MySQL学习10 - 多表查询

• 一、多表连接查询
  • 1.交叉连接
  • 2.内连接
  • 3.外连接之左连接
  • 4.外连接之右连接
  • 5.全外连接
• 二、符合条件连接查询
• 三、子查询
  • 1.带in关键字的子查询
  • 2.带比较运算符的子查询
  • 3.带EXISTS关键字的子查询

本节重点：

• 多表连接查询
• 符合条件连接查询
• 子查询

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、多表连接查询

两张表的准备工作已完成，比如现在我要查询的员工信息以及该员工所在的部门。从该题中，我们看出既要查员工又要查该员工的部门，肯定要将两张表进行连接查询，多表连接查询。

重点：外链接语法

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;

1.交叉连接

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（关于笛卡尔积的含义，大家百度自行补脑）。

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

内连接：只连接匹配的行

#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;

3.外连接之左连接

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.外连接之右连接

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

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

示例1：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;

示例2：以内连接的方式查询employee和department表，并且以age字段的升序方式显示。

select employee.id,employee.name,employee.age,department.name from employee,department
    where employee.dep_id = department.id
    and age > 25
    order by age asc;

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)

小练习：

查询每个部门最新入职的那位员工

#创建表
create table employee(
id int not null unique auto_increment,
name varchar(20) not null,
sex enum('male','female') not null default 'male', #大部分是男的
age int(3) unsigned not null default 28,
hire_date date not null,
post varchar(50),
post_comment varchar(100),
salary double(15,2),
office int, #一个部门一个屋子
depart_id int
);

#查看表结构
mysql> desc employee;
+--------------+-----------------------+------+-----+---------+----------------+
| Field        | Type                  | Null | Key | Default | Extra          |
+--------------+-----------------------+------+-----+---------+----------------+
| id           | int(11)               | NO   | PRI | NULL    | auto_increment |
| name         | varchar(20)           | NO   |     | NULL    |                |
| sex          | enum('male','female') | NO   |     | male    |                |
| age          | int(3) unsigned       | NO   |     | 28      |                |
| hire_date    | date                  | NO   |     | NULL    |                |
| post         | varchar(50)           | YES  |     | NULL    |                |
| post_comment | varchar(100)          | YES  |     | NULL    |                |
| salary       | double(15,2)          | YES  |     | NULL    |                |
| office       | int(11)               | YES  |     | NULL    |                |
| depart_id    | int(11)               | YES  |     | NULL    |                |
+--------------+-----------------------+------+-----+---------+----------------+

#插入记录
#三个部门：教学，销售，运营
insert into employee(name,sex,age,hire_date,post,salary,office,depart_id) values
('egon','male',18,'20170301','老男孩驻沙河办事处外交大使',7300.33,401,1), #以下是教学部
('alex','male',78,'20150302','teacher',1000000.31,401,1),
('wupeiqi','male',81,'20130305','teacher',8300,401,1),
('yuanhao','male',73,'20140701','teacher',3500,401,1),
('liwenzhou','male',28,'20121101','teacher',2100,401,1),
('jingliyang','female',18,'20110211','teacher',9000,401,1),
('jinxin','male',18,'19000301','teacher',30000,401,1),
('成龙','male',48,'20101111','teacher',10000,401,1),

('歪歪','female',48,'20150311','sale',3000.13,402,2),#以下是销售部门
('丫丫','female',38,'20101101','sale',2000.35,402,2),
('丁丁','female',18,'20110312','sale',1000.37,402,2),
('星星','female',18,'20160513','sale',3000.29,402,2),
('格格','female',28,'20170127','sale',4000.33,402,2),

('张野','male',28,'20160311','operation',10000.13,403,3), #以下是运营部门
('程咬金','male',18,'19970312','operation',20000,403,3),
('程咬银','female',18,'20130311','operation',19000,403,3),
('程咬铜','male',18,'20150411','operation',18000,403,3),
('程咬铁','female',18,'20140512','operation',17000,403,3)
;

小练习答案：

select * from employee as t1
inner join
(select post,max(hire_date) as new_date from employee group by post) as t2
on t1.post=t2.post
where t1.hire_date=t2.new_date;