LeetCode-SQL每日一练—184. 部门工资最高的员工

Posted 2022-03-07 莫若心

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文章目录

🐴 1.SQL题目概述

LeetCode原题链接

表： Employee
+--------------+---------+
| 列名          | 类型    |
+--------------+---------+
| id           | int     |
| name         | varchar |
| salary       | int     |
| departmentId | int     |
+--------------+---------+
id是此表的主键列。
departmentId是Department表中ID的外键。
此表的每一行都表示员工的ID、姓名和工资。它还包含他们所在部门的ID。
表： Department
+-------------+---------+
| 列名         | 类型    |
+-------------+---------+
| id          | int     |
| name        | varchar |
+-------------+---------+
id是此表的主键列。
此表的每一行都表示一个部门的ID及其名称。

🚀题目:
编写SQL查询以查找每个部门中薪资最高的员工。
按任意顺序返回结果表。
查询结果格式如下例所示。

输入：
Employee 表:
+----+-------+--------+--------------+
| id | name  | salary | departmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Jim   | 90000  | 1            |
| 3  | Henry | 80000  | 2            |
| 4  | Sam   | 60000  | 2            |
| 5  | Max   | 90000  | 1            |
+----+-------+--------+--------------+
Department 表:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+
输出：
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Jim      | 90000  |
| Sales      | Henry    | 80000  |
| IT         | Max      | 90000  |
+------------+----------+--------+
解释：Max 和 Jim 在 IT 部门的工资都是最高的，Henry 在销售部的工资最高。

🐴 2.解题思路

这个题中Employee和Department 存在主外键的关系，所以直接用内连接即可
因为工资的最高存在多个，所以可以考虑用group by 分组或者排序的方式

🐴 3.方法实现

🌈3.1 Oracle实现

m.Department as “Department”,
m.Employee as “Employee”,
m.Salary as “Salary”
from
(
SELECT b.name AS “DEPARTMENT”,
a.name AS “EMPLOYEE”,
a.Salary,
DENSE_RANK() OVER(partition by a.departmentId ORDER BY a.salary desc) rank
FROM Employee a,Department b
where a.departmentId = b.id
) m where rank = 1

🌈3.2 MySQL实现

select
m.Department,
m.Employee,
m.Salary
from (
SELECT b.name AS “Department”,
a.name AS “Employee”,
a.Salary,
DENSE_RANK() OVER w AS “rank”
FROM Employee a,Department b
where a.departmentId = b.id
WINDOW w AS (partition by a.departmentId ORDER BY a.salary desc)
) m where m.rank=1;