
62
relattonal database design,
2,
29-30
ce
rcifi
cacion summar
y,
48
cloud
co
mputin
g, 8-9
dem
onstrnri
on
schemas, 42-47
de,·elopme
nt
t
oo
ls
and
languages, 10
lab an:.wers,
54
lab questions,
52
Oracle Enterpri
se
Manage
r,
7
-8
Oracle
Se
r
ver
Archicecrure, 3-4
Oracle WebLogic Server,
5-7
Index
S4S
rehnional structures.
See
re
larional structures
self test, 50-52
self rest answers, 53-54
S
QL
Developer t
oo
l.
3
7-41
SQ
L language, 26-29
SQL
•Plus t
oo
l, 30-36
tw
o-
minure drill,
49
Relational Dacabase
Mana
gemenc Syscem (RDBM
S),
2
rclation;il darnba e tables,
62
relauon;il struc
tur
es
d
ata
modeling, 12
e
ntiti
es
and
relation
s,
1
2-
17
c
ntit
y·rclation
:.
hip
di
agra
m:.,
20--2
1
primary keys,
17
real-world
:.cc
nari
~.
11-1
2
refere
nu
al integr
it:y
and fo reign keys, 18-20
relationships,
17
- 18
rows
and
tables, 22-26
relational tables.
23
relari
om
1I
th
eory,
62
rel;itions, 5 7
rclmionships, 12-18
remainder
of
division operation,
17
9, 206-208
REPLAC E functio
n,
17
8,
1
99-201
re
served words,
64
rcstric
cin
g rows
ce
mfl
cn
ci
on summar
y,
164-1
65
lab an:.wer,
17
3-17
4
l
ab
question,
17
0
~
df
(
~
l
.
16
~
17
0
se
lf
rest answers, 1
71
- 1 72
tw
o-
minut
e drill, 1
66-
167
WHERE
clause.
See
WHERE
clause
5
46
OCA
Oracle Database I
2c:
SQL Fundamentals I
Exam
Guide
rig
ht
ourer jo
in
s, 345-346
RM
in
date
fo
rm
at masks, 243
ro
les
in
SQL
Deve
loper, 40
RO
LLBA
CK
co
mm
an
d,
451
,
453-454
,
456-457
ro
ll
oock
s,
4
32
Roman
num
erals
in
dat
e
fo
rm
at masks, 243
ro
und
brackets ()
IN
opera
to
r,
1
25
pr
ece
den
ce,
7
6,
139-140
sers, 398
R
OUND
function
dates, 179, 220-221
num
eri
c,
203--2
04
ro
w-l
eve
l res
tdt
s, 300
row order of set operators,
41
0-414
RO
WL
D data typ
e,
4
86
rows, 22-26
co
untin
g,
28
4-
28
5
deleting from t
ab
les,
446-44
8
fo
r D
ML
s
tate
ments, 373-3
74
inse
rt
ing
int
o tables, 4
33
-440
l
oc
king, 460
l
og
ical m
od
els,
13
mergin
g,
450-45 l
res
ui
c
tin
g.
See
res
tr
ic
tin
g
ro
ws
so
rti
ng
.
See
so
rLin
g
rabies,
57
upda
tin
g, 441-445
RPADfun
c
ti
on, 192-194
RR
in
date fo
rm
al masks,
210
, 242
rules
ch
ec
k
co
nstra
int
s, 500
relational
rn
bles,
23
R
un
me
nu
in
S
QL
D
eve
l
ope
r, 39
runcime b
indin
g, 151
s date formar masks, 2
43
number
fo
nna
t masks, 240
SAV
EPO
!N
T
co
mm
And , 45
1,
45
4,
4
58
-4
59
sav
in
g pa
ss
wo rds, 39
sca
l
ar
subqueries,
366
sca
lar
va
lues
with
S
ET
,
44
2-443
scale
of
num
bers,
59,
487
schemas, 478-4
79
demons
trati
on, 42
-4
3
o
bj
ect names,
47
9-481
cables,
57
-58
scie
nt
i
fi
c noration in
num
be
r form
ar
m<l
sks, 240
searched CASE e
xp
ressions, 262, 265-266
seco
nd n
on
nal
fo
rm
(
2N
f)
, 15
seco
nd
s
DAT
E da
La
type, 487
dtlte
fo
nn
ar masks, 2
44
SELE
CT
lisrs in subqueries, 373
SELE
CT
stateme
nt
ove
rview, 56-57
ca
pabilities, 5
6,
62--63
certific
Mi
on s
umm
ar
y,
95
co
nditi
onal expressions.
