uid fee utime
a 100 2005-10-10
b 223 2005-10-10
b 240 2005-9-10
b 90 2005-8-10
b 70 2005-7-10
a 70 2005-7-10得到
uid fee utime
a 30 2005-10-10(2005-10-10的fee减2005-7-10的fee)
b -17 2005-10-10(2005-10-10的fee减2005-9-10的fee)得到相同uid的相邻时间的fee的差谢谢
a 100 2005-10-10
b 223 2005-10-10
b 240 2005-9-10
b 90 2005-8-10
b 70 2005-7-10
a 70 2005-7-10得到
uid fee utime
a 30 2005-10-10(2005-10-10的fee减2005-7-10的fee)
b -17 2005-10-10(2005-10-10的fee减2005-9-10的fee)得到相同uid的相邻时间的fee的差谢谢
这个语句得到的应该不止两条记录!!uid 为b 的记录应该会有多条吧?
楼主给个答案,学习中...
from
(
select *,row_number() over(partition by uid order by utime desc) as rn
from t
) tt
group by tt
不知行不行!