Tài liệu OCA: Oracle Database 11g Administrator Certified Associate- P6 pptx

When using group functions or aggregate functions in a query, the columns that do not have any aggregate function applied to them must appear in the GROUP BY clause of the query.. For ex

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Utilizing Aggregate Functions 181

F single 47

M single 92

married 180

F married 63

M married 117

more Dba Queries In the “Exploring DBA Queries Using Aggregate Functions” sidebar, you saw some que-ries written to find out the space allocated by tablespace, the space allocated by schema, and the space allocated by tablespace and schema These were written using three dif-ferent SQL statements You can see the power of CUBE in the following SQL The results from all the three SQL statements you tried before are in this summary report, showing the different levels of aggregation SELECT tablespace_name, owner, SUM(bytes)/1048576 size_mb FROM dba_segments GROUP BY CUBE (tablespace_name, owner); TABLESPACE_NAME OWNER SIZE_MB - -

1564.8125 <- Grand Total HR 1.75 <- Subtotal HR schema IX 1.625 <- Subtotal IX schema OE 8.875 … … …

FLOWS 100.6875 <- Subtotal FLOWS schema USERS 21.25 <- Subtotal USERS tablespace USERS HR .1875 <- HR schema in USERS tablespace USERS OE 2.625 <- OE schema in USERS tablespace USERS SH 2

USERS SCOTT .375

USERS BTHOMAS 16.0625 SYSAUX 716.375 <- Subtotal SYSAUX tablespace … … …

SYSAUX FLOWS 100.6875 SYSTEM 701.625 <- Subtotal SYSTEM tablespace SYSTEM SYS 685.1875

Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.

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182 Chapter 3 N Using Group Functions

SYSTEM OUTLN 5625

SYSTEM SYSTEM 15.875 EXAMPLE 77.3125 EXAMPLE HR 1.5625 … … …

As you can see in the result, the space used by each schema in each tablespace is shown as well as the total space used in each tablespace and the total space used by each schema The total space used in the database (including all tablespaces) is also shown in the very first line Three functions come in handy with the ROLLUP and CUBE modifiers of the GROUP BY clause—GROUPING, GROUP_ID, and GROUPING_ID In the examples you have seen using the ROLLUP and CUBE modifiers, there was no way of telling which row is a subtotal and which row is a grand total You can use the GROUPING function to overcome this problem Review the following SQL example: SELECT gender, marital_status, count(*) num_rec, GROUPING (gender) g_grp, GROUPING (marital_status) ms_grp FROM oe.customers GROUP BY CUBE(marital_status, gender); G MARITAL_STATUS NUM_REC G_GRP MS_GRP -

319 1 1

F 110 0 1

M 209 0 1

single 139 1 0

F single 47 0 0

M single 92 0 0

married 180 1 0

F married 63 0 0

M married 117 0 0 The G_GRP column has a 1 for NULL values generated by the CUBE or ROLLUP modifier for GENDER column Similarly, the MS_GRP column has a 1 when NULL values are generated in the MARITAL_STATUS column Using a DECODE function on the result of the GROUPING function, you can produce a more meaningful result set, as in the following example:

SELECT DECODE(GROUPING (gender), 1, ‘Multi-Gender’,

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‘Multi-MaritalStatus’, marital_status) marital_status, count(*) num_rec

FROM oe.customers GROUP BY CUBE(marital_status, gender);

GENDER MARITAL_STATUS NUM_REC

-Multi-Gender Multi-MaritalStatus 319

F Multi-MaritalStatus 110

M Multi-MaritalStatus 209

Multi-Gender single 139

F single 47

M single 92

Multi-Gender married 180

F married 63

M married 117

You can use the GROUPING function in the HAVING clause to filter out rows You can display only the summary results using the GROUPING function in the HAVING clause The GROUPING_ID function returns the exact level of the group It is derived from the GROUPING function by concatenating the GROUPING levels together as bits, and gives the GROUPING_ID Review the following example closely to understand this: SELECT gender, marital_status, count(*) num_rec, GROUPING (gender) g_grp, GROUPING (marital_status) ms_grp, GROUPING_ID (gender, marital_status) groupingid FROM oe.customers GROUP BY CUBE(gender, marital_status); G MARITAL_STATUS NUM_REC G_GRP MS_GRP GROUPINGID -