See
co
nditi
onal
exp
ressi
ons
D
ESC
RIBE
co
mm
an
d,
57-62
exe
rc ises, 72-75
e
xpr
essions exercises, 92-95
gro
up
func
ti
ons.
See
gro
up
functions
HAVING
cl
au
se, 301-
305
jo
ins.
See
jo
ins
lab
an
swer,
104
- 1
08
lab q uestio
n,
100-101
~
O
T
NULL
and
NULLABLE
co
lumns, 88-90
NULL
co
n
ce
pt
, 87
-9
l
primicive,
64
-69
se If
test
, 98-100
se If test answers, I 02-103
SQL.
See
S
QL
expressions and Op<mllors
syncax rules, 6
9-
71
tw
o
-minu
te
drill, 96-97
SELE
CT
FO
R U
PDAT
E stateme
nt
, 4
59
-4
60
self-d
oc
ume
ntin
g
co
d
e,
434
self-joins, 338-341
self-refer
en
cing foreign keys,
44,
500
semi
co
lons (;) as statement terminacors, 70-
71
sepa
rat
ors
in
number
fo
rmat masks, 240
servi
ce
virtualizatio
n,
8
sessio ns
description, 4
th
r
ee
tier env
ironm
e
nt
,
6-
7
S
ET
AU
T
OCO
MMIT
sta tem
en
t,
45
9
S
ET
clause in
UPDATE
,
44
2
S
ET
DEFI
NE
co
mma
nd, 160-161
S
ET
UNUS
ED
co
mm
a
nd,
4
93
secs
and,
ET
opcmtors, 401-403
ceniflcauo
n summary,
415
co
mbining
qucric~.
4
00
co
mparison,
12
5-
127
compl
ex
examples,
4
05~
descrip
ti
on,
3%
IN
TER
ECT
opcrncor, 4
03
l:i
b answer, 421
l
ab
question,
41
9
MI
NUS
opernror, 403-404
ord
er
o(
rows
returned, 410-414
principles, 398-399
scalar values, 442-443
self r
es
t, 417-419
self test
an
swers,
420
scc·o
ri
e
n1
cd
languages, 28-29
two-minute drill, 416
UNION
ALL, 401-402
Ve
nn Diagrams, 396-397
working with, 399-400, 407-410
SI IOW
co
mmand,
57
SH
OW
u~r
command,
58
simple
CASE
exp
ressions, 262-264
single
ampe~and
ub:.1icu1ion, 150-152
single q uotes
(')
datc-W:.(.'<.I
cond
iti
on~.
117
dat
e format masks, 242
IN operato
r,
1
25-
1
26
lit
eral
s,
84-87
number fonnatting,
239
s
in
gle-row functions
ceniflcation summar
y,
223-224
characccrs.
Si?c
char
ac
ter (unctions
conversion.
See
co
nversion fun
ct
i
ons
dares.
Sec
d:ire (unc
ri
ons
l
ab
answer, 231-232
lnb
quc~tion,
229
numeric.
See
nume
ri
c functions
O\'erview, 180-182
se
lf
test, 227-229
~
If
cc
c an
wen.,
230-231
two-
minut
e drill, 225-226
single-row
~
u bquerie~.
