MySQL多表查询实例详解【链接查询、子查询等】

本文实例讲述了MySQL多表查询。分享给大家供大家参考，具体如下：

准备工作：准备两张表，部门表（department）、员工表（employee）

create table department(
id int,
name varchar(20)
);
create table employee(
id int primary key auto_increment,
name varchar(20),
sex enum('male','female') not null default 'male',
age int,
dep_id int
);

#插入数据
insert into department values
(200,'技术'),
(201,'人力资源'),
(202,'销售'),
(203,'运营');
insert into employee(name,sex,age,dep_id) values
('egon','male',18,200),
('alex','female',48,201),
('wupeiqi','male',38,201),
('yuanhao','female',28,202),
('nvshen','male',18,200),
('xiaomage','female',18,204)
;

# 查看表结构和数据
mysql> desc department;
+-------+-------------+------+-----+---------+-------+
| Field | Type    | Null | Key | Default | Extra |
+-------+-------------+------+-----+---------+-------+
| id  | int(11)   | YES |   | NULL  |    |
| name | varchar(20) | YES |   | NULL  |    |
+-------+-------------+------+-----+---------+-------+
2 rows in set (0.19 sec)

mysql> desc employee;
+--------+-----------------------+------+-----+---------+----------------+
| Field | Type         | Null | Key | Default | Extra     |
+--------+-----------------------+------+-----+---------+----------------+
| id   | int(11)        | NO  | PRI | NULL  | auto_increment |
| name  | varchar(20)      | YES |   | NULL  |        |
| sex  | enum('male','female') | NO  |   | male  |        |
| age  | int(11)        | YES |   | NULL  |        |
| dep_id | int(11)        | YES |   | NULL  |        |
+--------+-----------------------+------+-----+---------+----------------+
5 rows in set (0.01 sec)

mysql> select * from department;
+------+--------------+
| id  | name     |
+------+--------------+
| 200 | 技术     |
| 201 | 人力资源   |
| 202 | 销售     |
| 203 | 运营     |
+------+--------------+
4 rows in set (0.02 sec)

mysql> select * from employee;
+----+----------+--------+------+--------+
| id | name   | sex  | age | dep_id |
+----+----------+--------+------+--------+
| 1 | egon   | male  |  18 |  200 |
| 2 | alex   | female |  48 |  201 |
| 3 | wupeiqi | male  |  38 |  201 |
| 4 | yuanhao | female |  28 |  202 |
| 5 | nvshen  | male  |  18 |  200 |
| 6 | xiaomage | female |  18 |  204 |
+----+----------+--------+------+--------+
6 rows in set (0.00 sec)

ps:观察两张表，发现department表中id=203部门在employee中没有对应的员工，发现employee中id=6的员工在department表中没有对应关系。

一多表链接查询

SELECT 字段列表
FROM 表1 INNER|LEFT|RIGHT JOIN 表2
ON 表1.字段 = 表2.字段;

（1）先看第一种情况交叉连接：不适用任何匹配条件。生成笛卡尔积.—>重复最多

mysql> select * from employee,department;
+----+----------+--------+------+--------+------+--------------+
| id | name   | sex  | age | dep_id | id  | name     |
+----+----------+--------+------+--------+------+--------------+
| 1 | egon   | male  |  18 |  200 | 200 | 技术     |
| 1 | egon   | male  |  18 |  200 | 201 | 人力资源   |
| 1 | egon   | male  |  18 |  200 | 202 | 销售     |
| 1 | egon   | male  |  18 |  200 | 203 | 运营     |
| 2 | alex   | female |  48 |  201 | 200 | 技术     |
| 2 | alex   | female |  48 |  201 | 201 | 人力资源   |
| 2 | alex   | female |  48 |  201 | 202 | 销售     |
| 2 | alex   | female |  48 |  201 | 203 | 运营     |
| 3 | wupeiqi | male  |  38 |  201 | 200 | 技术     |
| 3 | wupeiqi | male  |  38 |  201 | 201 | 人力资源   |
| 3 | wupeiqi | male  |  38 |  201 | 202 | 销售     |
| 3 | wupeiqi | male  |  38 |  201 | 203 | 运营     |
| 4 | yuanhao | female |  28 |  202 | 200 | 技术     |
| 4 | yuanhao | female |  28 |  202 | 201 | 人力资源   |
| 4 | yuanhao | female |  28 |  202 | 202 | 销售     |
| 4 | yuanhao | female |  28 |  202 | 203 | 运营     |
| 5 | nvshen  | male  |  18 |  200 | 200 | 技术     |
| 5 | nvshen  | male  |  18 |  200 | 201 | 人力资源   |
| 5 | nvshen  | male  |  18 |  200 | 202 | 销售     |
| 5 | nvshen  | male  |  18 |  200 | 203 | 运营     |
| 6 | xiaomage | female |  18 |  204 | 200 | 技术     |
| 6 | xiaomage | female |  18 |  204 | 201 | 人力资源   |
| 6 | xiaomage | female |  18 |  204 | 202 | 销售     |
| 6 | xiaomage | female |  18 |  204 | 203 | 运营     |

（2）内连接：只连接匹配的行，以双方为基准

#找两张表共有的部分，相当于利用条件从笛卡尔积结果中筛选出了匹配的结果
#department没有204这个部门，因而employee表中关于204这条员工信息没有匹配出来
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id;
+----+---------+------+--------+--------------+
| id | name  | age | sex  | name     |
+----+---------+------+--------+--------------+
| 1 | egon  |  18 | male  | 技术     |
| 2 | alex  |  48 | female | 人力资源   |
| 3 | wupeiqi |  38 | male  | 人力资源   |
| 4 | yuanhao |  28 | female | 销售     |
| 5 | nvshen |  18 | male  | 技术     |
+----+---------+------+--------+--------------+
5 rows in set (0.00 sec)

