Employee
表包含所有员工信息,每个员工有其对应的 Id, salary 和 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 | +----+-------+--------+--------------+
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,找出每个部门工资前三高的员工。例如,根据上述给定的表格,查询结果应返回:
+------------+----------+--------+ | Department | Employee | Salary | +------------+----------+--------+ | IT | Max | 90000 | | IT | Randy | 85000 | | IT | Joe | 70000 | | Sales | Henry | 80000 | | Sales | Sam | 60000 | +------------+----------+--------+
例1:
- # Write your MySQL query statement below
- Select d.Name as Department, e.Name as Employee, e.Salary
- from Department d, Employee e
- where b.DepartmentId = d.Id and (
- Select count(distinct Salary) From Employee where DepartmentId=d.Id and Salary > e.Salary
- )<3
- order by Department
例2:
- # Write your MySQL query statement below
- SELECT d.Name AS Department, e1.Name AS Employee, e1.Salary
- FROM Employee e1 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);
以例2:基本思路是通过两个employee的表,进行对比,约束条件是 departmentid 相等and salary 最大的。
而小于三则是对数量的约束。
例3:
- # Write your MySQL query statement below
- SELECT D1.Name Department, E1.Name Employee, E1.Salary
- FROM Employee E1, Employee E2, Department D1
- WHERE E1.DepartmentID = E2.DepartmentID
- AND E2.Salary >= E1.Salary
- AND E1.DepartmentID = D1.ID
- GROUP BY E1.Name
- HAVING COUNT(DISTINCT E2.Salary) <= 3
- ORDER BY D1.Name, E1.Salary DESC;