MySQL部门工资前三高的所有员工

「爱情、让人受尽委屈。」 2023-10-06 18:50 7阅读 0赞

#### 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;