#上述sql等同于
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where employee.dep_id=department.id;

（3）外链接之左连接：优先显示左表全部记录

#以左表为准，即找出所有员工信息，当然包括没有部门的员工
#本质就是：在内连接的基础上增加左边有，右边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id;
+----+----------+--------------+
| id | name   | depart_name |
+----+----------+--------------+
| 1 | egon   | 技术     |
| 5 | nvshen  | 技术     |
| 2 | alex   | 人力资源   |
| 3 | wupeiqi | 人力资源   |
| 4 | yuanhao | 销售     |
| 6 | xiaomage | NULL     |
+----+----------+--------------+
6 rows in set (0.00 sec)

（4）外链接之右连接：优先显示右表全部记录

#以右表为准，即找出所有部门信息，包括没有员工的部门
#本质就是：在内连接的基础上增加右边有，左边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id=department.id;
+------+---------+--------------+
| id  | name  | depart_name |
+------+---------+--------------+
|  1 | egon  | 技术     |
|  2 | alex  | 人力资源   |
|  3 | wupeiqi | 人力资源   |
|  4 | yuanhao | 销售     |
|  5 | nvshen | 技术     |
| NULL | NULL  | 运营     |
+------+---------+--------------+
6 rows in set (0.00 sec)

（5）全外连接：显示左右两个表全部记录(了解)

#外连接：在内连接的基础上增加左边有右边没有的和右边有左边没有的结果
#注意：mysql不支持全外连接 full JOIN
#强调：mysql可以使用此种方式间接实现全外连接

语法：select * from employee left join department on employee.dep_id = department.id
union all
select * from employee right join department on employee.dep_id = department.id;

 mysql> select * from employee left join department on employee.dep_id = department.id
     union
    select * from employee right join department on employee.dep_id = department.id
      ;
+------+----------+--------+------+--------+------+--------------+
| id  | name   | sex  | age | dep_id | id  | name     |
+------+----------+--------+------+--------+------+--------------+
|  1 | egon   | male  |  18 |  200 | 200 | 技术     |
|  5 | nvshen  | male  |  18 |  200 | 200 | 技术     |
|  2 | alex   | female |  48 |  201 | 201 | 人力资源   |
|  3 | wupeiqi | male  |  38 |  201 | 201 | 人力资源   |
|  4 | yuanhao | female |  28 |  202 | 202 | 销售     |
|  6 | xiaomage | female |  18 |  204 | NULL | NULL     |
| NULL | NULL   | NULL  | NULL |  NULL | 203 | 运营     |
+------+----------+--------+------+--------+------+--------------+
7 rows in set (0.01 sec)

#注意 union与union all的区别：union会去掉相同的纪录

二、符合条件连接查询

以内连接的方式查询employee和department表，并且employee表中的age字段值必须大于25，即找出年龄大于25岁的员工以及员工所在的部门

select employee.name,department.name from employee inner join department
　　on employee.dep_id = department.id
　　where age > 25;

三、子查询

#1：子查询是将一个查询语句嵌套在另一个查询语句中。
#2：内层查询语句的查询结果，可以为外层查询语句提供查询条件。
#3：子查询中可以包含：IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等关键字
#4：还可以包含比较运算符：= 、 !=、> 、<等

（1）带in关键字的子查询

#查询平均年龄在25岁以上的部门名
select id,name from department
  where id in
    (select dep_id from employee group by dep_id having avg(age) > 25);
# 查看技术部员工姓名
select name from employee
  where dep_id in
    (select id from department where name='技术');
#查看不足1人的部门名
select name from department
  where id not in
    (select dep_id from employee group by dep_id);

（2）带比较运算符的子查询

#比较运算符：=、!=、>、>=、<、<=、<>
#查询大于所有人平均年龄的员工名与年龄
mysql> select name,age from employee where age > (select avg(age) from employee);
+---------+------+
| name  | age |
+---------+------+
| alex  |  48 |
| wupeiqi |  38 |
+---------+------+
#查询大于部门内平均年龄的员工名、年龄

思路：

（1）先对员工表(employee)中的人员分组(group by)，查询出dep_id以及平均年龄。
(2)将查出的结果作为临时表,再对根据临时表的dep_id和employee的dep_id作为筛选条件将employee表和临时表进行内连接。
(3)最后再将employee员工的年龄是大于平均年龄的员工名字和年龄筛选。

mysql> select t1.name,t1.age from employee as t1
       inner join
      (select dep_id,avg(age) as avg_age from employee group by dep_id) as t2
      on t1.dep_id = t2.dep_id
      where t1.age > t2.avg_age;
+------+------+
| name | age |
+------+------+
| alex |  48 |

(3)带EXISTS关键字的子查询

#EXISTS关字键字表示存在。在使用EXISTS关键字时，内层查询语句不返回查询的记录。而是返回一个真假值。True或False
#当返回True时，外层查询语句将进行查询；当返回值为False时，外层查询语句不进行查询
#department表中存在dept_id=203，Ture
mysql> select * from employee where exists (select id from department where id=200);
+----+----------+--------+------+--------+
| id | name   | sex  | age | dep_id |
+----+----------+--------+------+--------+
| 1 | egon   | male  |  18 |  200 |
| 2 | alex   | female |  48 |  201 |
| 3 | wupeiqi | male  |  38 |  201 |
| 4 | yuanhao | female |  28 |  202 |
| 5 | nvshen  | male  |  18 |  200 |
| 6 | xiaomage | female |  18 |  204 |
+----+----------+--------+------+--------+
#department表中存在dept_id=205，False
mysql> select * from employee where exists (select id from department where id=204);
Empty set (0.00 sec)

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