376-377, 382-383
sl
ashes(/)
d.ite fonnat masks. 2
44
cvnlw11ion order,
76
precedence,
14
0
sta
tements, 70
smalle
st
items in group , 287-289
so
rting
ce
rc
i
fl
carion summary, 164-1
65
co
mposite, 146-147
exercise,
14
8-149
labanswer, 173-174
lab question,
17
0
ORDER
BY
c
hrn
se,
14
3-
14
9
positional,
14
6
se
If
rest, 1
68-
17
0
sci( test
an~wers.
171-1
72
rwo-minure drill, 1
66-
1
67
so
ur
ce
tables (or joins, 317
S P in date fonrnn
masks,
2
44
span
ning multiple
line
',
7l
spelling errors,
79,
43
0
S
PTH
in
date
fo
rmat masks, 2
44
In
dex
547
SQL
(Structur
ed
Q uery
umguage).
2,
10,
26
co
mm
ands, 27-28
joim,
322-323
set-o
ri
entat
ion, 28-29
standard~,
27
SQL
[)e,·eloper t
oo
l, 4,
37
AUTOCOMMIT, 459
datab.,se
co
nnections, 39-41
installing and launc
hin
g, 37
-38
user interface, 38-39
SQL
cxpu.>t>sio
r
~
.im
l oper.uor
s,
75-76
mithme
ri
c,
76-78
chamcter
and
string
co
n
ca
tenation,
82-83
co
l
umn
al
i:is
ing, 79-81
li
ccrn
ls, 83-84
single quotes and a
lt
ernative quote, 84-87
sqldevclopcr.exc
fil
e,
38
sqldeveloper.sh script,
38
SQL•Loa
der utilit
y,
4
25
SQL•P
l
us,
4
AUTOCOMMIT, 4
59
database
co
nn
ections, 34-36
on Linux, 30-32
on
Wind
ows, 32-34
sqlplus.exe file, 32, 34
sqlplusw.exe file,
32
S
48
OC
A Oracle Database I 2
c:
SQL Fundamenta
ls
I
Ex
am Guide
SS
in dnte form
at
m
ns
ks, 210, 2
44
SSSSS in
dat
e
fo
rmnt mas
ks
, 2
44
s
t.an
dan
J
dev
i
atio
n, 2 2
STAR_
TR
ANSF-ORMA
TI
ON_ENABLED parameter, 3
72
r:i
r rrnnsfo
rm
nrions,
37
1-372
s
ro
r
r.s
of cransactions, 4
53
-4
55
s
tat
eme
nt
tcm1inators, 7
0-
71
STDDEV function,
282
strin
gs
co
nc
ar
enanon, 82-8
3,
189- 191
replaceme
nt,
199
structure of mbles, 482-484
Stm
c
tu
m l Q uery Language.
See
S
QL
(Structured
Qu
e
ry
La
nguage)
subqueries,
366
certifi
ca
ti
on sum
m<ir
y, 386
fo
r
co
mp
J
ri
~ons,
369-3
71
co
mp
lex, 3
74
-3
75
correlated, 3
77
-
37
8
dcscr
ipc
ion, 366-367
in
lm
e views, 372-373
IN
SE
RT, 436-4
38
lab
an
swer,
39
4
l
a
b~
1
u
e~c
i
on,
392
multiple-row, 376-3
77
, 3 2-383
fo
r proj
ec
tions, 373
re
li
nble nnd user friendl
y,
383-385
rows for DML s
tat
ements,
37
3
-3
74
se
lf
t
es
t, 388-391
self t
es
t
a
n
~
w
en>
,
393-394
s
in
gl
e-
ro
w,
376
-3
77,
38
2- 3
83
s
ta
r trans
fo
rmations, 371
-3
72
ia
hl
cs
fr
om, 4
90
-491
tw
o-
m
inute
dr
ill, 387
types, 3
67
-3
6
8,
379-381
su
bs
titu
ti
on v;ir
in
bles, 1
50
co
lumn
names, 15
4-
1
55
double
amp
ersand subs
ti
tuti
on, 1
52
- 1
54
exercise, 1
63-
164
expressio ns and t
ex
t,
155-1
57
s
in
gle a
mp
c
n.
