MySQL部门工资前三高的所有员工-蒲公英云

MySQL部门工资前三高的所有员工

SQL架构
题目描述
题解
- 方法：使用 JOIN 和子查询
- 解题思路

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));
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');
insert into Department (Id, Name) values ('1', 'IT');
insert into Department (Id, Name) values ('2', 'Sales');

题目描述

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。

题解

方法：使用 JOIN 和子查询

算法

公司里前 3 高的薪水意味着有不超过 3 个工资比这些值大。

select e1.Name as 'Employee', e1.Salary
from Employee e1
where 3 >
(
    select count(distinct e2.Salary)
    from Employee e2
    where e2.Salary > e1.Salary
);

在这个代码里，我们统计了有多少人的工资比 e1.Salary 高，所以样例的输出应该如下所示。

+----------+--------+
| Employee | Salary |
+----------+--------+
| Joe      |  85000 |
| Henry    |  80000 |
| Max      |  90000 |
| Randy    |  85000 |
+----------+--------+
4 rows in set (0.00 sec)

然后，我们需要把表 Employee 和表 Department 连接来获得部门信息。

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
        );
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Joe      |  85000 |
| Sales      | Henry    |  80000 |
| Sales      | Sam      |  60000 |
| IT         | Max      |  90000 |
| IT         | Randy    |  85000 |
| IT         | Will     |  70000 |
+------------+----------+--------+
6 rows in set (0.00 sec)
SELECT
    Department.NAME AS Department,
    e1.NAME AS Employee,
    e1.Salary AS Salary 
FROM
    Employee AS e1,Department 
WHERE
    e1.DepartmentId = Department.Id 
    AND 3 > (SELECT  count( DISTINCT e2.Salary ) 
             FROM    Employee AS e2 
             WHERE    e1.Salary < e2.Salary     AND e1.DepartmentId = e2.DepartmentId     ) 
ORDER BY Department.NAME,Salary DESC;

解题思路

select 
    d.name as Department, e1.name as employee, e1.salary as salary
from 
    Department d 
    join Employee e1 on d.id = e1.departmentid
    join Employee e2 on e1.departmentid = e2.departmentid and e1.salary<=e2.salary
group by 
    d.name, e1.name
having 
    count(distinct e2.salary)<=3
order by 
    d.name, e1.salary desc;
+------------+----------+--------+
| Department | employee | salary |
+------------+----------+--------+
| IT         | Max      |  90000 |
| IT         | Joe      |  85000 |
| IT         | Randy    |  85000 |
| IT         | Will     |  70000 |
| Sales      | Henry    |  80000 |
| Sales      | Sam      |  60000 |
+------------+----------+--------+
6 rows in set (0.00 sec)
SELECT     
   d.NAME department, t.NAME employee, salary 
FROM
   ( 
      SELECT *, 
      @r := IF(@pD = departmentid, IF(@pS = salary, @r, @r + 1 ), 1 ) AS 'rank',
      @pD := departmentid,
      @pS := salary 
     FROM 
      Employee, ( SELECT @pS := NULL, @pD := NULL, @r := 0 ) init 
     ORDER BY
      departmentid, salary DESC 
        ) t
    JOIN Department d ON t.departmentid = d.id 
WHERE
   t.rank <=3;