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MySQL多表查询

MySQL之多表查询

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);

 

posted @ 2018-10-16 17:32  别来无恙-  阅读(1926)  评论(0编辑  收藏  举报