求一SQL,拜谢

shuai45 2011-12-28 05:19:32
表:test
字段
id score class
1 10 1
2 11 1
3 30 2
4 59 2

求一SQL,要查询每个class的成绩前两个的信息数据。
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lwlianglove2 2011-12-29
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不明白这个前两个是什么意思,按成绩降序排列去前两个吗?
黄_瓜 2011-12-28
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这个200分?


中国风 2011-12-28
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當成績相同時,可用DENSE_RANK,如下效果
use Tempdb
go
--> -->

if not object_id(N'Test') is null
drop table Test
Go
Create table Test([id] int,[score] int,[class] int)
Insert Test
select 1,10,1 union all
select 2,11,1 union all
select 3,30,2 union all
select 4,59,2 UNION ALL
select 5,59,2 UNION ALL
select 2,10,1

Go
SELECT [id],[score],[class],名次=rn
FROM (Select *,DENSE_RANK()OVER(PARTITION BY [class] ORDER BY [score] desc)as rn from Test )t
WHERE rn<=2
/*
id score class 名次
2 11 1 1
1 10 1 2
2 10 1 2
4 59 2 1
5 59 2 1
3 30 2 2
*/
--小F-- 2011-12-28
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---------------------------------
-- Author: liangCK 小梁
-- Title : 查每个分组前N条记录
-- Date : 2008-11-13 17:19:23
---------------------------------

--> 生成测试数据: #T
IF OBJECT_ID('tempdb.dbo.#T') IS NOT NULL DROP TABLE #T
CREATE TABLE #T (ID VARCHAR(3),GID INT,Author VARCHAR(29),Title VARCHAR(39),Date DATETIME)
INSERT INTO #T
SELECT '001',1,'邹建','深入浅出SQLServer2005开发管理与应用实例','2008-05-10' UNION ALL
SELECT '002',1,'胡百敬','SQLServer2005性能调校','2008-03-22' UNION ALL
SELECT '003',1,'格罗夫Groff.J.R.','SQL完全手册','2009-07-01' UNION ALL
SELECT '004',1,'KalenDelaney','SQLServer2005技术内幕存储引擎','2008-08-01' UNION ALL
SELECT '005',2,'Alex.Kriegel.Boris.M.Trukhnov','SQL宝典','2007-10-05' UNION ALL
SELECT '006',2,'飞思科技产品研发中心','SQLServer2000高级管理与开发','2007-09-10' UNION ALL
SELECT '007',2,'胡百敬','SQLServer2005数据库开发详解','2008-06-15' UNION ALL
SELECT '008',3,'陈浩奎','SQLServer2000存储过程与XML编程','2005-09-01' UNION ALL
SELECT '009',3,'赵松涛','SQLServer2005系统管理实录','2008-10-01' UNION ALL
SELECT '010',3,'黄占涛','SQL技术手册','2006-01-01'

--SQL查询如下:

--按GID分组,查每个分组中Date最新的前2条记录


--1.字段ID唯一时:
SELECT * FROM #T AS T WHERE ID IN(SELECT TOP 2 ID FROM #T WHERE GID=T.GID ORDER BY Date DESC)

--2.如果ID不唯一时:
SELECT * FROM #T AS T WHERE 2>(SELECT COUNT(*) FROM #T WHERE GID=T.GID AND Date>T.Date)

--SQL Server 2005 使用新方法

--3.使用ROW_NUMBER()进行排位分组
SELECT ID,GID,Author,Title,Date
FROM
(
SELECT rid=ROW_NUMBER() OVER(PARTITION BY GID ORDER BY Date DESC),*
FROM #T
) AS T
WHERE rid<=2

--4.使用APPLY
SELECT DISTINCT b.*
FROM #T AS a
CROSS APPLY
(
SELECT TOP(2) * FROM #T WHERE a.GID=GID ORDER BY Date DESC
) AS b


