182、列出各个部门中工资高于本部门的平均工资的员工数和部门号，并按部门号排序

johnny

182、列出各个部门中工资高于本部门的平均工资的员工数和部门号，并按部门号排序

创建表：

mysql> create table employee921(id int primary key auto_increment,name varchar(5

0),salary bigint,deptid int);

 

插入实验数据：

mysql> insert into employee921 values(null,'zs',1000,1),(null,'ls',1100,1),(null

,'ww',1100,1),(null,'zl',900,1) ,(null,'zl',1000,2), (null,'zl',900,2) ,(null,'z

l',1000,2) , (null,'zl',1100,2);

 

编写sql语句：

 

（）select avg(salary) from employee921 group by deptid;

（）mysql> select employee921.id,employee921.name,employee921.salary,employee921.dep

tid tid from  employee921 where salary > (select avg(salary) from employee921 where  deptid = tid);

效率低的一个语句，仅供学习参考使用（在group by之后不能使用where，只能使用having，在group by之前可以使用where，即表示对过滤后的结果分组）：

mysql> select employee921.id,employee921.name,employee921.salary,employee921.dep

tid tid from  employee921 where salary > (select avg(salary) from employee921 group by deptid having deptid = tid);

（）select count(*) ,tid

from (

select employee921.id,employee921.name,employee921.salary,employee921.deptid tid

from        employee921

where salary >

(select avg(salary) from employee921 where  deptid = tid)

) as t

group by tid ;

 

另外一种方式：关联查询

select a.ename,a.salary,a.deptid

from emp a,

(select deptd,avg(salary) avgsal from emp group by deptid ) b

where a.deptid=b.deptid and a.salary>b.avgsal