选择每个薪水高于其部门平均水平的员工 [英] SELECT every employee that has a higher salary than the AVERAGE of his department

问题描述

我的数据库中只有1个名为 EMPLOYEE 的表，下面是3个列:

I have only 1 table named EMPLOYEE on my database with the 3 following collumns:

Employee_Name, Employee_Salary, Department_ID

现在，我必须选择每位薪水高于其部门平均水平的员工. 我该怎么做?

Now I have to SELECT every employee that has a higher salary than the AVERAGE of his department. How do I do that?

我遇到的主要问题是，将每个Employee_Salary与

The main problem that I have is that when comparing each Employee_Salary with

SELECT  AVG(department_ID) FROM employee GROUP BY Department_ID

内部队列的返回集返回多行.

the return set of the inner queue returns multiple rows.

我认为我需要执行加入操作，但是我不知道如何做.

I think I need to perform a join operation but I do not know how.

推荐答案

请尝试以下查询

Select * from employee a where Employee_Salary > (select avg(Employee_Salary) from 
employee b group by Department_ID having b.Department_ID = a.Department_ID)

或

Select * from employee a where Employee_Salary> (select avg(Employee_Salary) from 
employee b where b.Department_ID = a.Department_ID group by Department_ID)

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