a
nd
su
b:.
ci
cu
cion,
15
0-
152
SUBS
TR
funcu
on, 1
78,
181, 198-199
suhtrac
ci
on
dates, 211
ev
aluat
io
n or
de
r,
76
uffixes
in
Jate for
mat
masks, 244
SU
M
fu
nc
ti
on, 2 l, 285-286
syncax c
rr
on., 4
30
sy
ntax
rules
in
SEL
ECT
,
69-
71
SY
SO
ATE function, 1
79
, 211
sysdba
co
nne
ction,
40
system dac
e,
211
SYSTEM schema,
47
9
T
ab
le tab, 4
95
cables,
22
-26, 57
nlinses,
32
4
co
lumn spe
ci
fi
cations,
48
9
-4
90
co
nscrainc
s,
4
25
cr
ea
ting, 494-496
definitions, 4
92
-493
deleting rows from, 44
6-
4
50
dropping
an
d
trun
ca
tin
g,
4
93
-4
94
in
serting rows int
o,
433-440
merging row:.
int
o, 4
50-45
1
neste
d,
4 2
po
pula
tin
g. 425-426
read-onl
y,
4
93
s
tm
c
tur
e,
4
82
-4
4
fr
om s
ul'X}u
eries, 490-491
updat
in
g rows in, 441-445
targer
tab
le in join
s,
31
7
TC L (T
mn
s.,ction
Co
ntr
ol Language)
co
mmnnds, 28
termi
nJ
ton>,
sta
tement, 70-71
testing dnrnbase
co
nn
ec
tion
s,
34-36
te
xt
dnte formm masks, 2
44
subs
titut
ing, 155-1
57
T H in date
fo
rmar m
;is
ks
, 244
third-gen
era
tion l
an
guages (3
GLs),
10
third normal f
or
m, 1
5,
333
thr
ee-tier en
\l
iro
nm
c
nt
,
6-
7
TH
SP
in
date
fo
rmat masks, 2
44
tim
e,
system, 211
T IMESTAMP da
rn
typ
e,
59,
4 5
TI
M
ES
T
AMP
WITI I L
OCA
L TIMEZO
NE
da
ta
type, 485
T IM EST AMP
\'V
IT I I T IMEZO
NE
d
:i
ra type,
485
TN
S (Transpare
nt
etwork Subs
trat
e)
layer,
36
TNS
Co
nn
ec
ri
on Type se
nin
g,
40
msnames.ora file, 35
T
O_C
HAR
function, 237
dmes, 241-246
description,
181
numbers,
238-2
41
TO_DATE function, 212, 237
characters, 247-248
description,
181
TO_
UMBER function,
23
7
characters, 248-249
description, 181
Tool
fo
r Application Developer
:.
(TOAD),
4
Tools me
nu
in
SQ
L Developer, 39
totals in
co
lumns, 285-286
trailing characters, rrimming, 195
Tran ac
ti
on
Co
nu
ol Language (TCL) commands, 28
tran
s.
