select id, sum(a1) as a1, sum(a2) as a2, (select top 1 a3 from [table] where id=t.id order by rq desc) from [table] t where rq <@ldate group by id
select t.* from tb t where rq = (select max(rq) from tb where id = t.id)select t.* from tb t where not exists (select 1 from tb where id = t.id and rq > t.rq)
select sum(a1) as a1,sum(a2) as a2, a3 from table where rq <@ldate and rq =max(rq)group by id
select id,sum(a1) as a1,sum(a2) as a2,(select top 1 a3 from [table] as b where [table].id=b.id order by b.rq desc) from [table] where rq <@ldate group by id
select id,sum(a1) as a1,sum(a2) as a2,
max(a3)
from table where rq <@ldate group by id这样?
a3=(select top 1 a3 from [table] where id=a.id order by rq desc)
from [table] a where rq <@ldate group by id
http://topic.csdn.net/u/20091130/21/fb718680-98ff-4afb-98d8-cff2f8293ed5.html?24281
id1 1 2010-1-1
id2 3 2010-2-1
id1 3 2010-2-1
id1 2 2010-3-1得出来的结果应该是id a3
id1 2 (2010-3-1的a3)
id2 3
id,
sum(a1) as a1,
sum(a2) as a2,
(select top 1 a3 from [table] where id=t.id order by rq desc)
from [table] t
where rq <@ldate
group by id
a3
from table where rq <@ldate and rq =max(rq)group by id
select id,sum(a1) as a1,sum(a2) as a2,(select top 1 a3 from [table] as b where [table].id=b.id order by b.rq desc)
from [table] where rq <@ldate group by id