数据库查询问题？？？高手指教!!!!
id   uersId     userName       content     datetime      

1         34         dd                   sfsfsd         2000-5-6
2         43         aa                   sfsfsd         2001-2-6
3         34         dd                   sfsfsd         2005-5-6
4         34         dd                   sfsfsd         2000-5-6
5         43         aa                   sfsfsd         2007-9-14
6         43         aa                   sfsfsd         2002-8-6

如表:我想查出按时间排序,uersId中34,43中的前两条记录.怎么做啊?


------解决方案--------------------
userid等于34和43范围内的:
select top 2 [id],userid,content,[datetime] from 表名
where userid in( '34 ', '43 ') order by [datetime]
------解决方案--------------------
是这个意思吗？
Select Top 2 * From TableName Where UserID In (34, 43) Order By datetime
------解决方案--------------------

select top 2 * from 表 where uersId in (select distinct uersId from 表) order by datetime

子查询就可搞定了! 简单
------解决方案--------------------
select top 2 * from 表名
where userid in( '34 ', '43 ') order by [datetime]
------解决方案--------------------
方法多多，
select top 2 * from Table where userid=34 or userid=43 order by datetime desc
------解决方案--------------------
select top 2 * from 表 where userid in(select DISTINCT userid from 表)order by[datetime]
------解决方案--------------------
應該是取每組的前兩位吧，

try

--方法一：
Select * From TableName A
Where (Select Count(*) From TableName Where uersId = A.uersId And [datetime] > A.[datetime]) < 2
Order By uersId, [datetime]

--方法二：
Select * From TableName A
Where Exists (Select Count(*) From TableName Where uersId = A.uersId And [datetime] > A.[datetime] Having Count(*) < 2)
Order By uersId, [datetime]

--方法三：
Select * From TableName A
Where [datetime] In (Select TOP 2 [datetime] From TableName Where uersId = A.uersId Order By [datetime] Desc)
Order By uersId, [datetime]
------解决方案--------------------
试试这个，希望没理解错，

SELECT * FROM test
WHERE [datetime] IN (SELECT MAX([datetime]) FROM test GROUP BY userId)