MySQL行(记录)的操作(二) -- 多表查询

数据的准备

#建表
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),
('liwenzhou','male',18,200),
('jingliyang','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 | |
+-------+-------------+------+-----+---------+-------+

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 | |
+--------+-----------------------+------+-----+---------+----------------+

mysql> select * from department;
+------+--------------+
| id | name |
+------+--------------+
| 200 | 技术 |
| 201 | 人力资源 |
| 202 | 销售 |
| 203 | 运营 |
+------+--------------+

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 | liwenzhou | male | 18 | 200 |
| 6 | jingliyang | female | 18 | 204 |
+----+------------+--------+------+--------+

表department与employee

1. 多表连接查询

# 重点: 外连接表的语法

select 字段列表
    from 表1 inner|left|right|join 表2
    on 表1.字段 = 表2.字段;

1.1 笛卡尔积连接: 不适用任何匹配条件,将多张表所有的数据一一对于对应,生产一张大表.

图示笛卡尔积:

代码表示:

mysql> select * from employee,department;
结果:
+----+------------+--------+------+--------+------+--------------+
| 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 | liwenzhou  | male   |   18 |    200 |  200 | 技术         |
|  5 | liwenzhou  | male   |   18 |    200 |  201 | 人力资源     |
|  5 | liwenzhou  | male   |   18 |    200 |  202 | 销售         |
|  5 | liwenzhou  | male   |   18 |    200 |  203 | 运营         |
|  6 | jingliyang | female |   18 |    204 |  200 | 技术         |
|  6 | jingliyang | female |   18 |    204 |  201 | 人力资源     |
|  6 | jingliyang | female |   18 |    204 |  202 | 销售         |
|  6 | jingliyang | female |   18 |    204 |  203 | 运营         |
+----+------------+--------+------+--------+------+--------------+

但是我们知道这样的表太过于冗余,根本不是我们想看到的样子.

它应该时这样的:

我们的目标就是将两个分散出去的表，按照两者之间有关系的字段，能对应上的字段，把两者合并成一张表，这就是多表查询的一个本质。

怎么做呢?

来个where条件来筛选不就行了.

select * from employee, department where employee.dep_id = department.id;
结果:
+----+-----------+--------+------+--------+------+--------------+
| id | name      | sex    | age  | dep_id | id   | name         |
+----+-----------+--------+------+--------+------+--------------+
|  1 | egon      | 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         |
|  5 | liwenzhou | male   |   18 |    200 |  200 | 技术         |
+----+-----------+--------+------+--------+------+--------------+
这就是我们想要的展现形式.

但是有两个问题:

1. 我们左表employee表中的dep_id为204的那个数据没有了，右表department表的id为203的数据没有了，因为我们现在要的就是两表能对应上的数据一起查出来，那个204和203双方对应不上,有些场景可以符合,但还有一些场景就不符合了,如: 我想得到所有人的信息,就算它不属于任何一个职位.

2. 可读性不好,看图:

1.2 内连接: 只连接匹配的行

mysql> select * from employee inner join department on employee.dep_id = department.id  where department.name = '技术';
+----+-----------+------+------+--------+------+--------+
| id | name      | sex  | age  | dep_id | id   | name   |
+----+-----------+------+------+--------+------+--------+
|  1 | egon      | male |   18 |    200 |  200 | 技术   |
|  5 | liwenzhou | male |   18 |    200 |  200 | 技术   |
+----+-----------+------+------+--------+------+--------+

现在解决了第二个问题,把连表的操作和查询的操作分开了,可读性提高了...

1.3 外连接之左连接: 优先显示左表的全部记录

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 | liwenzhou  | 技术         |
|  2 | alex       | 人力资源     |
|  3 | wupeiqi    | 人力资源     |
|  4 | yuanhao    | 销售         |
|  6 | jingliyang | NULL         |
+----+------------+--------------+

现在解决第一个问题, 解决一些需要像上方一样的显示.

1.4 外连接之右连接: 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 | liwenzhou | 技术         |
| NULL | NULL      | 运营         |
+------+-----------+--------------+

1.5 全外连接: 显示左右两表全部记录

全外连接：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
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
;
#查看结果
+------+------------+--------+------+--------+------+--------------+
| id   | name       | sex    | age  | dep_id | id   | name         |
+------+------------+--------+------+--------+------+--------------+
|    1 | egon       | male   |   18 |    200 |  200 | 技术         |
|    5 | liwenzhou  | 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 | jingliyang | female |   18 |    204 | NULL | NULL        |
| NULL | NULL       | NULL   | NULL |   NULL |  203 | 运营         |
+------+------------+--------+------+--------+------+--------------+

#注意 union与union all的区别：union会去掉相同的纪录,而union all 不会去掉相同的记录.
如:
+------+------------+--------+------+--------+------+--------------+
| id   | name       | sex    | age  | dep_id | id   | name         |
+------+------------+--------+------+--------+------+--------------+
|    1 | egon       | 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         |
|    5 | liwenzhou  | male   |   18 |    200 |  200 | 技术         |
| NULL | NULL       | NULL   | NULL |   NULL |  203 | 运营         |
|    1 | egon       | male   |   18 |    200 |  200 | 技术         |
|    5 | liwenzhou  | 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 | jingliyang | female |   18 |    204 | NULL | NULL         |
+------+------------+--------+------+--------+------+--------------+

2. 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 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(子查询得到的是有人的部门id)
select name from department where id not in (select distinct dep_id from employee);

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 emp where age > (select avg(age) from emp);
+---------+------+
| name    | age  |
+---------+------+
| alex    | 48   |
| wupeiqi | 38   |
+---------+------+
2 rows in set (0.00 sec)


#查询大于部门内平均年龄的员工名、年龄
select t1.name,t1.age from emp t1
inner join 
(select dep_id,avg(age) avg_age from emp group by dep_id) t2
on t1.dep_id = t2.dep_id
where t1.age > t2.avg_age; 

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.

返回True时,外层查询语句可以运行查询,如果是False时,外层查询语句不运行查询.

#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 | liwenzhou  | male   |   18 |    200 |
|  6 | jingliyang | 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)