--结果
/*

ID GID Author Title Date
---- ----------- ----------------------------- --------------------------------------- -----------------------
003 1 格罗夫Groff.J.R. SQL完全手册 2009-07-01 00:00:00.000
004 1 KalenDelaney SQLServer2005技术内幕存储引擎 2008-08-01 00:00:00.000
005 2 Alex.Kriegel.Boris.M.Trukhnov SQL宝典 2007-10-05 00:00:00.000
007 2 胡百敬 SQLServer2005数据库开发详解 2008-06-15 00:00:00.000
009 3 赵松涛 SQLServer2005系统管理实录 2008-10-01 00:00:00.000
010 3 黄占涛 SQL技术手册 2006-01-01 00:00:00.000

(6 行受影响)
*/















--得到每组前几条数据
--假設每組Col1中, Col3不會重復

--建立測試環境
Create Table TEST
(Col1 Varchar(10),
Col2 Varchar(10),
Col3 Int)
--插入數據
Insert TEST Select 'BD1V','Label', 4
Union All Select 'BD1V', 'BATT', 2
Union All Select 'BD1V', 'ODD', 3
Union All Select 'BD1V', 'HDD', 5
Union All Select 'BD1V', 'LCD', 1
Union All Select 'BD1W','HDD', 3
Union All Select 'BD1W','RAM', 8
Union All Select 'BD1W','TP CABLE', 5
Union All Select 'BD1W','LCD', 6
Union All Select 'BD1W','Label', 2
Union All Select 'BL3', 'LCD CABLE', 7
Union All Select 'BL3', 'LABEL', 6
Union All Select 'BL3', 'LCD', 5
Union All Select 'BL3', 'RAM', 1
Union All Select 'BL3D', 'Label', 4
GO
--測試
--方法一:
Select Col1, Col2, Col3 From TEST A
Where (Select Count(*) From TEST Where Col1 = A.Col1 And Col3 > A.Col3) < 3
Order By Col1, Col3 Desc
--方法二:
Select Col1, Col2, Col3 From TEST A
Where Exists (Select Count(*) From TEST Where Col1 = A.Col1 And Col3 > A.Col3 Having Count(*) < 3)
Order By Col1, Col3 Desc
--方法三:
Select Col1, Col2, Col3 From TEST A
Where Col3 In (Select TOP 3 Col3 From TEST Where Col1 = A.Col1 Order By Col3 Desc)
Order By Col1, Col3 Desc
GO
--刪除測試環境
Drop Table TEST
--結果
/*
Col1 Col2 Col3
BD1V HDD 5
BD1V Label 4
BD1V ODD 3
BD1W RAM 8
BD1W LCD 6
BD1W TP CABLE 5
BL3 LCD CABLE 7
BL3 LABEL 6
BL3 LCD 5
BL3D Label 4
*/
叶子 2011-12-28
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declare @test table (id int,score int,class int)
insert into @test
select 1,10,1 union all
select 2,11,1 union all
select 3,30,2 union all
select 4,59,2 union all
select 5,22,1 union all
select 6,40,2

select * from @test a where id in
(select top 2 id from @test where class =a.class order by score desc)
/*
id score class
----------- ----------- -----------
2 11 1
4 59 2
5 22 1
6 40 2
*/
勿勿 2011-12-28
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if OBJECT_ID('test') is not null drop table test
go
create table test (id int,score int,class int)
insert into test values(1,10,1)
insert into test values(2,11,1)
insert into test values(3,30,2)
insert into test values(4,59,2)

select * from
(select row_number() over(order by score desc) as mk,* from test) t
where mk<=2

mk id score class
-------------------- ----------- ----------- -----------
1 4 59 2
2 3 30 2

(2 行受影响)


快溜 2011-12-28
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排序再取数据
-晴天 2011-12-28
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如果考虑并列,则:
select * from(
select *,rank()over(partition by class order by score desc)rn
)t where rn<=2
_0筱筱0_ 2011-12-28
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SELECT *
FROM TEST T1
WHERE (
SELECT COUNT(1) FROM TEST T2
WHERE T2.CLASS=T1.CLASS AND T2.SCORE>=T1.SCORE
)<=2
-晴天 2011-12-28
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select * from(
select *,row_number()over(partition by class order by score desc)rn
)t where rn<=2
Monkey__D__Luffy 2011-12-28
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select * from 
(select row_number() over(order by score desc) as mk,* from test) t
where mk<3
-晴天 2011-12-28
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前两个是指?
liangCK 2011-12-28
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这个200分?

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