LeetCode（185）：部门工资前三高的所有员工 Department Top Three Salaries（SQL）

2020.12.23 LeetCode 从零单刷个人笔记整理（持续更新）

github：https://github.com/ChopinXBP/LeetCode-Babel

传送门：部门工资前三高的所有员工

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。

---

#每个部门前三高的薪水说明该部门不超过三个人比这一薪水值大
SELECT e1.Salary
FROM Employee e1
WHERE 3 > (
    SELECT COUNT(DISTINCT e2.Salary)
    FROM Employee e2
    WHERE e1.Salary < e2.Salary AND e1.DepartmentId = e2.DepartmentId
);

#再将该查询得到的薪水与原表连接即可
SELECT d.Name AS 'Department', e1.Name AS 'Employee', e1.Salary
FROM Employee e1
INNER JOIN Department d
ON e1.DepartmentId = d.Id
WHERE 3 > (
    SELECT COUNT(DISTINCT e2.Salary)
    FROM Employee e2
    WHERE e2.Salary > e1.Salary AND e1.DepartmentId = e2.DepartmentId
);

#自定义变量法
SELECT dep.Name Department, emp.Name Employee, emp.Salary
FROM (## 自定义变量RANK, 查找出 每个部门工资前三的排名
        SELECT te.DepartmentId, te.Salary,
               CASE
                    WHEN @pre = DepartmentId THEN @rank:= @rank + 1
                    WHEN @pre := DepartmentId THEN @rank:= 1
               END AS RANK
        FROM (SELECT @pre:=null, @rank:=0)tt,
             (## (部门,薪水)去重,根据 部门(升),薪水(降) 排序
                 SELECT DepartmentId,Salary
                 FROM Employee
                 GROUP BY DepartmentId,Salary
                 ORDER BY DepartmentId,Salary DESC
             )te
       )t
INNER JOIN Department dep ON t.DepartmentId = dep.Id
INNER JOIN Employee emp ON t.DepartmentId = emp.DepartmentId and t.Salary = emp.Salary and t.RANK <= 3
ORDER BY t.DepartmentId, t.Salary DESC ## t 结果集已有序,根据该集合排序

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