Question

你好，有人可以告诉我如何根据给定的日期范围（不是按年或月的周数）获得一周的数字，但是日期范围。

(defvar *acceptor* (make-instance 'easy-acceptor :port 4242))
(start *acceptor*)
(define-easy-handler (test :uri "/test") () "Pass")

结果

Eg : From Date : '6/26/2017' and To Date: '7/23/2017'

Answer 1

如果您对TVF（表格值函数）开放

我经常使用udf来创建动态日期/时间范围。它比递归CTE更快并且是参数驱动的。您提供日期/时间范围，DatePart和增量。

示例

Select WeekNbr = 'Week '+cast(RetSeq as varchar(10)) ,WeekBeg = cast(RetVal as date) ,WeekEnd = cast(RetVal+6 as date) From [dbo].[udf-Range-Date]('2017-06-26','2017-07-23','WK',1)

<强>返回

WeekNbr WeekBeg WeekEnd Week 1 2017-06-26 2017-07-02 Week 2 2017-07-03 2017-07-09 Week 3 2017-07-10 2017-07-16 Week 4 2017-07-17 2017-07-23

感兴趣的UDF

CREATE FUNCTION [dbo].[udf-Range-Date] (@R1 datetime,@R2 datetime,@Part varchar(10),@Incr int) Returns Table Return ( with cte0(M) As (Select 1+Case @Part When 'YY' then DateDiff(YY,@R1,@R2)/@Incr When 'QQ' then DateDiff(QQ,@R1,@R2)/@Incr When 'MM' then DateDiff(MM,@R1,@R2)/@Incr When 'WK' then DateDiff(WK,@R1,@R2)/@Incr When 'DD' then DateDiff(DD,@R1,@R2)/@Incr When 'HH' then DateDiff(HH,@R1,@R2)/@Incr When 'MI' then DateDiff(MI,@R1,@R2)/@Incr When 'SS' then DateDiff(SS,@R1,@R2)/@Incr End), cte1(N) As (Select 1 From (Values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) N(N)), cte2(N) As (Select Top (Select M from cte0) Row_Number() over (Order By (Select NULL)) From cte1 a, cte1 b, cte1 c, cte1 d, cte1 e, cte1 f, cte1 g, cte1 h ), cte3(N,D) As (Select 0,@R1 Union All Select N,Case @Part When 'YY' then DateAdd(YY, N*@Incr, @R1) When 'QQ' then DateAdd(QQ, N*@Incr, @R1) When 'MM' then DateAdd(MM, N*@Incr, @R1) When 'WK' then DateAdd(WK, N*@Incr, @R1) When 'DD' then DateAdd(DD, N*@Incr, @R1) When 'HH' then DateAdd(HH, N*@Incr, @R1) When 'MI' then DateAdd(MI, N*@Incr, @R1) When 'SS' then DateAdd(SS, N*@Incr, @R1) End From cte2 ) Select RetSeq = N+1 ,RetVal = D From cte3,cte0 Where D<=@R2 ) /* Max 100 million observations -- Date Parts YY QQ MM WK DD HH MI SS Syntax: Select * from [dbo].[udf-Range-Date]('2016-10-01','2020-10-01','YY',1) Select * from [dbo].[udf-Range-Date]('2016-01-01','2017-01-01','MM',1) */

从SQL Server的日期范围中获取周

1 个答案: