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SQL中的去重操作
阅读量:5363 次
发布时间:2019-06-15

本文共 1971 字,大约阅读时间需要 6 分钟。

if not object_id('Tempdb..#T') is null    drop table #TGoCreate table #T([ID] int,[Name] nvarchar(1),[Memo] nvarchar(2))Insert #Tselect 1,N'A',N'A1' union allselect 2,N'A',N'A2' union allselect 3,N'A',N'A3' union allselect 4,N'B',N'B1' union allselect 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
=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
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 A14 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!

这个是我在CSDN论坛中找到的资料,原文在:http://bbs.csdn.net/topics/390315926

以上内容就在2楼

转载于:https://www.cnblogs.com/qiywtc/p/4506721.html

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