Lesson05 从多表中查询数据

Oracle OCP 考试系列培训
之
1Z0-007 Lesson5
www.OracleOnLinux.cn

5-1 Copyright © 2012, http://OracleOnLinux.cn. Part rights reserved.

5
Displaying Data
from Multiple Tables


Objectives

After completing this lesson, you should be able to do
the following:
• Write SELECT statements to access data from
more than one table using equijoins and non-
equijoins
• Join a table to itself by using a self-join
• View data that generally does not meet a join
condition by using outer joins
• Generate a Cartesian product of all rows from two
or more tables


Obtaining Data from Multiple Tables

EMPLOYEES DEPARTMENTS

…

…


Oracle 的多表查询

使用连接从多个表中查询数据;

SELECT last_name,department_name
SELECT last_name,department_name
last_name,department_name
last_name,department_name
FROM employees,departments;
FROM employees,departments;

产生笛卡尔乘积.


连接的类型

Oracle 以前的连接 (8i SQL: 1999
and prior 90SQL):
90SQL): 适应性连接:
等值连接 -- Equijoin 交叉连接 -- Cross joins
非等值连接 -- Non-equijoin 自然连接 -- Natural joins
外连接 -- Outer join 使用Using子句的连接
自连接 -- Self join 完全外连接或者左、右外连接


使用 Oracle 的语法连接多个表

使用连接从多个表中查询数据.
SELECT
SELECT table1.column, table2.column
table1.column, table2.column
FROM
FROM table1, table2
table1, table2
WHERE
WHERE table1.column1 = table2.column2;
table1.column1 = table2.column2;

• 在 WHERE 子句中写入连接条件.
• 当多个表中有重名列时，在列的名字前加上表名作为
前缀.


什么是等值连接?
?

… …

Foreign key Primary key


使用等值连接查询数据

SELECT employees.employee_id, employees.last_name,
employees.department_id, departments.department_id,
departments.location_id
FROM employees, departments
WHERE employees.department_id = departments.department_id;

…


使用 AND 操作符增加查询条件


… …


使用 AND 操作符增加查询条件

employees.department_id, departments.department_id,
FROM employees, departments
WHERE employees.department_id = departments.department_id
AND employees.salary>10000 ;


使用表的别名

• 使用表的别名简化了查询.
• 提供虚拟数据源，自连接中更能体现.

SELECT e.employee_id, e.last_name, e.department_id,
d.department_id, d.location_id
FROM employees e , departments d
WHERE e.department_id = d.department_id;

• 表别名不能与表名混用
d.department_id, departments.location_id
FROM employees e , departments d
WHERE e.department_id = d.department_id;


对多表作等值连接查询
EMPLOYEES DEPARTMENTS LOCATIONS

…

为了连接n个表, 至少需要n-1个连接条件.例如,为了连接三
个表,至少需要两个连接条件.


对多表作等值连接查询
EMPLOYEES DEPARTMENTS LOCATIONS JOBS
JOBS
col job_title for a30
col last_name for a20
col department_name for a20
set linesize 120
SELECT e.last_name,d.department_name,j.job_title,l.city
FROM employees e,departments d,jobs j,locations l
WHERE e.department_id=d.department_id
AND e.job_id=j.job_id
AND d.location_id=l.location_id;
LAST_NAME DEPARTMENT_NAME JOB_TITLE CITY
--------------- ------------------------------ ----------------------------------- ------------------------------
King Executive President Seattle
De Haan Executive Administration Vice President Seattle
Kochhar Executive Administration Vice President Seattle
......


非等值连接
EMPLOYEES JOB_GRADES

在 EMPLOYEES 表中所有薪水
位于JOB_GRADES
JOB_GRADES
JOB_GRADES表最低薪水
… 和最高薪水之间雇员的薪水信
息.
.
除等值连接之外的所有连接都认为是非等值连接。


使用非等值连接查询数据

SELECT e.last_name, e.salary, j.grade_level
FROM employees e, job_grades j
WHERE e.salary
BETWEEN j.lowest_sal AND j.highest_sal;

…


外连接

DEPARTMENTS EMPLOYEES

…
没有雇员属于190部门.