319 1 1 3

single 139 1 0 2

married 180 1 0 2

F 110 0 1 1

F single 47 0 0 0

F married 63 0 0 0

M 209 0 1 1

M single 92 0 0 0

M married 117 0 0 0

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In this example, you can clearly identify the level of grouping using the GROUPING_ID function The GROUP_ID function is used to distinguish the duplicate groups In the follow-ing example, the GROUP_ID() value is 1 for duplicate groups When writing complex aggre-gates, you can filter out the duplicate rows by using the HAVING GROUP_ID = 0 clause in the SELECT statement

SELECT gender, marital_status, count(*) num_rec, GROUPING_ID (gender, marital_status) groupingid, GROUP_ID() groupid

FROM oe.customers GROUP BY gender, CUBE(gender, marital_status);

G MARITAL_STATUS NUM_REC GROUPINGID GROUPID

-

-F single 47 0 0

F married 63 0 0

M single 92 0 0

M married 117 0 0

F single 47 0 1

F married 63 0 1

M single 92 0 1

M married 117 0 1

F 110 1 0

M 209 1 0

F 110 1 1

M 209 1 1

Nesting Functions

Functions can be nested so that the output from one function is used as input to another

Operators have an inherent precedence of execution such as * before +, but function precedence is based on position only Functions are evaluated innermost to outermost and left to right This nesting technique is common with some functions, such as DECODE (covered in Chapter 2), where it can be used to implement limited IF…THEN…ELSE logic within a SQL statement

For example, the V$SYSSTAT view contains one row for each of three interesting sort statistics If you want to report all three statistics on a single line, you can use DECODE com-bined with SUM to filter out data in the SELECT clause This filtering operation is usually done in the WHERE or HAVING clause, but if you want all three statistics on one line, you can issue this command:

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SUM (DECODE (name,’sorts (disk)’, value,0)) on_disk, SUM (DECODE

(name,’sorts (rows)’, value,0)) rows_sortedFROM v$sysstat;

IN_MEMORY ON_DISK ROWS_SORTED - - -

728 12 326714What happens in the previous statement is a single pass through the V$SYSSTAT table

The presummary result set would have the same number of rows as V$SYSSTAT (232, for instance) Of these 232 rows, all rows and columns have zeros, except for one row in each column that has the data of interest Table 3.3 shows the data that was used in this example

The summation operation then adds all the zeros to your interesting data and gives you the results you want

ta b l e 3 3 Presummarized Result Set

Nesting Single-Row Functions with Group Functions

Nested functions can include single-row functions nested within group functions, as you’ve just seen, or group functions nested within either single-row functions or other group func-tions For example, suppose you need to report on the departments in the EMP table, show-ing either the number of jobs or the number of managers, whichever is greater You would enter the following:

SELECT deptno, GREATEST(

COUNT(DISTINCT job), COUNT(DISTINCT mgr)) cnt, COUNT(DISTINCT job) jobs,

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COUNT(DISTINCT mgr) mgrs FROM scott.emp

GROUP BY deptno;

DEPTNO CNT JOBS MGRS

10 3 3 2

20 4 3 4

30 3 3 2

Nesting Group Functions You can also nest group functions within group functions Only one level of nesting is allowed when nesting a group function within a group function To report the maximum number of jobs in a single department, you would query the following: SELECT MAX(COUNT (DISTINCT job_id)) FROM employees GROUP BY department_id; MAX(COUNT(DISTINCTJOB_ID))

3

Group functions can be nested only one level If you try to nest more than one level of nested group functions, you will encounter an error Also, there is no reason to do so Here is an example to show the error, though the SQL does not mean much: SELECT MIN (MAX (COUNT (DISTINCT job_id))) FROM employees GROUP BY department_id; SELECT MIN (MAX (COUNT (DISTINCT job_id))) *

ERROR at line 1:

ORA-00935: group function is nested too deeply

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Exam Essentials 187

Summary

Though this chapter is small in terms of OCA certification exam content, this chapter is very important for the test It is important to understand the concept of grouping data, where GROUP BY and HAVING clauses can be used, and the rules associated with using these clauses

I started this chapter by discussing the group-function fundamentals and reviewed the group functions by concentrating on the functions that are important for the test

I also discussed how group functions can be used in the SELECT, HAVING, and ORDER BYclauses of SELECT statements Most group functions can be applied to all data values or only to the distinct data values Except for COUNT(*), group functions ignore NULLs Pro-grammer-written functions cannot be used as group functions COUNT, SUM, and AVG are the most commonly used group functions

