日期:2014-05-18  浏览次数:20610 次

group问题
表A
ID Name Num
1 a 1  
2 b 2  
1 e 5
1 d 4
1 c 3
2 f 6
想要的结果(根据ID分组,取Num最大的,并要保留最大Num的Name列)
ID Name Num
1 e 5
2 f 6

------解决方案--------------------
抢分啊

select *
from tb t
where not exists(select 1 from tb where id=t.id and num>t.num)

select *
from tb t
where num=(select max(num) from tb where id=t.id)

------解决方案--------------------
更多
参考
http://topic.csdn.net/u/20120201/10/59cea135-9e32-43ca-8218-bd4219713617.html
------解决方案--------------------
SQL code
--> 测试数据:#
if object_id('tempdb.dbo.#') is not null drop table #
create table #(ID int, Name varchar(8), Num int)
insert into #
select 1, 'a', 1 union all
select 2, 'b', 2 union all
select 1, 'e', 5 union all
select 1, 'd', 4 union all
select 1, 'c', 3 union all
select 2, 'f', 6

select * from # t where not exists (select 1 from # where ID=t.ID and Num>t.Num)
/*
ID          Name     Num
----------- -------- -----------
1           e        5
2           f        6
*/

------解决方案--------------------
探讨

请问这种情况不能使用group by得出结果吗?

------解决方案--------------------
子查询group by取ID、max(Num)再关联
------解决方案--------------------
SQL code
--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 jo