sql获取两个日期之间5分钟的行间隔

Question

有没有办法在日期之间获得像 5,10 或 15 分钟这样的分钟间隔？这是一个例子：

我有两个约会：

2016-07-25 14:00:00.000 || 2016-07-25 18:00:00.000

我想要一个函数，它可以根据我给出的参数间隔返回一组行：

当间隔参数为5分钟时：

2016-07-25 14:05:00.000
2016-07-25 14:10:00.000
2016-07-25 14:15:00.000
2016-07-25 14:20:00.000
2016-07-25 14:25:00.000
2016-07-25 ...
2016-07-25 18:00:00.000

当间隔参数为10分钟时：

2016-07-25 14:10:00.000
2016-07-25 14:20:00.000
2016-07-25 14:30:00.000
2016-07-25 14:40:00.000
2016-07-25 14:50:00.000
2016-07-25 ...
2016-07-25 18:00:00.000

谢谢！

Answer 1

Demo Here

使用递归 CTE，这可以插入到一个函数中..

declare @start datetime='2016-07-25 14:00:00.000'
declare @end datetime='2016-07-25 18:00:00.000'

;With cte(starttime,endtime)
as
(
select @start,@end
union all
select dateadd(minute,5,starttime),endtime
from cte c
where datediff(minute,c.starttime,c.endtime)>0
)
select * from cte

Answer 2

您可以使用 Numbers/Tally 表，但我使用 UDF 创建动态日期范围

Select * from [dbo].[udf-Create-Range-Date]('2016-07-25 14:00:00.000','2016-07-25 18:00:00.000','MI',5)

注意不仅是start/end参数，还有DatePart和Increment

返回

RetVal
2016-07-25 14:00:00.000
2016-07-25 14:05:00.000
2016-07-25 14:10:00.000
...
2016-07-25 17:55:00.000
2016-07-25 18:00:00.000

UDF

CREATE FUNCTION [dbo].[udf-Create-Range-Date] (@DateFrom datetime,@DateTo datetime,@DatePart varchar(10),@Incr int)

Returns 
@ReturnVal Table (RetVal datetime)

As
Begin
    With DateTable As (
        Select DateFrom = @DateFrom
        Union All
        Select Case @DatePart
               When 'YY' then DateAdd(YY, @Incr, df.dateFrom)
               When 'QQ' then DateAdd(QQ, @Incr, df.dateFrom)
               When 'MM' then DateAdd(MM, @Incr, df.dateFrom)
               When 'WK' then DateAdd(WK, @Incr, df.dateFrom)
               When 'DD' then DateAdd(DD, @Incr, df.dateFrom)
               When 'HH' then DateAdd(HH, @Incr, df.dateFrom)
               When 'MI' then DateAdd(MI, @Incr, df.dateFrom)
               When 'SS' then DateAdd(SS, @Incr, df.dateFrom)
               End
        From DateTable DF
        Where DF.DateFrom < @DateTo
    )

    Insert into @ReturnVal(RetVal) Select DateFrom From DateTable option (maxrecursion 32767)

    Return
End

-- Syntax Select * from [dbo].[udf-Create-Range-Date]('2016-10-01','2020-10-01','YY',1) 
-- Syntax Select * from [dbo].[udf-Create-Range-Date]('2016-10-01','2020-10-01','DD',1) 
-- Syntax Select * from [dbo].[udf-Create-Range-Date]('2016-10-01','2016-10-31','MI',15) 
-- Syntax Select * from [dbo].[udf-Create-Range-Date]('2016-10-01','2016-10-02','SS',1) 

Answer 3

基本上，我认为您正在寻找的是这个脚本：

CREATE FUNCTION dbo.udf1 (@interval int, @start datetime, @end datetime)
RETURNS @out TABLE (date1 datetime)
BEGIN
--declare @start datetime='2016-07-25 14:00:00.000', @end datetime='2016-07-25 18:00:00.000', @curr datetime

set @curr = @start
while (dateadd(mi,@interval,@curr)<=@end)
begin
    set @curr = dateadd(mi,@interval,@curr)
    insert into @out
    select @curr
end
return;
END

select *
from dbo.udf1(10,'2016-07-25 14:00:00.000','2016-07-25 18:00:00.000')

Answer 4

感谢您的回答，我知道如何使用它

 create function fn_daterange
   (
   @MinDate as datetime,
   @MaxDate as datetime,
   @intval  as datetime
   )
returns table
--**************************************************************************
-- Procedure: fn_daterange()
--    Author: Ron Savage
--      Date: 12/16/2008
--
-- Description:
-- This function takes a starting and ending date and an interval, then
-- returns a table of all the dates in that range at the specified interval.
--
-- Change History:
-- Date        Init. Description
-- 12/16/2008  RS    Created.
-- **************************************************************************
as
return
   WITH times (startdate, enddate, intervl) AS
      (
      SELECT @MinDate as startdate, @MinDate + @intval as enddate, @intval as intervl
         UNION ALL
      SELECT startdate + intervl as startdate, enddate + intervl as enddate, intervl as intervl
      FROM times
      WHERE startdate + intervl <= @MaxDate
      )
   select startdate, enddate from times;

go