寻求一SQL，牛人帮忙，给思路也行。

[shuai45]

id name1 name2 createTime
1 a 2011-04-04 14:50
2 a 2011-04-04 14:52
3 a 2011-04-04 14:20
4 b 2011-04-04 11:12
5 b 2011-04-04 11:11
6 f 2011-04-03
7 g 2011-04-02
8 k 2011-04-01



要求结果：
id name1 createTime
1 a 2011-04-04 14:50
4 b 2011-04-04 11:11
6 f 2011-04-03
7 g 2011-04-02
8 k 2011-04-01

我文字表带不清楚，所以 就列出来。对了是MYSQL的数据库。不过可以给个实现思路啥的。

[tony312ws]

select distinct ......

[shuai45]

[Quote=引用 3 楼 tony312ws 的回复:]
select distinct ......
[/Quote]晕，你压根没看我写的。。。。

[shuai45]

id name1 name2 createTime
1 a 2011-04-04 14:50
2 a 2011-04-04 14:52
3 a 2011-04-04 14:20
4 b 2011-04-04 11:12
5 b 2011-04-04 11:11
6 f 2011-04-03
7 g 2011-04-02
8 k 2011-04-01



结果：
id name1 createTime
1 a 2011-04-04 14:50
4 b 2011-04-04 11:12
6 f 2011-04-03
7 g 2011-04-02
8 k 2011-04-01

[--小F--]

select * from tb t where id=(select min(id) from tb where name1=.name1 and createTime
=t.createTime)

[cd731107]

--sql2000的语法，不知道能不能用
select from tb a where id in (select top 1 id from tb b where a.name1=b.name1 order by b.id) 

[yzf86211861]

select id,name,min(convert(datetime,createTime)) from table 
group by  id,name
order by id, name1, createTime desc

[mustudent]

Select * From # s Where Not Exists(Select 1 From #  where s.name1=#.name1 and s.id>#.id)
--Exists子查询
--你要是想只取ID最小的记录就用这个,要是取时间就用CreateTime

[dovei]

楼主这个问题...有点麻烦
二笔取小的..三笔取中间的.那有四笔呢 取哪个?

[叶子]

declare @t table (id int,name1 varchar(1),name2 varchar(10),createTime varchar(6))
insert into @t
select 1,'a','2011-04-04','14:50' union all
select 2,'a','2011-04-04','14:52' union all
select 3,'a','2011-04-04','14:20' union all
select 4,'b','2011-04-04','11:12' union all
select 5,'b','2011-04-04','11:11' union all
select 6,'f','2011-04-03',null union all
select 7,'g','2011-04-02',null union all
select 8,'k','2011-04-01',null

select * from @t t where id=
(select min(id) from @t where name2=t.name2 and name1=t.name1)
order by id
/*
id          name1 name2      createTime
----------- ----- ---------- ----------
1           a     2011-04-04 14:50
4           b     2011-04-04 11:12
6           f     2011-04-03 NULL
7           g     2011-04-02 NULL
8           k     2011-04-01 NULL
*/

[xiaol028]

每天回帖即可获得10分可用分！小技巧
id name1 createTime
1 a 2011-04-04 14:50
4 b 2011-04-04 11:12
6 f 2011-04-03
7 g 2011-04-02
8 k 2011-04-01
name2?

[shuai45]

结果：
id name1 createTime
1 a 2011-04-04 14:50
4 b 2011-04-04 11:12
6 f 2011-04-03
7 g 2011-04-02
8 k 2011-04-01

[kingtiy]

SELECT a.* FROM tb a JOIN (
SELECT min(id) id,name1 FROM tb 
GROUP BY name1
) b 
ON a.id=b.id
ORDER BY a.id

[hanrenwowo]

楼主是想在id不变的情况下删除id为2，3，5三条记录，然后查询出来

[hanrenwowo]

题意为id不变，删除name2字段，再删除id为2，3，5条记录，然后查询出来

[xuandalin]

name 2 呢