1
cr
ions,
451
co
ntr
ol
s
tat
e
men
ts, 45
5-456
database, 451-455
isolation,
461
su:i
r
ts
and e
nd
s, 453-455
wi
th
TRUNCATE, 429
Transpare
nt
Network Sub
sm
1t
e (TNS) laye
r,
36
uan
sposition errors, 430
TRIM function, 178, 194- 196
TR
UNC
function, 1
79
elates, 222-223, 487
numbers,
205-206
TRUN
CA
TE
co
mmand, 429, 438, 44
6,
449
-4
50
TRUN
CA
TE
TABLE
co
mm
ii
nd, 493
truncating tables,
493-4
94
tntLh cahles
AND, 134
OT, 1
37-138
OR, 136
wples in logical model
s,
13
tw
o-dimensional model, 482
-4
83
tw
o-tier processing, 4
rype
ciistin
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a
uc
oma
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48
7-4
88
errors from, 430-432
with row insertion, 434
U in number format m
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240
UM
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unhounded data t)•pc , 483
UND
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58-
161
In
dex
549
underscore characters (_) in pattern comparisons,
1
27-
128
umon
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UN
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operator
description, 396, 399
working with, 401-403
UN
I
ON
ALL operator
description, 396,
399
wo
rking with, 401
-402
unique constraints, 497
-4
98
umque keys in logical models, 13
Univelbal Markup Language
(U
ML), 12
updating and U PDATE
co
mmand
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15
overvi
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42
6-4
27
rows in cables,
441
-445
UP
PE
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77,
185-1
86
up
percase
co
nv
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85-
1
86
SELECT Stateme
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,
69
UPSERT
co
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US7
ASC
II
database character set, 122
user interface in SQL Developer,
38-39
USER_ O BJECTS ,·i
ew,
478
user processes
in
cl
i
ent
ti
er, 4
User SYS,
47
9
use
mam
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databases, 34
SQ
L Develope
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39
users, 4
2-43
clie
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tier, 4
schemas, 4
78-479
US
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m
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j
oi
ns, 318
JOIN 0 clause, 330-33 l
JOIN USI
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cl11
usc, 328-330
SS
O OCA Oracle Data
ba
se I 2c: SQL Fundamentals I
Ex
am Guide
V in number format
masks,
240
VALUES
clause
INSERT
INTO, 436, 438
s
ubqu
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es,
374
VARCHAR data type. 484
VARCHAR2
dnca
rypc,
59
COU T,
280
de
sc
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n,
4 4
TO_CHAR, 238
v11
ri
nh
l
es,
substitution, 150
column n
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1
54-
155
double
ampe
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sa
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52-
154
exe
rcise, 1
63-
164
cxprcs:.io
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xt,
1
55-
1
57
single ampersand substitution, 150-152
VAR
IA CE
fun
ctio
n,
282
Venn
Dia
g
ram
s. 396-397
VER
I
FY
command.
162
Versioni
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f!
menu
in
QL Developer, 39
View
menu
in
SQL Developer,
39
views,
inline, 372-373
W in dtue
fo
r
mnt
m
asb
descripuon,
243
precision,
22
1
web
11pplica1ion
s, 5-6
week
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fo
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ma
sks
description,
243
precisio
n,
22
1
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.
..
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eme
nc
s,
264
WHERE
clause
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lean operator
s,
133-139
cha
ra
cter
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condition
s,
11
4-116
comparison opcrarors.
Sec
comparison operators
date·based conditions,
117
-
11
9
DELETE,
427
group
functions, 294
gro
up
rcsulr , 300
numeric·based conditions,
111
-
113
overview, l 1
0-l
l I
precedence
mies,
1
39-
14
2
MM
tran formation
:.,
37
1
s
ubqu
e
ri
es,
366-367, 376, 492
UPDATE.
426-427, 441-443
width m
nu
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ma
sks,
240
wi
ldcurJ charact
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pa
ri
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1
27-
13
1
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s,
SQL *Plus on,
32
-3
4
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in
date fonMt
ma~ks,
243
Yin
dace
fomrnl
masks,
242
years
DATE
daw
type.
487
date format
mask
s, 210, 242
precision.
221
YY
in date fonnat
mask
s,
210,
242
YYY
in
date
form
,
1t
m
t1
b,
242
YYYY
in
dare
format
mask
s
de
sc
ription, 210,
242
pre
cisio
n,
22
1
Zachman
framewo
rk
, 12
zeroes
vs.
ULL,
87
numher
fo
rim1r
1
m1sks,
240
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