Description:
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 | 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 |
+----+-------+--------+--------------+
The Department table holds all departments of the company.
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
Write a SQL query to find employees who earn the top three salaries in each of the department. For the above tables, your SQL query should return the following rows (order of rows does not matter).
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Randy | 85000 |
| IT | Joe | 85000 |
| IT | Will | 70000 |
| Sales | Henry | 80000 |
| Sales | Sam | 60000 |
+------------+----------+--------+
Explanation:
In IT department, Max earns the highest salary, both Randy and Joe earn the second highest salary, and Will earns the third highest salary. There are only two employees in the Sales department, Henry earns the highest salary while Sam earns the second highest salary.
Solutions: From the Official Solution Section
MySQL: 生成两个初始DB，互相比较放到WHERE条件里，然后Join On
SELECT Department.Name AS Department, e1.Name AS Employee, e1.Salary
FROM Employee e1 JOIN Department
ON e1.DepartmentId = Department.Id
WHERE 3 > (
SELECT COUNT(DISTINCT e2.Salary)
FROM Employee e2
WHERE e2.Salary > e1.Salary AND e1.DepartmentId = e2.DepartmentId
)
;
MS SQL: 用命令转化一个DB到有结果的版本，然后再Join On
SELECT Department.Name AS Department, a.Name AS Employee, a.Salary
FROM (
SELECT e.*, DENSE_RANK() OVER (PARTITION BY e.DepartmentId ORDER BY e.Salary DESC) AS Rank
FROM Employee e) AS a JOIN Department
ON a.DepartmentId = Department.Id
WHERE a.Rank <= 3
;
