MySQL查询部门工资前三高的所有员工

0x01.需求

Employee 表包含所有员工信息，每个员工有其对应的工号 Id，姓名 Name，工资 Salary 和部门编号 DepartmentId 。

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 85000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
| 5  | Janet | 69000  | 1            |
| 6  | Randy | 85000  | 1            |
| 7  | Will  | 70000  | 1            |
+----+-------+--------+--------------+

Department 表包含公司所有部门的信息。

+----+----------+
| Id | Name     |
+----+----------+
| 1  | IT       |
| 2  | Sales    |
+----+----------+

编写一个 SQL 查询，找出每个部门获得前三高工资的所有员工。例如，根据上述给定的表，查询结果应返回：

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Randy    | 85000  |
| IT         | Joe      | 85000  |
| IT         | Will     | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+

解释：

IT 部门中，Max 获得了最高的工资，Randy 和 Joe 都拿到了第二高的工资，Will 的工资排第三。销售部门（Sales）只有两名员工，Henry 的工资最高，Sam 的工资排第二。

SQL架构：

Create table If Not Exists Employee (Id int, Name varchar(255), Salary int, DepartmentId int)
Create table If Not Exists Department (Id int, Name varchar(255))
Truncate table Employee
insert into Employee (Id, Name, Salary, DepartmentId) values ('1', 'Joe', '85000', '1')
insert into Employee (Id, Name, Salary, DepartmentId) values ('2', 'Henry', '80000', '2')
insert into Employee (Id, Name, Salary, DepartmentId) values ('3', 'Sam', '60000', '2')
insert into Employee (Id, Name, Salary, DepartmentId) values ('4', 'Max', '90000', '1')
insert into Employee (Id, Name, Salary, DepartmentId) values ('5', 'Janet', '69000', '1')
insert into Employee (Id, Name, Salary, DepartmentId) values ('6', 'Randy', '85000', '1')
insert into Employee (Id, Name, Salary, DepartmentId) values ('7', 'Will', '70000', '1')
Truncate table Department
insert into Department (Id, Name) values ('1', 'IT')
insert into Department (Id, Name) values ('2', 'Sales')

0x02.需求分析和解决思路

我们从需求中可以得知：

• 我们的最终表包含三个列，部门，名字，工资
• 这些信息都是各个部门工资最高的前三位

所以，核心问题应该是找出部门工资最高的前三位，思路就是，去重后比这个人工资高的人少于3，这里可以使用两表联结。注意，是各个部门的前三位，所以比较的时候要保证部门ID相等

要将这些人的信息汇总，所以我们可以使用内联结，条件是部门人员信息表中的部门ID等于部门信息表中的ID。

0x03.SQL语句

SELECT 
    d.Name AS 'Department',e1.Name AS 'Employee',e1.Salary
FROM 
    Employee e1
        JOIN 
    Department d ON e1.DepartmentId = d.Id
WHERE 
    3 > (SELECT 
            COUNT(DISTINCT e2.Salary)
        FROM
            Employee e2
        WHERE   
            e2.Salary > e1.Salary
            AND e1.DepartmentId = e2.DepartmentId
    )
;

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