外连接的语法

• 为了看到那些不匹配的数据，必须使用外连接.
• 外连接的操作符为 (+).
(+).
• 不能把连接操作符(+)同时运用在两端(99SQL支持).
• 等值连接"丢失"记录，外连接则显示所有记录.
FROM
FROM table1, table2
table1, table2
WHERE table1.column(+) = table2.column;
WHERE table1.column(+) = table2.column;
table1.column(+)
table1.column(+)

SELECT
FROM
FROM table1, table2
table1, table2
WHERE
WHERE table1.column = table2.column(+);
table1.column = table2.column(+) ;
(+);
table2.column(+)
table2.column(+)
(+);


使用外连接

SELECT e.last_name, e.department_id, d.department_name
FROM employees e, departments d
WHERE e.department_id(+) = d.department_id ;

…


自连接

EMPLOYEES (WORKER) EMPLOYEES (MANAGER)

… …

WORKER表中的MANAGER_ID 等于 MANAGER 表中的
WORKER MANAGER_ID
EMPLOYEE_ID.


对一张表作自连接查询

SELECT worker.last_name || ' works for '
|| manager.last_name
FROM employees worker, employees manager
WHERE worker.manager_id = manager.employee_id ;

…
员工的管理者编号就是管理者作为员工的员工编号


使用 SQL: 1999 语法执行表的连接

使用连接从多个表中查询数据.

SELECT
FROM
FROM table1
table1
[CROSS JOIN table2] |
[CROSS JOIN table2] |
[NATURAL JOIN table2] |
[JOIN table2 USING (column_name)] |
[JOIN table2
[JOIN table2
ON(table1.column_name = table2.column_name)] |
ON(table1.column_name = table2.column_name)] |
[LEFT|RIGHT|FULL OUTER JOIN table2
ON (table1.column_name = table2.column_name)];
ON (table1.column_name = table2.column_name)];


建立交叉连接

• CROSS JOIN 子句产生两个表的交叉连接.
• 产生的结果等于两个表执行笛卡尔乘积.

SELECT last_name, department_name
FROM employees
CROSS JOIN departments ;

…


建立自然连接

• NATURAL JOIN 子句基于两个表中列名完全相同的多
个列产生连接.
• 从两个表中选出连接列的值相等的所有行.
• 如果两个列的名称相同,但是具有不同的数据类型，则
查询会返回一个错误.
– 字段名同，含义不同；（不想连接却连上了）
– 字段名不同，含义同；（想连却连不上）
– 字段名同，含义同；（不可以显示指定不连）
– 不能做非等值连接；


用自然连接查询数据

SELECT department_id, department_name,
location_id, city
FROM departments
NATURAL JOIN locations ;


使用 USING 子句建立连接

• 如果某些列有相同的名称但数据类型不匹配,自然连接
将出错，可以在自然连接的 NATURAL JOIN 子句上
使用 USING 子句来设置用于等值连接的列.
• 不要在引用列上使用表名或者别名作为前缀.
• NATURAL JOIN 与 USING
USING子句是相互独立的.
– 能解决同名不同意的问题；
– 解决不了同意不同名的问题；


使用 USING 子句查询数据

SELECT e.employee_id, e.last_name, d.location_id
FROM employees e JOIN departments d
USING (department_id) ;

…


使用ON
ON
ON子句建立连接

• 自然连接的条件是基于表中所有同名列的等值连接.
• 为了设置任意的连接条件或者指定连接的列，需要使
用ON
ON子句.
ON
• 连接条件与其它的查询条件分开书写.
• 使用ON 子句使查询语句更容易理解.
ON


使用 ON 子句查询数据

ON (e.department_id = d.department_id);

…


使用ON
ON子句建立Three-Way 连接
ON Three-Way

SELECT employee_id, city, department_name
FROM employees e
JOIN departments d
ON d.department_id = e.department_id
JOIN locations l
ON d.location_id = l.location_id;

…


INNER 连接与OUTER 连接的对比
OUTER

• 在SQL: 1999中, 两个表的连接只返回匹配行的被叫做
内连接.
• 两个表的连接结果既包括了内连接，又包括了不匹配
左（右）边表的结果集，也就是左（右）外连接.
• 两个表的连接结果既包含了内连接的结果，也包含了
左右外连接的结果，被叫做完全连接.


左外连接

FROM employees e
LEFT OUTER JOIN departments d
ON (e.department_id = d.department_id) ;

…


右外连接

FROM employees e
RIGHT OUTER JOIN departments d

…


完全外连接

FROM employees e
FULL OUTER JOIN departments d

…

注意完全外连接的执行顺序？


增加其他的查询条件

ON (e.department_id = d.department_id)
AND e.manager_id = 149 ;


Types of Joins

Joins that are compliant with the SQL:1999 standard
include the following:
• Cross joins
• Natural joins
• USING clause
• Full (or two-sided) outer joins
• Arbitrary join conditions for outer joins


Joining Tables Using SQL:1999 Syntax

Use a join to query data from more than one table:
FROM table1
[JOIN table2
ON (table1.column_name = table2.column_name)]|
ON (table1.column_name = table2.column_name)]|
[CROSS JOIN table2];


Creating Natural Joins

• The NATURAL JOIN clause is based on all columns
in the two tables that have the same name.
• It selects rows from the two tables that have equal
values in all matched columns.
• If the columns having the same names have
different data types, an error is returned.


