求解比大小数据

[tianyasong]

表1
date
------------------------------------------------
1 2009-09-14 10:23:49.567
1 2009-09-03 12:00:35.346
1 2009-09-03 12:00:35.345
1 2009-09-03 12:00:35.343
1 2009-09-03 12:00:26.310
1 2009-09-03 11:58:29.310
表2
date
------------------------------------------------
1 2009-09-14 10:23:49.567
1 2009-09-03 12:00:35.343
1 2009-09-03 11:58:29.310
希望得到比表2每个时间大的表1最小数据
希望结果：2009-09-03 12:00:35.345
2009-09-03 12:00:26.310

[dawugui]

select * from tb1 where date > (select max(date) from tb2)

[jwdream2008]

没明白！

[lihan6415151528]

select * from 表1 
where date > (select max(date) from 表2)

[dawugui]

你的结果好象看不懂.

[guguda2008]

看不懂，撤退

[华夏小卒]

要求有问题

[chuifengde]

SELECT a,b,c=(SELECT MIN(b) FROM 表1 WHERE b>b.b) FROM 表2 b 

[tianyasong]

意思是，表二中2009-09-03 11:58:29.310
表一比它时间大的最小时间2009-09-03 12:00:26.310
表二中2009-09-03 12:00:35.343
表一取得：2009-09-03 12:00:35.345
表二中2009-09-14 10:23:49.567
表一没有对应

[xiequan2]

--> 测试数据：@tb1
declare @tb1 table([id] int,[date] datetime)
insert @tb1
select 1,'2009-09-14 10:23:49.567' union all
select 1,'2009-09-03 12:00:35.346' union all
select 1,'2009-09-03 12:00:35.345' union all
select 1,'2009-09-03 12:00:35.343' union all
select 1,'2009-09-03 12:00:26.310' union all
select 1,'2009-09-03 11:58:29.310'
--> 测试数据：@tb2
declare @tb2 table([id] int,[date] datetime)
insert @tb2
select 1,'2009-09-14 10:23:49.567' union all
select 1,'2009-09-03 12:00:35.343' union all
select 1,'2009-09-03 11:58:29.310'

 
 

select * from
(
select dif=(select min(date) from @tb1 where date>t.date) from @tb2 t  
) m where dif is not null

/*
dif
-----------------------
2009-09-03 12:00:35.347
2009-09-03 12:00:26.310

(2 行受影响)


*/

[华夏小卒]

[Quote=引用 9 楼 xiequan2 的回复:]
SQL code--> 测试数据：@tb1declare@tb1table([id]int,[date]datetime)insert@tb1select1,'2009-09-14 10:23:49.567'unionallselect1,'2009-09-03 12:00:35.346'unionallselect1,'2009-09-03 12:00:35.345'unionallselect¡­
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