帮朋友发贴:找出每个部门获得前三高工资的所有员工

吉普赛的歌 2019-11-19 07:55:39
/*
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。
*/


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吉普赛的歌 2019年11月24日
来个人, 结分结贴吧。
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吉普赛的歌 2019年11月19日
按题目来看, 似乎是 MySQL 的数据, 不过 MySQL 从 8.0 开始也有与 SqlServer 一样的排名函数, 所以就不赘述了。
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吉普赛的歌 2019年11月19日
USE tempdb
GO
IF OBJECT_ID('dbo.[Employee]') IS NOT NULL 
	DROP TABLE dbo.[Employee]
GO
CREATE TABLE dbo.[Employee](
[Id] int
,[Name] NVARCHAR(10)
,[Salary] int
,[DepartmentId] int
)
GO
SET NOCOUNT ON
INSERT INTO dbo.[Employee] VALUES(N'1',N'Joe',N'85000',N'1')
INSERT INTO dbo.[Employee] VALUES(N'2',N'Henry',N'80000',N'2')
INSERT INTO dbo.[Employee] VALUES(N'3',N'Sam',N'60000',N'2')
INSERT INTO dbo.[Employee] VALUES(N'4',N'Max',N'90000',N'1')
INSERT INTO dbo.[Employee] VALUES(N'5',N'Janet',N'69000',N'1')
INSERT INTO dbo.[Employee] VALUES(N'6',N'Randy',N'85000',N'1')
INSERT INTO dbo.[Employee] VALUES(N'7',N'Will',N'70000',N'1')
GO
IF OBJECT_ID('Department') IS NOT NULL
	DROP TABLE Department
GO
CREATE TABLE Department(
	Id INT,
	[Name] NVARCHAR(10)	
)
GO
SET NOCOUNT ON
INSERT INTO Department VALUES(1,'IT')
INSERT INTO Department VALUES(2,'Sales')
----------- 以上为测试表和测试数据 --------------

SELECT Department
	,Employee
	,Salary
FROM (
	SELECT DENSE_RANK() OVER (PARTITION BY e.DepartmentId ORDER BY e.Salary DESC) AS rid
		,d.Name AS Department
		,e.Name AS Employee
		,e.Salary
	FROM Employee AS e INNER JOIN Department AS d 
		ON e.[DepartmentId]=d.Id
) AS t
WHERE t.rid<=3
ORDER BY Department,rid
/*
Department Employee   Salary
---------- ---------- -----------
IT         Max        90000
IT         Joe        85000
IT         Randy      85000
IT         Will       70000
Sales      Henry      80000
Sales      Sam        60000
*/
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