mysql之多表查询

#部门表
create table department(
id int,
name varchar(20)
);
#员工表
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
);

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),
(‘liwenzhou‘,‘male‘,18,200),
(‘jingliyang‘,‘female‘,18,204)
 

select 字段 from 表1 inner|left|right join 表2 on 表1.字段 = 表2.字段;

select * from department,employee;
select * from employee,department;

select * from employee,department where employee.dep_id = department.id;
select employee.name from employee,department where employee.dep_id = deparment.id and department.name = ‘技术‘;

select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id = department.id;

select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id = department.id;

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;

select employee.name,department.name from employee inner join department on employee.dep_id = department.id where age > 25;

select employee.id,employee.name,employee.age,department.name from employee,department where employee.dep_id = department.id and age > 25 order by age desc;

select * from employee inner join department on employee.dep_id = department.id;
select employee.name from employee inner join department on employee.id = department.id where department.name = ‘技术‘;

select id from department where name = ‘技术‘;
select name from employee where dep_id = (select id from department where name = ‘技术‘);

select id,name from department where id in (select dep_id from employee group by dep_id having avg(age) > 25);
select department.name from department inner join employee on department.id = employee.dep_id group by department.name having avg(age)>25;
select name from employee where dep_id in (select id from department where name = ‘技术‘);
select name from department where id not in (select distinct dep_id from employee);

select name,age from employee where age > (select avg(age) from employee);

select * from employee where exists (select id from department where id = 200);
select * from employee where exists (select id from department where id = 204);