MySQL怎么完成多表查询？MySQL多表查询的语句

创建表

# 创建表
create table department(id int,name varchar(20));
create table employee1(
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 employee1(name,sex,age,dep_id) values('egon','male',18,200),('alex','female',48,201),('tom','male',38,201),('yuanhao','female',28,202),('lidawei','male',18,200),('jinkezhou','female',18,204);

# 查看表
mysql> select * from employee1;
+----+-----------+--------+------+--------+
  id   name        sex      age    dep_id  
+----+-----------+--------+------+--------+
   1   egon        male       18      200  
   2   alex        female     48      201  
   3   tom         male       38      201  
   4   yuanhao     female     28      202  
   5   lidawei     male       18      200  
   6   jinkezhou   female     18      204  
+----+-----------+--------+------+--------+
6 rows in set (0.00 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 (0.00 sec)

多表连接查询

交叉连接

交叉连接：不适用任何匹配条件。生成笛卡尔积

mysql> select * from employee1 ,department;

内连接

内连接：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正确的结果。（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
mysql> select * from employee1,department where employee1.dep_id=department.id;

#上面用where表示的可以用下面的内连接表示，建议使用下面的那种方法
mysql> select * from employee1 inner join department on employee1.dep_id=department.id;

# 也可以这样表示哈
mysql> select employee1.id,employee1.name,employee1.age,employee1.sex,department.name from employee1,department where employee1.dep_id=department.id;

左连接left

优先显示左表全部记录。

#左链接：在按照on的条件取到两张表共同部分的基础上，保留左表的记录
mysql> select * from employee1 left join department on department.id=employee1.dep_id;

mysql> select * from department left join  employee1 on department.id=employee1.dep_id;

右连接right

优先显示右表全部记录。

#右链接：在按照on的条件取到两张表共同部分的基础上，保留右表的记录
mysql> select * from employee1 right join department on department.id=employee1.dep_id;
mysql> select * from department right join employee1 on department.id=employee1.dep_id;

全部连接join

mysql> select * from department full join employee1;

符合条件多表查询

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,
即找出公司所有部门中年龄大于25岁的员工

mysql> select * from employee1 inner join department on employee1.dep_id=department.id and 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

mysql> select * from employee1 inner join department on employee1.dep_id=department.id and age>25 and age>25 order by age asc;

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

示例：

# 查询平均年龄在25岁以上的部门名
mysql> select name from department where id in ( select dep_id from employee1 group by dep_id having avg(age) > 25 );

# 查看技术部员工姓名
mysql> select name from employee1 where dep_id = (select id from department where name='技术');

# 查看小于2人的部门名
mysql> select name from department where id in (select dep_id from employee1 group by dep_id having count(id) < 2) union select name from department where id not in (select distinct dep_id from employee1);
# 提取空部门                              
#有人的部门
mysql> select * from department where id not in (select distinct dep_id from employee1);

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