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SQL distinct 返回多列结果

zedan 2009-06-18 10:57:43
表中有sn,datetime.
select distinct sn from table
返回只能显示一列sn记录
我想显示出全部列的记录
sn datetime
1 09-01-20
1 09-02-10
2 09-03-09
2 09-03-10
3 09-02-10

结果为:
1 09-03-20 其中这里为(09-02-10,第二行的时间也可以)
2 09-03-10 (如上)
3 09-02-10

是要去除重复的部分,然后把结果显示出来,包括所有列。但distinct不基于其它列。


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wolfxin2010 2011-05-06
Select sn ,distinct from table Where date in (select max(distinct) from table group by sn)忘记加个Where
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wolfxin2010 2011-05-06
Select sn ,distinct from table date in (select max(distinct) from table group by sn) 可以试试这个
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datongye 2011-04-20
为什么不能发个例程?全是纸上谈兵。
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zedan 2009-06-18
大家太热情了。
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永生天地 2009-06-18
我也是这么做
[Quote=引用 2 楼 jwdream2008 的回复:]
SQL codeselect [sn],max([datetime]) from table wheren [sn] in( select distinct sn from table)
[/Quote]
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--小F-- 2009-06-18
--参考
--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)
*/



--2、删除重复记录有大小关系时,保留大或小其中一个记录


--> --> (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),保留最小一条
方法1:
delete a from #T a where exists(select 1 from #T where Name=a.Name and ID<a.ID)

方法2:
delete a from #T a left join (select min(ID)ID,Name from #T group by Name) b on a.Name=b.Name and a.ID=b.ID where b.Id is null

方法3:
delete a from #T a where ID not in (select min(ID) from #T where Name=a.Name)

方法4(注:ID为唯一时可用):
delete a from #T a where ID not in(select min(ID)from #T group by Name)

方法5:
delete a from #T a where (select count(1) from #T where Name=a.Name and ID<a.ID)>0

方法6:
delete a from #T a where ID<>(select top 1 ID from #T where Name=a.name order by ID)

方法7:
delete a from #T a where ID>any(select ID from #T where Name=a.Name)



select * from #T

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

(2 行受影响)
*/


--II、Name相同ID保留最大的一条记录:

方法1:
delete a from #T a where exists(select 1 from #T where Name=a.Name and ID>a.ID)

方法2:
delete a from #T a left join (select max(ID)ID,Name from #T group by Name) b on a.Name=b.Name and a.ID=b.ID where b.Id is null

方法3:
delete a from #T a where ID not in (select max(ID) from #T where Name=a.Name)

方法4(注:ID为唯一时可用):
delete a from #T a where ID not in(select max(ID)from #T group by Name)

方法5:
delete a from #T a where (select count(1) from #T where Name=a.Name and ID>a.ID)>0

方法6:
delete a from #T a where ID<>(select top 1 ID from #T where Name=a.name order by ID desc)

方法7:
delete a from #T a where ID<any(select ID from #T where Name=a.Name)


select * from #T
/*
ID Name Memo
----------- ---- ----
3 A A3
5 B B2

(2 行受影响)
*/
回复
叶子 2009-06-18
[Quote=引用 10 楼 zedan 的回复:]
嗯,我的疏忽,
datetime只是临时给出的,其实另一(些)列是字符串列,不能用max,min之类的,
再帮帮忙......
[/Quote]

我15楼代码就是字符串的,也可以用min和max呀
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叶子 2009-06-18

declare @table table (sn int,[datetime] varchar(20))
insert into @table
select 1,'09-01-20' union all
select 1,'09-02-10' union all
select 2,'09-03-09' union all
select 2,'09-03-10' union all
select 3,'09-02-10'

select sn,min([datetime]) as [datetime] from @table group by sn
/*
sn datetime
----------- --------------------
1 09-01-20
2 09-03-09
3 09-02-10
*/

select sn,max([datetime]) as [datetime] from @table group by sn
/*
sn datetime
----------- --------------------
1 09-02-10
2 09-03-10
3 09-02-10
*/
回复
feixianxxx 2009-06-18
这个也行~
create table t (sn int,[datetime] datetime)
insert t
select 1,'09-01-20' union all
select 1,'09-02-20' union all
select 2,'09-03-09' union all
select 2,'09-03-10' union all
select 3,'09-02-20'

select sn,[datetime] from T t1
where
not exists(select * from t where t1.sn=t.sn and datediff(dd,t.datetime,t1.datetime)>0)
回复
nalnait 2009-06-18
[Quote=引用 8 楼 feixianxxx 的回复:]
SQL codeselect sn,[datetime] from Table1 t1 where not exists(select * from table1 where t1.sn=table1.sn and sn<t1.sn)
[/Quote]
这个可以满足你的需求
回复
feixianxxx 2009-06-18
[Quote=引用 10 楼 zedan 的回复:]
嗯,我的疏忽,
datetime只是临时给出的,其实另一(些)列是字符串列,不能用max,min之类的,
再帮帮忙......
[/Quote]
LZ试试我在九楼写的~用exists
回复
feixianxxx 2009-06-18
create table t (sn int,[datetime] datetime) 
insert t
select 1,'09-01-20' union all
select 1,'09-02-20' union all
select 2,'09-03-09' union all
select 2,'09-03-10' union all
select 3,'09-02-20'

select MIN([datetime])as datetime from t group by sn
/*----------------
2009-01-20 00:00:00.000
2009-03-09 00:00:00.000
2009-02-20 00:00:00.000
-------------------*/
回复
zedan 2009-06-18
嗯,我的疏忽,
datetime只是临时给出的,其实另一(些)列是字符串列,不能用max,min之类的,
再帮帮忙......
回复
feixianxxx 2009-06-18
[Quote=引用 8 楼 feixianxxx 的回复:]
SQL codeselect sn,[datetime] from Table1 t1 where not exists(select * from table1 where t1.sn=table1.sn and sn<t1.sn)
[/Quote]
修改下;
create table t (sn int,[datetime] datetime)
insert t
select 1,'09-01-20' union all
select 1,'09-02-20' union all
select 2,'09-03-09' union all
select 2,'09-03-10' union all
select 3,'09-02-20'

select sn,[datetime] from T t1
where
not exists(select * from t where t1.sn=t.sn and t.datetime<t1.datetime)

/*-----------------
1 2009-01-20 00:00:00.000
2 2009-03-09 00:00:00.000
3 2009-02-20 00:00:00.000
-------------*/
回复
feixianxxx 2009-06-18
select sn,[datetime] from Table1 t1 where  not exists(select * from table1 where t1.sn=table1.sn and sn<t1.sn) 
回复
jwdream2008 2009-06-18
学习,顶了!
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jwdream2008 2009-06-18

1.select sn,min(datetime)from tb group by sn
2.select sn,max(datetime)from tb group by sn
回复
olddown 2009-06-18
[Quote=引用 4 楼 olddown 的回复:]
SQL codeselectsn,max(datetime)asdatetimefromtablegroupbydatetime
[/Quote]
select sn,max(datetime) as datetime from table group by sn

回复
olddown 2009-06-18

select sn,max(datetime) as datetime from table group by datetime
回复
feixianxxx 2009-06-18
 select MIN([datetime])as datetime from 表名 group by sn
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