SQL复杂查询，百思不得其解

【求助】SQL复杂查询，百思不得其解
建表SQL：


CREATE TABLE [dbo].[t_user](
[id] [nchar](10) NOT NULL,
[name] [nchar](10) NOT NULL,
[money] [int] NOT NULL,
[time] [datetime] NOT NULL,
 CONSTRAINT [PK_t_user] PRIMARY KEY CLUSTERED 
(
[id] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]
) ON [PRIMARY]

GO

ALTER TABLE [dbo].[t_user] ADD  CONSTRAINT [DF_t_user_time]  DEFAULT (getdate()) FOR [time]
GO

数据：


id        name        moneytime
1         zhangsan  1002013-03-04 16:45:55.760
2         lisi      1102013-03-04 16:46:09.917
3         wangwu    1202013-04-04 16:56:42.343
4         zhangsan  802013-05-04 16:56:54.123
5         zhangsan  2632013-06-04 16:57:02.230
6         lisi      642013-06-04 16:57:11.950
7         wangwu    2652013-07-04 16:57:21.850
8         lisi      2642013-07-04 16:57:32.730

问题：
需要按照【name】查询每个人最近一次的消费记录，每个人一条记录，结果应如下：


id        name        moneytime
5         zhangsan  2632013-06-04 16:57:02.230
7         wangwu    2652013-07-04 16:57:21.850
8         lisi      2642013-07-04 16:57:32.730

SQL学的很烂，求助各位，帮忙解决！谢谢！
SQL疑难问题

[解决办法]
;with tmp
as(select *,rn=row_number()over(partition by name order by time desc) from t_user)
select * from tmp
where rn=1
[解决办法]

引用:

Quote: 引用:

Quote: 引用:

;with tmp
as(select *,rn=row_number()over(partition by name order by time desc) from t_user)
select * from tmp
where rn=1

谢谢！能够正常获取，至于语法还在研究，话说partition by见都没见过。。。
再次谢谢wufeng4552，速度好快！

--SQL SERVER 2000 可以这样：
--> 测试数据：@T
declare @T table([id] int,[name] varchar(8),[money] int,[time] datetime)
insert @T
select 1,'zhangsan',100,'2013-03-04 16:45:55.760' union all
select 2,'lisi',110,'2013-03-04 16:46:09.917' union all
select 3,'wangwu',120,'2013-04-04 16:56:42.343' union all
select 4,'zhangsan',80,'2013-05-04 16:56:54.123' union all
select 5,'zhangsan',263,'2013-06-04 16:57:02.230' union all
select 6,'lisi',64,'2013-06-04 16:57:11.950' union all
select 7,'wangwu',265,'2013-07-04 16:57:21.850' union all
select 8,'lisi',264,'2013-07-04 16:57:32.730'

select * from @T t
where [time]=(select max([time]) from @T where name=t.name)

/*
id name money time
----------- -------- ----------- -----------------------
5 zhangsan 263 2013-06-04 16:57:02.230
7 wangwu 265 2013-07-04 16:57:21.850
8 lisi 264 2013-07-04 16:57:32.730

谢谢回复！问题同qy1116，如果有时间一样的，就会多出一条，谢谢。

select * from @T t
where id=(select top 1 id from @T where name=t.name order by [time] desc)

这样就不多了，信不信？

是的不多了，但是少了很多：


select * from t_user where id=(select top 1 id from t_user where name=t_user.name order by [time] desc)

结果：


idnamemoneytime
8       lisi    2642013-07-04 16:57:32.730

你把我的改了，是这样的：


select * from t_user t 
where id=(select top 1 id from t_user where name=t.name order by [time] desc)

[解决办法]

引用:

Quote: 引用:

select *  from [t_user]
where time in (select max(time) from [t_user]
group by name )

这个可以，不过如果时间相同，比如如下数据：


id        name        moneytime
1         zhangsan  1002013-03-04 16:45:55.760
2         lisi      1102013-03-04 16:46:09.917
3         wangwu    1202013-04-04 16:56:42.343
4         zhangsan  802013-05-04 16:56:54.123
5         zhangsan  2632013-06-04 16:57:02.230
6         lisi      642013-06-04 16:57:11.950
7         wangwu    2652013-07-04 16:57:21.850
8         lisi      2642013-07-04 16:57:32.730
9         lisi      2412013-07-04 16:57:32.730

使用


select *  from [t_user]
where time in (select max(time) from [t_user]
group by name )

得到的结果如下：


id        name        moneytime
5         zhangsan  2632013-06-04 16:57:02.230
7         wangwu    2652013-07-04 16:57:21.850
8         lisi      2642013-07-04 16:57:32.730
9         lisi      2412013-07-04 16:57:32.730

不过，谢谢回复！

谢谢提醒！！害我下班没走还杀死了多少细胞又想到一个方法。。。还有最近消费记录一般是不会出现重复的时间吧，lz考虑的多了哦嘿嘿

select * from  [t_user]
where id in(
select MAX(id)  from [t_user]
where time in (select  max(time) from [t_user]
group by name 
)
group by name)

SQL复杂查询百思不得其解

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