[LeetCode][SQL]Department Top Three Salaries

Department Top Three Salaries

The Employee table holds all employees. Every employee has an Id, and there is also a column for the department Id.

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
| 5  | Janet | 69000  | 1            |
| 6  | Randy | 85000  | 1            |
+----+-------+--------+--------------+

The Department table holds all departments of the company.

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

Write a SQL query tofind employees who earn the top three salaries in each of the department. For the above tables, your SQL query should return the following rows.

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Randy    | 85000  |
| IT         | Joe      | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+

https://leetcode.com/problems/department-top-three-salaries/

一开始想得很简单，就像一个循环一样，外层塞一个DepartmentId给内层，内层去找这个部门下的top3的，轻松愉快。

问题是MySQL不支持这语法Orz.

‘This version of MySQL doesn‘t yet support ‘LIMIT & IN/ALL/ANY/SOME subquery‘

1 select D.Name, E1.Name, E1.Salary from Employee E1, Department D where E1.Id in
2 (
3     select E2.Id from Employee E2
4     where E1.DepartmentId = E2.DepartmentId
5     order by E2.Salary desc
6 )
7 and E1.DepartmentId = D.Id
8 order by E1.DepartmentId, E1.Salary desc

然后换了个曲线救国的写法，还是外层塞DepartmentId进去，里层数有几个。

搞了很久都挂在这个case上：

1 insert into Employee values(‘1‘, ‘Joe‘, ‘70000‘, ‘1‘);
2 insert into Employee values(‘2‘, ‘Henry‘, ‘80000‘, ‘1‘);
3 insert into Employee values(‘3‘, ‘Sam‘, ‘80000‘, ‘1‘);
4 insert into Employee values(‘4‘, ‘Max‘, ‘90000‘, ‘1‘);
5
6 insert into Department values(‘1‘, ‘IT‘);

Case期望4条都选出来....

本来是count(*)，最后改成count(distinct(E2.Salary)) 就过了

1 select D.Name, E1.Name, E1.Salary from Employee E1, Department D
2 where (
3     select count(distinct(E2.Salary)) from Employee E2
4     where E1.DepartmentId = E2.DepartmentId
5     and E1.Id <> E2.ID
6     and E1.Salary < E2.Salary
7 ) < 3
8 and E1.DepartmentId = D.Id
9 order by E1.DepartmentId, E1.Salary desc