When using group functions or aggregate functions in a query, the columns that do not have any aggregate function applied to them must appear in the GROUP BY clause of the query

The HAVING clause is used to filter out data after the aggregates are calculated Group tions cannot be used in the WHERE clause

func-You can create superaggregates using the CUBE and ROLLUP modifiers in the GROUP BY clause

Exam Essentials

Understand the usage of DISTINCT in group functions When DISTINCT is specified, only one of each non-NULL value is applied to the function To apply all non-NULL values, the keyword ALL should be used

Know where group functions can be used Group functions can be used in GROUP BY, ORDER BY, and HAVING clauses They cannot be used in WHERE clauses

Know how MIN and MAX sort date and character data Older dates evaluate to lower values,

while newer dates evaluate to higher values Character data, even if it contains numbers, is sorted according to the NLS_SORT specification

Know which expressions in a SELECT list must appear in a GROUP BY clause If any

group-ing is performed, all nongroup function expressions and nonconstant expressions must appear in the GROUP BY clause

Know the order of precedence for evaluating nested functions You may need to

evalu-ate an expression containing nested functions Make sure you understand the left-to-right order of precedence used to evaluate these expressions

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A Statement 1 will fail because it is missing a GROUP BY clause.

B Statement 2 will return one row, and statement 1 may return more than one row.

C Statement 2 will fail because it does not have the columns used in the GROUP BY clause

in the SELECT clause

D Statement 1 will display two columns, and statement 2 will display two values for each

state

2 Using the SALES table described here, you need to report the following:

Gross, net, and earned revenue for the second and third quarters of 1999

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Gross, net, and earned revenue for sales in the states of Illinois, California, and

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Texas (codes IL, CA, and TX)

Will all the requirements be met with the following SQL statement?

SELECT state_code, SUM(ALL gross), SUM(net), SUM(earned)FROM sales_detail

WHERE TRUNC(sales_date,’Q’) BETWEEN TO_DATE(’01-Apr-1999’,’DD-Mon-YYYY’) AND TO_DATE(’01-Sep-1999’,’DD-Mon-YYYY’)

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Review Questions 189

A The statement meets all three requirements.

B The statement meets two of the three requirements.

C The statement meets one of the three requirements.

D The statement meets none of the three requirements.

E The statement will raise an exception.

3 Which line in the following SQL has an error?

1 SELECT department_id, SUM(salary)

4 John is trying to find out the average salary of employees in each department He noticed

that the SALARY column can have NULL values, and he does not want the NULLs included when calculating the average Identify the correct SQL that will produce the desired results

A SELECT department_id, AVG(salary)

FROM employeesGROUP BY department_id;

B SELECT department_id, AVG(NVL(salary,0))

C SELECT department_id, NVL(AVG(salary), 0)

D SELECT department_id, AVG(salary)

FROM employeesGROUP BY department_idHAVING salary IS NOT NULL;

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190 Review Questions

5 Review the following two SQL statements, and choose the appropriate option.

1 SELECT department_id, COUNT(*)FROM employees

HAVING COUNT(*) > 10GROUP BY department_id;

2 SELECT department_id, COUNT(*)FROM employees

WHERE COUNT(*) > 10GROUP BY department_id;

A Statement 1 and statement 2 will produce the same results.

B Statement 1 will succeed, and statement 2 will fail.

C Statement 2 will succeed, and statement 1 will fail.

D Both statements fail.

6 Read the following SQL carefully, and choose the appropriate option The JOB_ID column

shows the various jobs

SELECT MAX(COUNT(*))FROM employees

GROUP BY job_id, department_id;

A Aggregate functions cannot be nested.

B The columns in the GROUP BY clause must appear in the SELECT clause for the query

to work

C The GROUP BY clause is not required in this query.

D The SQL will produce the highest number of jobs within a department

7 Identify the SQL that produces the correct result.

A SELECT department_id, SUM(salary)

FROM employeesWHERE department_id <> 50GROUP BY department_idHAVING COUNT(*) > 30;

B SELECT department_id, SUM(salary) sum_sal

FROM employeesWHERE department_id <> 50GROUP BY department_idHAVING sum_sal > 3000;

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C SELECT department_id, SUM(salary) sum_sal

FROM employeesWHERE department_id <> 50AND sum_sal > 3000

GROUP BY department_id;

D SELECT department_id, SUM(salary)

FROM employeesWHERE department_id <> 50 AND SUM(salary) > 3000GROUP BY department_id;

8 Consider the following SQL, and choose the most appropriate option.

SELECT COUNT(DISTINCT SUBSTR(first_name, 1,1))FROM employees;

A A single-row function nested inside a group function is not allowed.

