oracle 查询每一笔数据里面的最早数据

[jayleehom2000]

id hm sj
1 1 2011/08/29 14:56:34
1 2 2011/08/30 15:56:34
1 2 2011/08/17 20:55:34
1 3 2011/08/20 14:00:34
1 4 2011/08/17 14:20:34
2 1 2011/08/29 14:56:34
2 2 2011/08/30 15:56:34
3 2 2011/08/17 20:55:34
2 3 2011/08/20 14:00:34
2 4 2011/08/17 14:20:34
3 1 2011/08/29 14:56:34
3 2 2011/08/30 15:56:34
3 2 2011/08/17 20:55:34
3 3 2011/08/20 14:00:34
3 4 2011/08/17 14:20:34


我想要的结果就是 根据id 查询出每一笔记录的最早的那个数据
结果就是这样
1 4 2011/08/17 14:20:34
2 4 2011/08/17 14:20:34
3 4 2011/08/17 14:20:34


麻烦问一下 这个怎么写啊？ 谢了

[007-x]

可以试试row_number

select id, hm, sj from (
select id, hm, sj,row_number()over(partition by id order by sj) n from table_name) where n=1

[xiaoyu871225]

SELECT t.id,t.hm,t.sj FROM (SELECT id,hm,sj,row_number() over(PARTITION BY ID ORDER BY sj) rn FROM ) t
WHERE t.rn = 1

[xiaoyu871225]

SELECT t.id,t.hm,t.sj FROM (SELECT id,hm,sj,row_number() over(PARTITION BY ID ORDER BY sj) rn FROM ) t
WHERE t.rn = 1

[scrack]

select id, hm, sj from (
select id, hm, sj,row_number()over(partition by id order by sj) n from table_name) where n=1

[xiaoyu871225]

[code=SQL]SELECT t.id,t.hm,t.sj FROM (SELECT id,hm,sj,row_number() over(PARTITION BY ID ORDER BY sj) rn FROM test_tab) t
WHERE t.rn = 1

[/code]

[cosio]

select id, hm, sj from (
select id, hm, sj,row_number()over(partition by id order by sj) n from table_name) where n=1

[monkey_li]

用min(a.date)来处理

[cainiao1923]

就用row_number()

[NLP爱好者]

大家很给力啊 ，学习了 ，啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