mysql 多表查询

查询数据表

#创建相关表
create table department(
id int,
name varchar(20) 
);

create table employee(
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,'HR'),
(202,'销售'),
(203,'运营');

insert into employee(name,sex,age,dep_id) values
('jack','male',18,200),
('toms','female',48,201),
('rusa','male',38,201),
('lura','female',28,202),
('huos','male',18,200),
('aule','female',18,204)
;

#查看表结构与表数据
mysql> desc department;
mysql> desc employee;
mysql> select * from department;
mysql> select * from employee;

多表连接查询

交叉连接:不适用任何匹配条件,生成笛卡尔积。

mysql> select * from employee,department; 

内连接: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 employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id; 
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

语法:SELECT 字段列表 FROM 表1 LEFT|RIGHT JOIN 表2 ON 表1.字段 = 表2.字段;

外链接之左连接: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,其本质就是: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; 

外链接之右连接: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其本质就是: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;

全外连接:显示左右两个表全部记录

是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但是注意:mysql不支持全外连接 full JOIN。mysql可以使用此种方式间接实现全外连接

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;
ps:union与union all的区别:union会去掉相同的纪录

符合条件连接查询

#例一: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;

#例二: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,department
where employee.dep_id = department.id
and age > 25
order by age asc;

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 和 NOT EXISTS等关键字
  • 还可以包含比较运算符:= 、 !=、> 、< 等

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);

#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);

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);

#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;

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);

#department表中存在dept_id=205,False
mysql> select * from employee where exists (select id from department where id=204);

 

posted @ 2019-02-18 14:48  Coolc  阅读(163)  评论(0编辑  收藏  举报