Question

我的开始日期是2009 - 04 - 01（2009年4月4日），结束日期是getdate（）。

我需要计算从开始日期到今天某些事情的每个日期。我还需要将其分为两种情况：

案例1：如果是星期五，我需要回到347天，例如

select datediff(dw, '2010-11-22','2011-11-04')

每个星期五，然后执行一些选择。

案例2：如果它是从星期一到星期四再返回349天，例如

   select datediff(dw, '2010-11-19','2011-11-03')

我怎么写这个，我需要执行的只是把它当作：

declare @startDate date
declare @endDate date
declare @dateHolder date
declare @tsID int
set @startDate = '2009-04-01'
set @endDate = getdate()
set datefirst 1;


while (select count (tsID) from #tempI)>0
            begin
            select top 1 @tsID = tsID from #tempI
                while (select count (rateDate) from #tempI)>0
                begin
                    select top 1 @dateHolder = rateDate from #tempI
                      case (select datename  (dw, @dateHolder) = '5' then  someColumn = @dateHolder - 347 as dateIneedToUseForMyFormula
                      case (select datename (dw, @dateHolder) = '1' or '2' or '3' or '4' then someColumn = @dateHolder - 349 as dateIneedToUseForMyFormula
-- here i don't know how to write the code, so i'll write pseudo
-- i have tsID rateDate and rate
-- i need to put in a new column (the value obtained from taking the value
-- from the column rate corresponding to the  @dateHolder - the value from
-- the column rate corresponding to the @dateHolder - 347 or 349 
-- depending on the case) * 100

                end
            delete from #tempI where @dateHolder = rateDate
        end
    delete from #tempI where @tsID = tsID
    end

修改我在评论中被问到要做什么，并且考虑到这个问题，我认为我并不清楚。我将复制粘贴我在评论中写的内容： “ 首先遍历列表中的所有tsID，在这样做之后迭代所有日期，从中减去349或347天的日期，然后检查放入新列的结果：'rate'的值与我的日期对应的列 - 'rate'列的值，对应于349或347天前的日期。这是针对每个id“

的所有日期执行的

编辑2：预期输出

 tsID rateDate     rate    calculated 
  1    2009-04-01  0.12     null 
  1    2009-04-02  0.14     null 
  1    2009-04-03  0.11     null 
  2    2009-04-01  0.01     null 
  2    2009-04-02  0.012    null
  2    2009-04-03  0.43     null 
. . . 347 days later or 349 depending
     on what 2009-04-01 was 
  1   2010-03-16   0.454   (0.454 - 0.12)*100
  1   2010-03-17   0.34    (0.34 - 0.14)*100 
  1   2010-03-18   0.9     (0.9 - 0.11)*100 
then same for id 2.3...4...

Answer 1

试试这个

Declare @start Date='2009-09-01', @end Date=getdate();
;With NumberSequence( Number ) as
(
    Select @start as Number
        union all
    Select DATEADD(d,1,Number)
        from NumberSequence
        where Number < @end
)
Select 
(CASE WHEN (datepart(dw,Number) =6)  THEN DATEADD(d,-347,Number)
     ELSE DATEADD(d,-349,Number)
     END ) AS Date
From NumberSequence Option (MaxRecursion 10000)

Answer 2

使用工作日的案例陈述：

Select Case 
    When WeekDAy(yourdate, 3) < 4 then datediff(dw, '2010-11-19','2011-11-03')
    when WeekDAy(yourdate, 3) = 4 then datediff(dw, '2010-11-22','2011-11-04')
    Else ...?... End

如果两个日期之间的日期差异为X，请执行此操作，如果是Y则执行其他操作

2 个答案: