LeetCode Database 184 部门工资最高的员工

1. 题目描述

Employee 表包含所有员工信息，每个员工有其对应的 Id, salary 和 department Id。

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 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，找出每个部门工资最高的员工。例如，根据上述给定的表格，Max 在 IT 部门有最高工资，Henry 在 Sales 部门有最高工资。

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| Sales      | Henry    | 80000  |
+------------+----------+--------+

2. 题解

2.1. MySQL（MAX子查询）

子查询，找到当前员工所在部门的最高工资，如果当前员工的工资就是最高工资，则输出

SELECT tb2.Name AS Department, tb1.Name AS Employee, tb1.Salary
FROM Employee tb1
JOIN Department tb2 ON tb1.DepartmentId = tb2.Id
WHERE tb1.Salary = (
    -- 找出当前DepartmentId下，最高的工资
    SELECT MAX(Salary) FROM Employee tb3
    WHERE tb1.DepartmentId = tb3.DepartmentId)

2.2. MySQL（GROUP BY MAX）

SELECT tb2.Name AS Department, tb1.Name AS Employee, tb1.Salary
FROM Employee tb1
JOIN Department tb2 ON tb1.DepartmentId = tb2.Id
WHERE (tb1.DepartmentId, tb1.Salary) IN (
    SELECT DepartmentId, MAX(Salary) FROM Employee
    GROUP BY DepartmentId)