在开始之前，我们需要建立表，做建表和数据的准备的工作。

1.建表
create table department(
	id int,
    name varchar(20)
);
create table employee(
	id int primary key auto_increment,
    name varchar(20),
    sex enmu('male','female') not null default = 'male',
    age int,
    dep_id int,
    constraint fk_id foreign key(dep_id) references department(id),
);

2.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
desc department;
desc employee;

# 查看我们插入的数据
select * from department;
select * from employee;

多表的查询，我们主要是分为两种查询，这两种查询分别是多表连接查询和子查询两种情况。

多表连接查询

1.交叉连接：不适用任何匹配条件。生成笛卡尔积。

select * from employee,department;

2.内连接

# 找两张表共有部分，相当于利用笛卡儿积结果中筛选出正确的结果。
# 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
select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id;

上面的sql语句等价于下面的sql语句
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：优先显示左表全部记录

# 以左表为标准，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而右边没有的结果
select employee.id,employee.name,department.name as depart_name from employee letf join department on employee.dep_id=department.id;

4.外连接之右连接：优先显示右表全部记录

# 以右表为标准，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而左边没有的结果
select employee.id,employee.name,department.name as depart_name from employee right join department on empolyee.dep_id=department.id; 

5.全外连接：显示左右两个表全部的记录

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;
#注意 union与union all的区别：union会去掉相同的纪录

6.符合条件连接查询

# 实例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 inner join department on employee.dep_id=department.id order by age asc;

子查询

# 1.子查询是将一个查询语句嵌套在另外一个查询语句中。
# 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);
这两个sql语句是等价的。
 select department.name from employee inner join department on employee.dep_id=department.id where employee.age > (select avg(age) from employee group by dep_id);
#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 | 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)