B The GROUP BY clause is required to successfully run this query.

C Removing the DISTINCT qualifier will fix the error in the query.

D The query will execute successfully without any modification.

9 The sales order number (ORDER_NO) is the primary key in the table SALES_ORDERS Which

query will return the total number of orders in the SALES_ORDERS table?

A SELECT COUNT(ALL order_no) FROM sales_orders;

B SELECT COUNT(DISTINCT order_no) FROM sales_orders;

C SELECT COUNT(order_no) FROM sales_orders;

D SELECT COUNT(NVL(order_no,0) FROM sales_orders;

E All of the above

F A and C

10 Sheila wants to find the highest salary within each department of the EMPLOYEES table

Which query will help her get what she wants?

A SELECT MAX(salary) FROM employees;

B SELECT MAX(salary BY department_id) FROM employees;

C SELECT department_id, MAX(salary) max_sal FROM employees;

D SELECT department_id, MAX(salary) FROM employees GROUP BY department_

id;

E SELECT department_id, MAX(salary) FROM employees USING department_id;

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11 Which assertion about the following queries is true?

SELECT COUNT(DISTINCT mgr), MAX(DISTINCT salary)FROM emp;

SELECT COUNT(ALL mgr), MAX(ALL salary)FROM emp;

A They will always return the same numbers in columns 1 and 2.

B They may return different numbers in column 1 but will always return the same

num-ber in column 2

C They may return different numbers in both columns 1 and 2.

D They will always return the same number in column 1 but may return different

num-bers in column 2

12 Which clauses in the SELECT statement can use single-row functions nested in aggregate

functions? (Choose all that apply.)

A SELECT

B ORDER BY

C WHERE

D GROUP BY

13 Consider the following two SQL statements Choose the most appropriate option

1. select substr(first_name, 1,1) fn, SUM(salary) FROM employees GROUP BY first_name;

2. select substr(first_name, 1,1) fn, SUM(salary) FROM employees GROUP BY substr(first_name, 1,1);

A Statement 1 and 2 will produce the same result.

B Statement 1 and 2 will produce different results.

C Statement 1 will fail.

D Statement 2 will fail, but statement 1 will succeed.

14 How will the results of the following two SQL statements differ?

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A Statement 1 will return one row, and statement 2 may return more than one row.

B Both statements will fail because they are missing a GROUP BY clause.

C Both statements will return the same results.

D Statement 2 may return a smaller COUNT value than statement 1.

15 Why does the following SELECT statement fail?

SELECT colorname Colour, MAX(cost)FROM itemdetail

WHERE UPPER(colorname) LIKE ‘%WHITE%’

GROUP BY colourHAVING COUNT(*) > 20;

A A GROUP BY clause cannot contain a column alias.

B The condition COUNT(*) > 20 should be in the WHERE clause.

C The GROUP BY clause must contain the group functions used in the SELECT list.

D The HAVING clause can contain only the group functions used in the SELECT list.

16 What will the following SQL statement return?

select max(prod_pack_size)from sh.products

where min(prod_weight_class) = 5;

A An exception will be raised.

B The largest PROD_PACK_SIZE for rows containing PROD_WEIGHT_CLASS of 5 or higher

C The largest PROD_PACK_SIZE for rows containing PROD_WEIGHT_CLASS of 5

D The largest PROD_PACK_SIZE in the SH.PRODUCTS table

17 Why will the following query raise an exception?

select dept_no, avg(distinct salary), count(job) job_count

from empwhere mgr like ‘J%’

or abs(salary) > 10having count(job) > 5order by 2 desc;

A The HAVING clause cannot contain a group function.

B The GROUP BY clause is missing.

C ABS() is not an Oracle function.

D The query will not raise an exception.

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18 Which clause will generate an error when the following query is executed?

SELECT department_id, AVG(salary) avg_salFROM employees

GROUP BY department_idHAVING TRUNC(department_id) > 50;

A The GROUP BY clause, because it is missing the group function.

B The HAVING clause, because single-row functions cannot be used.

C The HAVING clause, because the AVG function used in the SELECT clause is not used in

the HAVING clause

D None of the above The SQL statement will not return an error.