Retrieving Records with Natural Joins

SELECT department_id, department_name,
location_id, city
FROM departments
NATURAL JOIN locations ;


Creating Joins with the USING Clause

• If several columns have the same names but the
data types do not match, the NATURAL JOIN clause
can be modified with the USING clause to specify
the columns that should be used for an equijoin.
• Use the USING clause to match only one column
when more than one column matches.
• Do not use a table name or alias in the referenced
columns.
• The NATURAL JOIN and USING clauses are
mutually exclusive.


Joining Column Names


… …
Foreign key Primary key


Retrieving Records with the USING Clause

departments.location_id, department_id
FROM employees JOIN departments

…


Qualifying Ambiguous
Column Names

• Use table prefixes to qualify column names that
are in multiple tables.
• Use table prefixes to improve performance.
• Use column aliases to distinguish columns that
have identical names but reside in different tables.
• Do not use aliases on columns that are identified
in the USING clause and listed elsewhere in the
SQL statement.


Using Table Aliases

• Use table aliases to simplify queries.
• Use table aliases to improve performance.
SELECT e.employee_id, e.last_name,
d.location_id, department_id


Creating Joins with the ON Clause

• The join condition for the natural join is basically
an equijoin of all columns with the same name.
• Use the ON clause to specify arbitrary conditions
or specify columns to join.
• The join condition is separated from other search
conditions.
• The ON clause makes code easy to understand.


Retrieving Records with the ON Clause

ON (e.department_id = d.department_id);

…


Self-Joins Using the ON Clause

EMPLOYEES (WORKER) EMPLOYEES (MANAGER)

… …

MANAGER_ID in the WORKER table is equal to
EMPLOYEE_ID in the MANAGER table.


Self-Joins Using the ON Clause

SELECT e.last_name emp, m.last_name mgr
FROM employees e JOIN employees m
ON (e.manager_id = m.employee_id);

…


Applying Additional Conditions
to a Join

ON (e.department_id = d.department_id)
AND e.manager_id = 149 ;


Creating Three-Way Joins with the
ON Clause

SELECT employee_id, city, department_name
FROM employees e
JOIN departments d
ON d.department_id = e.department_id
JOIN locations l

…


Non-Equijoins

EMPLOYEES JOB_GRADES

Salary in the EMPLOYEES
table must be between
… lowest salary and highest
salary in the JOB_GRADES
table.


Retrieving Records
with Non-Equijoins

SELECT e.last_name, e.salary, j.grade_level
FROM employees e JOIN job_grades j
ON e.salary
BETWEEN j.lowest_sal AND j.highest_sal;

…


Outer Joins

DEPARTMENTS EMPLOYEES

…

There are no employees in
department 190.


INNER Versus OUTER Joins

• In SQL:1999, the join of two tables returning only
matched rows is called an inner join.
• A join between two tables that returns the results
of the inner join as well as the unmatched rows
from the left (or right) tables is called a left (or
right) outer join.
• A join between two tables that returns the results
of an inner join as well as the results of a left and
right join is a full outer join.


LEFT OUTER JOIN

FROM employees e LEFT OUTER JOIN departments d

…


RIGHT OUTER JOIN

FROM employees e RIGHT OUTER JOIN departments d

…


FULL OUTER JOIN

SELECT e.last_name, d.department_id, d.department_name
FROM employees e FULL OUTER JOIN departments d

…


Cartesian Products

• A Cartesian product is formed when:
– A join condition is omitted
– A join condition is invalid
– All rows in the first table are joined to all rows in the
second table
• To avoid a Cartesian product, always include a
valid join condition.


Generating a Cartesian Product

EMPLOYEES (20 rows) DEPARTMENTS (8 rows)

…

Cartesian product:
20 x 8 = 160 rows

…


Creating Cross Joins

• The CROSS JOIN clause produces the cross-
product of two tables.
• This is also called a Cartesian product between
the two tables.

SELECT last_name, department_name
FROM employees
CROSS JOIN departments ;

…


Summary

In this lesson, you should have learned how to use
joins to display data from multiple tables by using:
• Equijoins
• Non-equijoins
• Outer joins
• Self-joins
• Cross joins
• Natural joins
• Full (or two-sided) outer joins


Practice 5: Overview

This practice covers the following topics:
• Joining tables using an equijoin
• Performing outer and self-joins
• Adding conditions


Practice 5

Q1:Click the Exhibit button to examine the structures of
Q1:Click
the EMPLOYEES, DEPARTMENTS, and TAX tables.


