sql交叉腰200分(0)

表AAA BB CC DS123 20090801 20DS123 20090801 30DS123 20090802 40DS123 20090802 50DS128 20090801 15DS128 20090801 16DS128 20090802 13DS128 20090802 17怎么才能得到AA 20090801 20090802DS123 20 40DS123 30 50DS128 15 13DS128 16 17

这种类型的问题已经问过很多了，要用存储过程才能实现。

不用,用临时表与动态sql语句能达得到,并不复杂

zbdzjx，liangyinghua你好，我是新手，要怎么写？多谢指教！

select AA,min(case when BB='20090801' then BB else 0 end) as 20090801,min(case when BB='20090802' then BB else 0 end) as 20090802from agroup by AA

百度一下:交叉表,能找出一大把.

各位不好意思，我的用意如下:利用下面控件DateTimePicker1--查询开始日期DateTimePicker2--查询结束日期考勤表 rskqdatacardid date time 0258 20090801 07550258 20090801 17010258 20090802 07530258 20090802 1700 . .0258 20090831 17100259 20090801 07550259 20080801 1708 . .0259 20090831 1715 . .查询结果：cardid 20090801 20090802 ...200908310258 0755 0753 1701 1700 17100259 0755 1708 1715 . .