19 Which statements are true? (Choose all that apply.)

A A group function can be used only if the GROUP BY clause is present.

B Group functions along with nonaggregated columns can appear in the SELECT clause

as long as a GROUP BY clause and a HAVING clause are present

C The HAVING clause is optional when the GROUP BY clause is used.

D The HAVING clause and the GROUP BY clause are mutually exclusive; you can use only

one clause in a SELECT statement

20 Read the following two statements, and choose the best option.

1. HAVING clause should always appear after the GROUP BY clause

2. GROUP BY clause should always appear after the WHERE clause

A Statement 1 and 2 are false.

B Statement 1 is true, and statement 2 is false.

C Statement 1 is false, and statement 2 is true.

D Statements 1 and 2 are true.

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Answers to Review Questions 195

Answers to Review Questions

1 D Though you do not have a state column in the SELECT clause, having it in the GROUP

BY clause will group the results by state, so you end up getting two values (two columns) for each state

2 A All requirements are met The gross-, net-, and earned-revenue requirements are

satis-fied with the SELECT clause The second- and third-quarter sales requirement is satissatis-fied with the first predicate of the WHERE clause—the sales date will be truncated to the first day

of a quarter; thus, 01-Apr-1999 or 01-Jul-1999 for the required quarters (which are both between 01-Apr-1999 and 01-Sep-1999) The state codes requirement is satisfied by the second predicate in the WHERE clause This question is intentionally misleading, but so are some exam questions (and, unfortunately, some of the code in some shops)

3 C Since the department_id column does not have any aggregate function applied to it,

it must appear in the GROUP BY clause The ORDER BY clause in the SQL must be replaced with a GROUP BY clause to make the query work

4 A Since group functions do not include NULL values in their calculation, you do not have to

do anything special to exclude the NULL values Only COUNT(*) includes NULL values

5 B An aggregate function is not allowed in the WHERE clause You can have the GROUP BY

and HAVING clauses in any order, but they must appear after the WHERE clause

6 D The SQL will work fine and produce the result Since group functions are nested, a

GROUP BY clause is required

7 A It is perfectly alright to have one function in the SELECT clause and another function in

the HAVING clause of the query Options B and C are trying to use the alias name, which is not allowed Option D has a group function in the WHERE clause, which is also not allowed

8 D The query will return how many distinct alphabets are used to begin names in the

EMPLOYEES table You can nest a group function inside a single-row function, and vice versa

9 E All the queries will return the same result Since ORDER_NO is the primary key, there cannot

be NULL values in the column Hence, ALL and DISTINCT will give the same result

10 D Option A will display the highest salary of all the employees Options B and E use

invalid syntax keywords Option C does not have a GROUP BY clause

11 B The first column in the first query is counting the distinct MGR values in the table The first

column in the second query is counting all MGR values in the table If a manager appears twice, the first query will count her one time, but the second will count her twice Both the first query and the second query are selecting the maximum salary value in the table

12 A, B A group function is not allowed in GROUP BY or WHERE clauses, whether you use it as

nested or not

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13 B Both statements are valid The first statement will produce the number of rows equal to

the number of unique first_name values The second statement will produce the number of rows equal to the unique number of first characters in the first_name column

14 D COUNT(*) will count all rows in the table COUNT(salary) will count only the number

of salary values that appear in the table If there are any rows with a NULL salary, ment 2 will not count them

state-15 A A GROUP BY clause must contain the column or expressions on which to perform the

grouping operation It cannot use column aliasing

16 A You cannot place a group function in the WHERE clause Instead, you should use a

HAVING clause

17 B There is at least one column in the SELECT list that is not a constant or group function,

so a GROUP BY clause is mandatory

18 D The HAVING clause filters data after the group function is applied If an aggregate

func-tion is not used in the HAVING clause, the column used must be part of the SELECT clause

19 C The HAVING clause can be used in a SELECT statement only if the GROUP BY clause is

present The optional HAVING clause filters data after the rows are summarized

20 C The GROUP BY and HAVING clauses can appear in any order in the SELECT clause If a

WHERE clause is present, it must be before the GROUP BY clause

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Û N

table using equijoins and nonequijoinsJoin a table to itself by using a self-join