Practice 5
Q1: For which situation would you use a nonequijoin query?

A. to find the tax percentage for each of the employees

B. to list the name, job_id, and manager name for all the
employees

C. to find the name, salary, and the department name of
employees who are not working with Smith

D. to find the number of employees working for the
Administrative department and earning less than 4000

E. to display name, salary, manager ID, and department name
of all the employees, even if the employees do
not have a department ID assigned


Practice 5

Q2:Click the Exhibit button to examine the structures of
Q2:Click
the EMPLOYEES and TAX tables.


Practice 5
Q2: You need to find the percentage tax applicable for
each employee. Which SQL statement would you use?

A. SELECT employee_id, salary, tax_percent
FROM employees e JOIN tax t
ON e.salary BETWEEN t.min_salary AND t.max_salary;

B. SELECT employee_id, salary, tax_percent
WHERE e.salary > t.min_salary AND < t.max_salary;

C. SELECT employee_id, salary, tax_percent
ON (MIN(e.salary) = t.min_salary
AND MAX(e.salary) = t.max_salary);

D. You cannot find the information because there is no
common column between the two tables.


Practice 5

Q3:Examine the structure of the EMPLOYEES and
Q3:Examine
DEPARTMENTS tables:


Practice 5

Q3:You want to create a report displaying
Q3:You
employee last names, department names, and
locations. Which query should you use?
A. SELECT e.last_name, d. department_name, d.location_id
FROM employees e NATURAL JOIN departments D
USING department_id ;

B. SELECT last_name, department_name, location_id
FROM employees NATURAL JOIN departments
WHERE e.department_id =d.department_id;

C. SELECT e.last_name, d.department_name, d.location_id
FROM employees e NATURAL JOIN departments d;

D. SELECT e.last_name, d.department_name, d.location_id
FROM employees e JOIN departments d USING (department_id );


Practice 5

Q4:Click the Exhibit button to examine the structure of the
EMPLOYEES, DEPARTMENTS, and LOCATIONS tables.

Two new departments are added to your company as shown:

DEPARTMENT_ID DEPARTMENT_NAME MGR_ID LOCATION_ID
9998 Engineering 123
9999 Administrative Boston

You need to list the names of employees, the department IDs, the
department names, and the cities where the departments are, even if
there are no employees in the departments and even if the
departments are not yet assigned to a location. You need to join the
EMPLOYEES, DEPARTMENTS, and LOCATIONS tables to retrieve this
information.

Which statement do you execute to retrieve this information?


Practice 5

Q4:


Practice 5
A. SELECT e.last_name, d.department_id, d.department_name,
l.city FROM departments d RIGHT OUTER JOIN employees e ON
d.department_id = e.department_id RIGHT OUTER JOIN locations l

B. SELECT e.last_name, d.department_id, d.department_name,
l.city FROM departments d FULL OUTER JOIN employees e
ON d.department_id = e.department_id FULL OUTER JOIN
locations l ON d.location_id = l.location_id;

C. SELECT e.last_name, d.department_id, d.department_name,
l.city FROM departments d LEFT OUTER JOIN employees e
ON d.department_id = e.department_id LEFT OUTER JOIN
locations l ON d.location_id = l.location_id;

D. SELECT last_name, department_id, department_name, city
FROM departments d NATURAL JOIN employees e NATURAL
JOIN locations l;


Practice 5

Q5:Click the Exhibit button and examine the data in the
EMPLOYEES and DEPARTMENTS tables.


Practice 5

Q5:You want to retrieve all employees' last names, along
with their managers' last names and their department
names. Which query would you use?
A. SELECT last_name, manager_id, department_name FROM employees e
FULL OUTER JOIN departments d ON (e.department_id = d.department_id);

B. SELECT e.last_name, m.last_name, department_name FROM employees
e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id)
LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);

C. SELECT e.last_name, m.last_name, department_name FROM employees
e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id)
LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);


Practice 5

Q5:

D. SELECT e.last_name, m.last_name, department_name FROM employees
e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id)
RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);

E. SELECT e.last_name, m.last_name, department_name FROM employees
e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id)
RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);

F. SELECT last_name, manager_id, department_name FROM employees e
JOIN departments d ON (e.department_id = d.department_id) ;


Lesson05 从多表中查询数据

More Related Content

What's hot

Similar to Lesson05 从多表中查询数据

More from renguzi

Recently uploaded

Lesson05 从多表中查询数据