MySQL多表查询的方法主要有以下几种:
1、多表链接查询
(1)笛卡尔积
(2)内链接 inner join
(3)外链接之左链接 left join
(4)外链接之右链接 right join
(5)全外链接
2、符合条件链接查询
3、子查询
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
1、内链接 inner join
只链接匹配的行,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;
上面的查询结果等同于下面:
select employee.id,employee.name,employee.age,employee.sex,department.name
from employee,department
where employee.dep_id=department.id;
2、外链接之左连接left join
优先显示左表全部记录,本质就是:在内连接的基础上增加左边有,右边没有的结果
select employee.id,employee.name,department.name as depart_name
from employee left join department
on employee.dep_id = department.id;
3、外链接之右链接right join
优先显示右表全部记录,本质就是:在内连接的基础上增加右边有,左边没有的结果
select employee.id,employee.name, department.name as depart_name
from employee right join department
on employee.dep_id = department.id;
4、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
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;
或者:
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;
注意 union与union all的区别:union会去掉相同的纪录
5、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属的部门
mysql> select employee.name,employee.age,department.name
-> from employee inner join department
-> on employee.dep_id = department.id
-> where age > 25;
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
6、子查询
子查询是将一个查询语句嵌套在另一个查询语句中
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
子查询中可以包含:in、not in、any、all、exists、not exists 等关键字
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;
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);
# exists为False时:
mysql> select * from employee where exists (select id from department where id=204);