Û N

View data that generally does not meet a join condition by

Û N

using outer joinsGenerate a Cartesian product of all rows from two or

Û N

more tables

Using subqueries to solve queries

ÛÛDefine subqueries

Û N

Describe the types of problems that the subqueries can solve

Û N

List the types of subqueries

Û N

Write single-row and multiple-row subqueries

Û N

Using the Set operators

ÛÛDescribe set operators

Û N

Use a set operator to combine multiple queries into a

Û N

single queryControl the order of rows returned

Û N

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A database has many tables that store data In Chapter 1,

“Introducing SQL,” you learned how to write simple queries that select data from one table Although this information is essential to passing the certification exam, the ability to join two or more related tables and access information is the core strength of relational databases Using the SELECT statement, you can write advanced queries that satisfy user requirements

This chapter focuses on querying data from more than one table using table joins and subqueries When you use two or more tables or views in a single query, it is a join query

You’ll need to understand how the various types of joins and subqueries work, as well as the proper syntax, for the certification exam

Set operators in Oracle let you combine results from two or more SELECT statements

The results of each SELECT statement are considered a set, and Oracle provides UNION, INTERSECT, and MINUS operators to get the desired results You will learn how these opera-tors work in this chapter

Writing Multiple-Table Queries

In relational database management systems (RDBMSs), related data can be stored in tiple tables You use the power of SQL to relate the information and query data A SELECTstatement has a mandatory SELECT clause and FROM clause The SELECT clause can have

mul-a list of columns, expressions, functions, mul-and so on The FROM clause tells you in which table(s) to look for the required information In Chapter 1, you learned to query data using simple SELECT statements from a single table In this chapter, you will learn how to retrieve data from more than one table

To query data from more than one table, you need to identify common columns that relate the two tables Here’s how you do it:

1. In the SELECT clause, you list the columns you are interested in from all the related tables

2. In the FROM clause, you include all the table names separated by commas

3. In the WHERE clause, you define the relationship between the tables listed in the FROMclause using comparison operators

You can also specify the relationship using a JOIN clause instead of the WHERE clause The

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Writing Multiple-Table Queries 199

SQL1999 standard Throughout this section, you’ll see examples of queries using the cle native syntax as well as the ISO/ANSI SQL1999 standard A query from multiple tables without a relationship or common column is known as a Cartesian join or cross join and is discussed later in this chapter

Ora-A join is a query that combines rows from two or more tables or views Oracle performs

a join whenever multiple tables appear in the query’s FROM clause The query’s SELECT clause can have the columns or expressions from any or all of these tables

If multiple tables have the same column names, the duplicate column names should be qualified in the queries with their table name or table alias

Inner Joins

Inner joins return only the rows that satisfy the join condition The most common operator used to relate two tables is the equality operator (=) If you relate two tables using an equal-

ity operator, it is an equality join, also known as an equijoin This type of join combines

rows from two tables that have equivalent values for the specified columns

Simple Inner Joins

A simple inner join has only the join condition specified, without any other filtering tions For example, let’s consider a simple join between the DEPARTMENTS and LOCATIONStables of the HR schema The common column in these tables is LOCATION_ID You will query these tables to get the location ID, city name, and department names in that city:

condi-SELECT locations.location_id, city, department_nameFROM locations, departments

WHERE locations.location_id = departments.location_id;

Here, you are retrieving data from two tables—two columns from the LOCATIONS table and one column from the DEPARTMENTS table These two tables are joined in the WHERE clause using an equality operator on the LOCATION_ID column It is not necessary for the column names in both tables to have the same name to have a join Notice that the LOCATION_IDcolumn is qualified with its table name for every occurrence This is to avoid ambiguity; it is not necessary to qualify each column, but it increases the readability of the query If the same column name appears in more than one table used in the query, you must qualify the column name with the table name or table alias

To execute a join of three or more tables, Oracle takes these steps:

1. Oracle joins two of the tables based on the join conditions, comparing their columns

2. Oracle joins the result to another table, based on join conditions

3. Oracle continues this process until all tables are joined into the result

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200 Chapter 4 N Using Joins and Subqueries

Complex Inner Joins

Apart from specifying the join condition in the WHERE clause, you may have another condition

to limit the rows retrieved Such joins are known as complex joins For example, to

con-tinue with the example in the previous section, if you are interested only in the departments that are outside the United States, use this query:

SELECT locations.location_id, city, department_nameFROM locations, departments

WHERE locations.location_id = departments.location_idAND country_id != ‘US’;

