【MySQL】多表查询

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目录

多表链接查询

笛卡尔积

内链接 inner join

外链接之左链接 left join

外链接之右链接 right join

全外链接

符合条件链接查询

子查询

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先准备两张表：部门表（department)、员工表（employee)

# 部门表





create table department(





    id int primary key auto_increment,





    name varchar(20) not null





    );





 





# 员工表





create table employee(





    id int primary key auto_increment,





    name varchar(20) not null,





    sex enum('male', 'female') not null default 'male',





    age int not null,





    dep_id int not null





    );





 





# 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)





;





 





# 注意：





department表中id=203的部门在employee中没有对应的员工





employee表中id=6的员工在department表中没有对应的部门

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多表链接查询

# 外链接语法：





select 字段列表





    from 表1 inner|left|right join 表2





    on 表1.字段 = 表2.字段;

笛卡尔积：

即交叉链接

不适用任何匹配条件，生成笛卡尔积（关于笛卡尔积的含义，请自行百度）

select * from employee, department;    # 语法

内链接 inner join

只链接匹配的行

# 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;

外链接之左链接 left join

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)

外链接之右链接 right join

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)

全外链接

显示左右两个表全部记录

# 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





 





mysql> 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 (0.00 sec)





 





# 注意 union与union all的区别：union会去掉相同的纪录

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符合条件链接查询

# 示例1 以内链接的方式查询：找出年龄大于25岁的员工以及员工所在的部门





mysql> select employee.name,employee.age,department.name





    -> from employee inner join department





    -> on employee.dep_id = department.id





    -> where age > 25;





+---------+-----+--------------+





| name    | age | name         |





+---------+-----+--------------+





| alex    |  48 | 人力资源     |





| wupeiqi |  38 | 人力资源     |





| yuanhao |  28 | 销售         |





+---------+-----+--------------+





3 rows in set (0.00 sec)





 





 





# 示例2 以内链接的方式查询：以age字段的升序方式显示





mysql> select employee.name,employee.age,department.name





    -> from employee inner join department





    -> on employee.dep_id = department.id





    -> order by age asc;    # 升序排序





+---------+-----+--------------+





| name    | age | name         |





+---------+-----+--------------+





| egon    |  18 | 技术         |





| nvshen  |  18 | 技术         |





| yuanhao |  28 | 销售         |





| wupeiqi |  38 | 人力资源     |





| alex    |  48 | 人力资源     |





+---------+-----+--------------+





5 rows in set (0.00 sec)

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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、not exists 等关键字.
4. 还可以包含比较运算符：=、 !=、>、< 等.

示例1：带in关键字的子查询

# 查询平均年龄在25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 id,name from employee





    where dep_id in





        (select id from department where name="技术");





 





# 查无人的部门名





select name from department





    where id not in





        (select dep_id from employee);

示例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





select name,age from employee 





    where age > (select avg(age) from employee);





 





# 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。





 





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 |





+------+-----+





1 row in set (0.00 sec)

示例3：带exists关键字的子查询

# exists关键字表示存在。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





 





# exists为True时：





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 |





+----+----------+--------+-----+--------+





6 rows in set (0.00 sec)





 





# exists为False时：





mysql> select * from employee where exists (select id from department where id=204);





Empty set (0.00 sec)

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