求助：关于使用聚合函数Max的问题

[jskenson]

两个表：
主表:m
id time1 .......
-----------------------------------------
1 2008-06-01 00:00:00.000
2 2008-06-02 00:00:00.000
3 2008-06-03 00:00:00.000
4 2008-06-04 00:00:00.000
副表:d
id time1 price .............
-----------------------------------------------
1 2008-05-01 00:00:00.000 1
1 2008-05-02 00:00:00.000 2
1 2008-06-03 00:00:00.000 3
2 2008-05-03 00:00:00.000 4
4 2008-05-03 00:00:00.000 5
4 2008-05-01 00:00:00.000 6
要实现在查询如下：
select d.id,max(d.time1) from d,m where d.time1<m.time1 and m.id=d.id group by d.id
我的问题是两个表各有很多其它字段，要在结果中全部体现，求一个比较好的查询方式，谢谢！
(聚合函数如Max\Min可以怎么样用作行定位吗？)

[yrwx001]

--怕列舉很多列名？
select * from m,(select top 1 * from d  where d.time1 < m.time1 and  d.id = m.id  order by d.time1 desc)A
where A.id = mi.d

[中国风]

--取大小值
/******************************************************************************************************************************************************
Name相同ID最小或最大的记录

整理人：中国风(Roy)

日期:2008.06.06
******************************************************************************************************************************************************/

--> --> (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


--1、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:
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


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

(2 行受影响)
*/


--2、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:
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

生成结果2:
/*
ID          Name Memo
----------- ---- ----
3           A    A3
5           B    B2

(2 行受影响)
*/

[jskenson]

楼上的理解错了，我有用到group by，第二个表相同ID下只取日期最大的那一条

[灰太狼]

似乎没有办法，只能把列都写上，然后都group by

[jskenson]

2楼高明啊！
以自己为条件？好像不是很好理解
消化一下先。。。

[octwind]

select * from m inner join (select * from b b1 where not exists(select 1 from b where id=b1.id and time1>b1.time1)) t
on m.id=t.id
或者
select * from m inner join (select * from b b1 where b1.id=(select top 1 id from b where id=b1.id order by time1 desc)) t
on m.id=t.id

[jskenson]

6楼有点不对，还有一个条件没有注意到：
副表取的时间条件是：小于主表时间的最大时间（同一ID内）