LeetCode MySQL 185. 部门工资前三高的所有员工(dense_rank)

文章目录

1. 题目

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。

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/department-top-three-salaries
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2. 解题

# Write your MySQL query statement below
select Department, Employee, Salary
from
(
    select d.Name Department, e.Name Employee, Salary,
        dense_rank() over(partition by d.Name order by Salary desc) rnk
    from Employee e left join Department d
    on e.DepartmentId = d.Id
) t
where rnk <= 3 and Department is not null

or

# Write your MySQL query statement below
select Department, Employee, Salary
from
(
    select d.Name Department, e.Name Employee, Salary,
        dense_rank() over(partition by d.Name order by Salary desc) rnk
    from Employee e right join Department d
    on e.DepartmentId = d.Id
) t
where rnk <= 3 and Employee is not null

or 使用内连接,自动去除 null 的行

select Department, Employee, Salary
from
(
    select d.Name Department, e.Name Employee, Salary,
        dense_rank() over(partition by d.Name order by Salary desc) rnk
    from Employee e join Department d
    on e.DepartmentId = d.Id
) t
where rnk <= 3

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转载自blog.csdn.net/qq_21201267/article/details/107736699