返回每用户的最近订单记录

[O西瓜]

有一订单表结构如下
OrderId UserId CreateTime

OrderId是唯一的，UserId是可以重复的，CreateTime是指下单时间

现要求找出所有用户最近的一笔订单，求帮忙。
比如有用户A和B，A有两条记录,B有三条记录，找出A和B最近的一次交易记录，返回OrderId,UserId,CreateTime

[Felixzhaowenzhong]

select orderid,userid,max(createtime)createtime from tb_order
group by orderid,userid

[Felixzhaowenzhong]

[Quote=引用 1 楼 的回复:]
SQL code
select orderid,userid,max(createtime)createtime from tb_order
group by orderid,userid
[/Quote]

错了。有问题。 用下面这个

create table tb_order(orderid int,userid int,createtime datetime)
insert tb_order 
select 1,1,'2012-09-12 12:00:01'
union select 2,1,'2012-09-12 13:00:01' 
union select 3,1,'2012-09-13 13:00:01' 
union select 4,2,'2012-09-12 13:00:01' 
union select 5,2,'2012-09-13 13:00:01' 
union select 6,3,'2012-09-13 13:00:01' 
union select 7,3,'2012-09-14 13:00:01' 

select a.* from tb_order a join
(select userid,MAX(createtime)createtime from tb_order
group by userid) b on a.userid =b.userid and a.createtime =b.createtime 
/*
orderid	userid	createtime
7	3	2012-09-14 13:00:01.000
5	2	2012-09-13 13:00:01.000
3	1	2012-09-13 13:00:01.000
*/

[gogodiy]

select * from [订单表] as a where not exists (select 1 from [订单表] where UserId=a.UserId and CreateTime>a.CreateTime)

[gw6328]

;WITH cte AS (
SELECT rn=ROW_NUMBER() OVER(partition BY userid ORDER BY createtime desc),* FROM tb_order
)
SELECT * FROM cte WHERE rn=1;

[gw6328]

;WITH cte AS (
SELECT rn=ROW_NUMBER() OVER(partition BY userid ORDER BY createtime desc,orderid desc),* FROM tb_order
)
SELECT * FROM cte WHERE rn=1;

[O西瓜]

非常感谢各位的回复，我都做了一一尝试，zhaowenzhong和jinfengyiye的方案是可行的，性能哪个更好不清楚，gogodiy的方案运行结果不对。