急求一不算复杂的sql语句

[zjl8008]

我有一查询
select t2.kpbh,t1.id,t2.lsj
from yp_pdd_mx t1,v_ypml t2
where t1.bh=48
and t1.ypbh=t2.ypbh
结果集中 kpbh，lsj 都可能存在重复。kpbh相同的lsj也可能重复，也就是存在kpbh、lsj都相同的

我想进一步得到的结果是
上面查询中
lsj最大记录中每个kpbh中只保留一行记录(kpbh、lsj都相同时，任取一个t1.id就行)的t1.id的记录集

[--小F--]

--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)
*/

[geniuswjt]

lz请明确说一下什么和什么相同时，什么取最大或最小或随机？
从你讲的话中没看懂

[叶子]

小F给出的很全面了。

[zjl8008]

[Quote=引用 2 楼 geniuswjt 的回复:]
lz请明确说一下什么和什么相同时，什么取最大或最小或随机？
从你讲的话中没看懂
[/Quote]
kpbh 相同时 取lsj最大的一行记录，但lsj有重复，这时随机取一个就行，只要一行。也就是结果中kpbh,lsj全相同的只要一行

[唐诗三百首]

"kpbh、lsj都相同时，任取一个t1.id就行"

-- 取最大的t1.id
select t2.kpbh,max(t1.id),t2.lsj
from yp_pdd_mx t1,v_ypml t2
where t1.bh=48 and t1.ypbh=t2.ypbh
group by t2.kpbh,t2.lsj

-- 取最小的t1.id
select t2.kpbh,min(t1.id),t2.lsj
from yp_pdd_mx t1,v_ypml t2
where t1.bh=48 and t1.ypbh=t2.ypbh
group by t2.kpbh,t2.lsj

[zjl8008]

[Quote=引用 5 楼 ap0405140 的回复:]
"kpbh、lsj都相同时，任取一个t1.id就行"

SQL code


-- 取最大的t1.id
select t2.kpbh,max(t1.id),t2.lsj
from yp_pdd_mx t1,v_ypml t2
where t1.bh=48 and t1.ypbh=t2.ypbh
group by t2.kpbh,t2.lsj

-- 取最小的t1.id
se……
[/Quote]
这么简单么？等检验一下结果集正确不
我也写了一个

select min(id) id from
(select t2.kpbh,t1.id,t2.lsj from yp_pdd_mx t1,v_ypml t2 where t1.bh=48 and t1.ypbh=t2.ypbh) t1,
(select kpbh,max(lsj) lsj from
(select t2.kpbh,t1.id,t2.lsj from yp_pdd_mx t1,v_ypml t2 where t1.bh=48 and t1.ypbh=t2.ypbh) t
group by kpbh) t2
where t1.kpbh=t2.kpbh and t1.lsj=t2.lsj
group by t1.kpbh

[zjl8008]

5#的结果不对，kpbh相同时lsj 我要取最大的，不是每个lsj都要保留。我自己的好像是对了啊

[-晴天]

[Quote=引用 7 楼 zjl8008 的回复:]
5#的结果不对，kpbh相同时lsj 我要取最大的，不是每个lsj都要保留。我自己的好像是对了啊
[/Quote]
自己对就OK啦!

[zjl8008]

谢谢各位，结贴