2018-08-21 发布 ┊ 1482 人浏览 ┊ 0 人评论 ┊ 来源：原创 ┊
收藏┊
分享至
这应该算一道典型的复杂sql了，学习备忘
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 e.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;
-