LOCATION_ID CITY DEPARTMENT_NAME - - -

1800 Toronto Marketing

2400 London Human Resources

2700 Munich Public Relations

2500 Oxford Sales

Using Table Aliases

Like columns, tables can have alias names Table aliases increase the readability of the query

You can also use them to shorten long table names with shorter alias names Specify the

table alias name next to the table, separated with a space You can rewrite the query in the

previous section using alias names, as follows:

SELECT l.location_id, city, department_nameFROM locations l, departments d

WHERE l.location_id = d.location_idAND country_id != ‘US’;

When tables (or views or materialized views) are specified in the FROM clause, Oracle looks for the object in the schema (or user) connected to the database If the table belongs

to another schema, you must qualify it with the schema name (You may avoid this by using synonyms, which are discussed in Chapter 7, “Creating Schema Objects.”) You can use the schema owner to qualify a table; you can also use the table owner and schema owner to qualify a column Here is an example:

SELECT locations.location_id, hr.locations.city ,department_name

FROM hr.locations, hr.departmentsWHERE locations.location_id = departments.location_id;

Keep in mind that you can qualify a column name with its schema and table only when

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CITY with the schema HR This is possible only if you qualify the LOCATIONS table with the schema The following SQL will produce an error:

SELECT locations.location_id, hr.locations.city ,department_name

FROM locations, hr.departmentsWHERE locations.location_id = departments.location_id;

SELECT locations.location_id, hr.locations.city *

ERROR at line 1:

ORA-00904: “HR”.”LOCATIONS”.”CITY”: invalid identifierWhen you use table alias names, you must qualify the column names with the alias name only; qualifying the columns with the table name will produce an error, as in this example:

SELECT locations.location_id, city, department_nameFROM locations l, hr.departments d

WHERE locations.location_id = d.location_id;

WHERE locations.location_id = d.location_id *

ERROR at line 3:

ORA-00904: “LOCATIONS”.”LOCATION_ID”: invalid identifierThe correct syntax is to replace locations.location_id with l.location_id in the SELECT and WHERE clauses

If there are no common column names between the two tables used in the join (the FROMclause), you don’t need to qualify the columns However, if you qualify the columns, you are telling the Oracle database engine where exactly to find the column; hence, you are improving the performance of the query

If there are column names common to multiple tables used in a join query, you must qualify the column name with a table name or table alias This is true for column names appearing in SELECT, WHERE, ORDER BY, GROUP BY, and HAVING clauses When using the ANSI syntax, the rule is different The ANSI syntax is discussed in the next section

When joining columns using the traditional syntax or ANSI syntax, if the column datatypes are different, Oracle tries to perform an implicit data-type conversion This may affect your query performance It is better if the columns used in the join condition have the same datatype or if you use the explicit conversion functions you learned in Chapter 2, “Using Single-Row Functions.”

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Using the ANSI Syntax

The difference between traditional Oracle join syntax and the ANSI/ISO SQL1999 syntax

is that in ANSI, the join type is specified explicitly in the FROM clause Using the ANSI tax is clearer and is recommended over the traditional Oracle syntax Simple joins can have the following forms:

syn-<table name> NATURAL [INNER] JOIN syn-<table name>

<table name> [INNER] JOIN <table name> USING (<columns>)

<table name> [INNER] JOIN <table name> ON <condition>

The following sections discuss each of the syntax forms in detail In all three syntaxes, the keyword INNER is optional and is the default

NATURAL JOIN

The NATURAL keyword indicates a natural join, where the join is based on all columns that have

same name in both tables In this type of join, you should not qualify the column names with the table name or table alias name Let’s return to the example of querying the DEPARTMENTSand LOCATIONS tables using LOCATION_ID as the join column The new Oracle syntax is as follows:

SELECT location_id, city, department_nameFROM locations NATURAL JOIN departments;

The common column in these two tables is LOCATION_ID, and that column is used to join the tables When specifying NATURAL JOIN, the columns with the same name in both tables should also have same datatype The following query will return the same results:

SELECT location_id, city, department_nameFROM departments NATURAL JOIN locations;

Notice that even though the LOCATION_ID column is in both tables, you did not qualify this column in the SELECT clause You cannot qualify the column names used for the join when using the NATURAL JOIN clause The following query will result in an error:

SELECT l.location_id, city, department_nameFROM departments NATURAL JOIN locations l;

SELECT l.location_id, city, department_name *

ERROR at line 1:

ORA-25155: column used in NATURAL join cannot have qualifierThe following query will not return an error because the qualifier is used on a column that’s not part of the join condition:

