Question

我有下表：

ID   GroupID    oDate       oTime     oValue
1    A          2014-06-01  00:00:00  100
2    A          2014-06-01  01:00:00  200
3    A          2014-06-01  02:00:00  300
4    A          2014-06-02  00:00:00  400
5    A          2014-06-02  01:00:00  425
6    A          2014-06-02  02:00:00  475
7    B          2014-06-01  00:00:00  1000
8    B          2014-06-01  01:00:00  1500
9    B          2014-06-01  02:00:00  2000
10   B          2014-06-02  00:00:00  3000
11   B          2014-06-02  01:00:00  3100
12   A          2014-06-03  00:00:00  525
13   A          2014-06-03  01:00:00  600
14   A          2014-06-03  02:00:00  625

我希望得到以下结果：

GroupID   oDate         oResult
A         2014-06-01    300
A         2014-06-02    125
B         2014-06-01    2000

oResult来自：

下午00:00:00的值减去00:00:00所选日期的值。例如，我想知道2014-06-01的结果。然后，

2014-06-02 00:00:00 400 substract 2014-06-01 00:00:00 100
oResult = 400 - 100 = 300

如何在SQL语法中实现此目的？谢谢。

Answer 1

您可以使用Common Table Expression编写查询：

;with CTE as
( select row_number() over ( partition by GroupID, oDate order by oTime Asc) as rownum, 
         GroupID, oDate, oValue,oTime
  from Test
)
select CTE.GroupID,CTE1.oDate, (CTE.oValue - CTE1.oValue) as oResult
from CTE
inner join CTE as CTE1 on datediff (day,CTE1.oDate, CTE.oDate) = 1  
and CTE1.rownum= CTE.rownum 
and  CTE1.GroupID= CTE.GroupID
where CTE.rownum = 1

Check Demo here ...

Answer 2

您可以在此处使用cross apply运算符请检查一下，

select a.GroupID,a.oDate, (ab.oValue - a.oValue) oResult from T as a
cross apply 
(
    select top 1 * from T  as b
    where a.oDate < b.oDate
    and oTime = '00:00:00.0000000'
    and a.ID < b.ID

)as ab
where a.ID in(1,4,7)

Demo

选择要在当前日期SQL上计算的下一个日期的值

2 个答案: