【MySQL】多表查询 -- 2019-08-08 20:38:59
时间：2022-03-15 15:13
原文:
目录
# 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
多表链接查询
- # 外链接语法：
- select 字段列表
- from 表1 inner|left|right join 表2
- on 表1.字段 = 表2.字段;
笛卡尔积：
即交叉链接
不适用任何匹配条件，生成笛卡尔积（关于笛卡尔积的含义，请自行百度）
select * from employee, department; # 语法内链接 inner join
只链接匹配的行
# 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;
外链接之左链接 left join
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)
外链接之右链接 right join
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)
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
- # 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
-
- 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;
-
- 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 (0.00 sec)
-
- # 注意 union与union all的区别：union会去掉相同的纪录
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在的部门
- mysql> select employee.name,employee.age,department.name
- -> from employee inner join department
- -> on employee.dep_id = department.id
- -> where age > 25;
- +---------+-----+--------------+
- | name | age | name |
- +---------+-----+--------------+
- | alex | 48 | 人力资源 |
- | wupeiqi | 38 | 人力资源 |
- | yuanhao | 28 | 销售 |
- +---------+-----+--------------+
- 3 rows in set (0.00 sec)
-
-
- # 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
- +---------+-----+--------------+
- | name | age | name |
- +---------+-----+--------------+
- | egon | 18 | 技术 |
- | nvshen | 18 | 技术 |
- | yuanhao | 28 | 销售 |
- | wupeiqi | 38 | 人力资源 |
- | alex | 48 | 人力资源 |
- +---------+-----+--------------+
- 5 rows in set (0.00 sec)
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、not exists 等关键字.
4. 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;
+------+-----+
| name | age |
+------+-----+
| alex | 48 |
+------+-----+
1 row in set (0.00 sec)
示例3：带exists关键字的子查询