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If you use SELECT *, common columns are listed only once in the result set The following example demonstrates this The common column in the COUNTRIES table and the REGIONStable is the REGION_ID

SQL> DESCRIBE regions Name Null? Type - - - REGION_ID NOT NULL NUMBER REGION_NAME VARCHAR2(25)

SQL> DESCRIBE countries Name Null? Type - - - COUNTRY_ID NOT NULL CHAR(2) COUNTRY_NAME VARCHAR2(40) REGION_ID NUMBER

SELECT *FROM regions NATURAL JOIN countries;

REGION_ID REGION_NAME CO COUNTRY_NAME - - -

1 Europe UK United Kingdom

1 Europe NL Netherlands

1 Europe IT Testing Update

1 Europe FR France

… … …Here is another example, which joins three tables:

SELECT region_id, region_name, country_id, country_name, location_id, city

FROM regionsNATURAL JOIN countriesNATURAL JOIN locations;

When specifying more than two tables using NATURAL JOIN syntax, it is a good idea to use parentheses to increase readability The previous SQL can be interpreted in two ways:

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If you do not use parentheses, Oracle uses left associativity by pairing the tables from left to right (as in the first scenario) By using parentheses, you can make the query less ambiguous, as shown here:

SELECT region_id, region_name, country_id, country_name, location_id, city

FROM locationsNATURAL JOIN (regionsNATURAL JOIN countries);

The same query written in traditional Oracle syntax is as follows:

SELECT regions.region_id, region_name, countries.country_id, country_name, location_id, city

FROM regions, countries, locationsWHERE regions.region_id = countries.region_idAND countries.country_id = locations.country_id;

Though NATURAL JOIN syntax is easy to read and use, its usage should be discouraged in good coding practice Since NATURAL JOIN joins the tables

by all the identical column names, you could end up having a wrong join condition if you’re not careful It is always better to explicitly specify the join condition using the syntaxes available

JOIN…USING

If there are many columns that have the same names in the tables you are joining and they

do not have the same datatype, or you want to specify the columns that should be ered for an equijoin, you can use the JOIN…USING syntax The USING clause specifies the col-umn names that should be used to join the tables Here is an example:

consid-SELECT location_id, city, department_nameFROM locations JOIN departments USING (location_id);

The column names used in the USING clause should not be qualified with a table name or table alias The column names not appearing in the USING clause can be qualified If there are other common column names in the tables and if those column names are used in the query, they must be qualified

Let’s consider this syntax with joining more than two tables:

SELECT region_name, country_name, cityFROM regions

JOIN countries USING (region_id)JOIN locations USING (country_id);

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The following query will result in an error because there is no common column between the REGIONS and LOCATIONS tables:

JOIN locations USING (country_id)JOIN countries USING (region_id);

JOIN locations USING (country_id) *

ERROR at line 3:

ORA-00904: “REGIONS”.”COUNTRY_ID”: invalid identifierYou can add a WHERE clause to limit the number of rows and an ORDER BY clause to sort the rows retrieved along with any type of join operation:

JOIN countries USING (region_id)JOIN locations USING (country_id)WHERE country_id = ‘US’

ORDER BY 1;

Remember that you cannot use alias or table names to qualify the column names on the columns used in the join operation anywhere in the query when using the NATURAL JOIN or JOIN USING syntax You may see ques-tions in the certification exam testing this rule

JOIN…ON

When you do not have common column names between tables to make a join or if you want to specify arbitrary join conditions, you can use the JOIN…ON syntax This syntax specifically defines the join condition using the column names You can qualify column names with a table name or alias name If the column name is common to multiple tables involved in the query, those column names must be qualified

Using the JOIN ON syntax over the traditional join method separates the table joins from the other conditions Since this syntax explicitly states the join condition, it is easier to read and understand Here is the three-table example you used in the previous section, written using the JOIN…ON syntax Notice the use of qualifier on the COUNTRY_ID column; this is required because COUNTRY_ID appears in COUNTRIES and LOCATIONS tables

SELECT region_name, country_name, cityFROM regions r

JOIN countries c ON r.region_id = c.region_idJOIN locations l ON c.country_id = l.country_idWHERE c.country_id = ‘US’;

Tiêu đề	Utilizing Aggregate Functions
Trường học	University of the People
Chuyên ngành	Database Administration
Thể loại	Lecture Notes
Thành phố	Los Angeles

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