Question

我正在尝试解决以下问题：雇员表保存所有雇员。每个员工都有一个ID，一个薪水，并且还有一个部门ID列。

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
+----+-------+--------+--------------+

“部门”表包含公司的所有部门。

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

编写一个SQL查询以查找每个部门中薪水最高的员工。对于上表，Max在IT部门中的薪水最高，而Henry在销售部门中的薪水最高。但是我想知道为什么下面的mysql查询不能得到正确的结果。

    SELECT d.Name as Department, e.Name as Employee, e.Salary FROM Employee e
    left join Department d on e.DepartmentId=d.Id
    GROUP BY d.Name
    order by e.Salary desc limit 1

Answer 1

此查询将执行您想要的操作：

SELECT d.Name AS Department, e.Name AS Employee, e.Salary
FROM Department d
JOIN Employee e
ON e.DepartmentID = d.Id AND 
   e.Salary = (SELECT MAX(Salary) 
               FROM Employee e2
               WHERE e2.DepartmentId = d.Id);

它将部门表与该部门中薪水最高的该部门的雇员相连。

输出：

Department  Employee    Salary
Sales       Henry       80000
IT          Max         90000

Answer 2

尝试一下：

SET @curRank = 0;  

SELECT * FROM (

SELECT  @curRank := @curRank + 1 AS rank, salary, e.id, e.name, DepartmentId
FROM dept d
 JOIN emp e ON e.DepartmentId = d.Id 
 ORDER BY salary DESC 

) AS tbl 

GROUP BY  departmentID

这将起作用。

Leetcode ---部门最高薪水

2 个答案: