求高手解决SQL语句问题

theskyworld 2011-03-26 04:49:54
库中表的数据为:

AID BID BNum BName BType BEndDay BStartDay

1 7427 SITS-20110425-P01 英国伦敦 1 2011-05-01 2011-04-25
1 7426 SITS-20110418-P01 英国利物浦 1 2011-04-24 2011-04-18
2 7424 SITS-20110411-P01 英国曼彻斯特 1 2011-04-17 2011-04-11
2 7423 SITS-20110409-P01 英国朴次茅斯 1 2011-04-15 2011-04-09
1 7420 SITS-20110427-P02 英国考文垂 0 2011-05-07 2011-04-27


我想要的结果是:
AID BID BNum BName BType BEndDay BStartDay

1 7420 SITS-20110427-P02 英国考文垂 0 2011-05-07 2011-04-27
2 7424 SITS-20110411-P01 英国曼彻斯特 1 2011-04-17 2011-04-11

意思是:根据AID分组 取出最大的那个BEndDay的整条记录,谢谢了,求SQL 设 表名 为 ABTABLE
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--小F-- 2011-03-26
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--处理表重复记录(查询和删除)
/******************************************************************************************************************************************************
1、Num、Name相同的重复值记录,没有大小关系只保留一条
2、Name相同,ID有大小关系时,保留大或小其中一个记录
整理人:中国风(Roy)

日期:2008.06.06
******************************************************************************************************************************************************/

--1、用于查询重复处理记录(如果列没有大小关系时2000用生成自增列和临时表处理,SQL2005用row_number函数处理)

--> --> (Roy)生成測試數據

if not object_id('Tempdb..#T') is null
drop table #T
Go
Create table #T([ID] int,[Name] nvarchar(1),[Memo] nvarchar(2))
Insert #T
select 1,N'A',N'A1' union all
select 2,N'A',N'A2' union all
select 3,N'A',N'A3' union all
select 4,N'B',N'B1' union all
select 5,N'B',N'B2'
Go


--I、Name相同ID最小的记录(推荐用1,2,3),方法3在SQl05时,效率高于1、2
方法1:
Select * from #T a where not exists(select 1 from #T where Name=a.Name and ID<a.ID)

方法2:
select a.* from #T a join (select min(ID)ID,Name from #T group by Name) b on a.Name=b.Name and a.ID=b.ID

方法3:
select * from #T a where ID=(select min(ID) from #T where Name=a.Name)

方法4:
select a.* from #T a join #T b on a.Name=b.Name and a.ID>=b.ID group by a.ID,a.Name,a.Memo having count(1)=1

方法5:
select * from #T a group by ID,Name,Memo having ID=(select min(ID)from #T where Name=a.Name)

方法6:
select * from #T a where (select count(1) from #T where Name=a.Name and ID<a.ID)=0

方法7:
select * from #T a where ID=(select top 1 ID from #T where Name=a.name order by ID)

方法8:
select * from #T a where ID!>all(select ID from #T where Name=a.Name)

方法9(注:ID为唯一时可用):
select * from #T a where ID in(select min(ID) from #T group by Name)

--SQL2005:

方法10:
select ID,Name,Memo from (select *,min(ID)over(partition by Name) as MinID from #T a)T where ID=MinID

方法11:

select ID,Name,Memo from (select *,row_number()over(partition by Name order by ID) as MinID from #T a)T where MinID=1

生成结果:
/*
ID Name Memo
----------- ---- ----
1 A A1
4 B B1

(2 行受影响)
*/


--II、Name相同ID最大的记录,与min相反:
方法1:
Select * from #T a where not exists(select 1 from #T where Name=a.Name and ID>a.ID)

方法2:
select a.* from #T a join (select max(ID)ID,Name from #T group by Name) b on a.Name=b.Name and a.ID=b.ID order by ID

方法3:
select * from #T a where ID=(select max(ID) from #T where Name=a.Name) order by ID

方法4:
select a.* from #T a join #T b on a.Name=b.Name and a.ID<=b.ID group by a.ID,a.Name,a.Memo having count(1)=1

方法5:
select * from #T a group by ID,Name,Memo having ID=(select max(ID)from #T where Name=a.Name)

方法6:
select * from #T a where (select count(1) from #T where Name=a.Name and ID>a.ID)=0

方法7:
select * from #T a where ID=(select top 1 ID from #T where Name=a.name order by ID desc)

方法8:
select * from #T a where ID!<all(select ID from #T where Name=a.Name)

方法9(注:ID为唯一时可用):
select * from #T a where ID in(select max(ID) from #T group by Name)

--SQL2005:

方法10:
select ID,Name,Memo from (select *,max(ID)over(partition by Name) as MinID from #T a)T where ID=MinID

方法11:
select ID,Name,Memo from (select *,row_number()over(partition by Name order by ID desc) as MinID from #T a)T where MinID=1

生成结果2:
/*
ID Name Memo
----------- ---- ----
3 A A3
5 B B2

(2 行受影响)
*/
AcHerat 2011-03-26
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select *
from ABTABLE t
where not exists(select * from tb where t.aid = aid and bendday < t.bendday)
vivian_lanlan 2011-03-26
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select * 
from(select rowid=row_number() over(partition by aid order by bendday desc),* from tb) as a
where rowid=1
vivian_lanlan 2011-03-26
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select * from tb as a
where not exists(select * from tb where a.aid=aid and bendday>A.bendday)

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