MYSQL | 求各个部门里薪水最高的员工（可能有多个员工）

题目

有两张表Employee和Department，要求如题

• employee

±—±------±-------±-------------+
| 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

±—±---------+
| Id | Name |
±—±---------+
| 1 | IT |
| 2 | Sales |
±—±---------+

• ⚠️多个员工薪水都是最高，都需要输出

使用JOIN和IN从句

• 分组自然想到GROUP BY

SELECT
    DepartmentId, MAX(Salary)
FROM
    Employee
GROUP BY DepartmentId;

可能多个员工薪水都最高，但是这里只会输出一个员工
| DepartmentId | MAX(Salary) |
|--------------|-------------|
| 1 | 90000 |
| 2 | 80000 |

• 解法：join两个table，然后查询(DepartmentId, Salary)在temp table中

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