oracle去重
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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
----------- ---- ----
A A1
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
----------- ---- ----
A A3
B B2
(2 行受影响)
*/
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
----------- ---- ----
A A1
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
----------- ---- ----
A A3
B B2
(2 行受影响)
*/
原文地址:
http://www.cnblogs.com/langtianya/p/5997048.html
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