两个日期间的SQL业务分钟
两个日期间的SQL业务分钟
提问于 2016-09-30 15:57:17
我希望在minutes 2008或更高版本中创建一个函数，用于计算两个日期时间之间的业务分钟。我已经测试了在这里和其他网站上找到的20+答案，而且还找不到一个一致工作的答案。业务的开始和结束时间分别是08:30和17:30。我想知道2016-10-09 18:35和2016:-11-09 9:00之间的区别。那是9月10日下午6点45分和9月11日上午9点我希望能回来30分钟，但他们都是0。我真的不想一分钟一分钟地循环检查每分钟，只是一些基本的SQL“数学”。
回答 1
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回答已采纳
发布于 2016-09-30 16:36:03
下面的SQL可以转换为1函数，或者将它们保持为两个单独的函数。就我个人而言，我认为把它们分开比较有用。你会有较少的触点和可以服务器的多个主.
在Business函数中，我添加了一个选项/插图，以排除周末和/或假日。如果没有必要，只需移除即可。
日期范围函数用于生成动态日期范围.它比递归cte方法快得多。
Select [dbo].[udf-Business-Minutes]('2016-09-10 18:35','2016-09-11 9:00')返回
30业务分钟函数
ALTER Function [dbo].[udf-Business-Minutes] (@D1 Datetime,@D2 Datetime)
Returns int
Begin
Return (
Select Minutes=count(*)
From [dbo].[udf-Range-Date](case when @D1<=@D2 then @D1 else @D2 end,case when @D1<=@D2 then @D2 else @D1 end,'MI',1)
Where Cast(RetVal as time) between '08:31' and '17:30'
and RetVal >case when @D1<=@D2 then @D1 else @D2 end
and DatePart(DW,RetVal) not in (7,1)
and Cast(RetVal as Date) not in (Select Date from (Values
('2016-01-01','New Year''s Day'),
('2016-01-18','Martin Luther King, Jr,'),
('2016-02-15','Washington''s Birthday'),
('2016-03-25','Good Friday'),
('2016-05-30','Memorial Day'),
('2016-07-04','Independence Day'),
('2016-09-05','Labor Day'),
('2016-11-24','Thanksgiving'),
('2016-11-25','Black Friday'),
('2016-12-26','Christmas Day')
) as Holidays (Date,Name)
)
)
End泛型日期范围函数
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)
*/**编辑-做了一个小调整，允许多天更改8:30到8:31，以避免重复计数**
Select [dbo].[udf-Business-Minutes]('2016-09-10 05:00','2016-09-10 19:00')回报540
同时(跨越两天)
Select [dbo].[udf-Business-Minutes]('2016-09-10 05:00','2016-09-11 19:00')返回1080
Select [dbo].[udf-Business-Minutes]('2015-12-31 12:30','2016-01-03 11:30:00') --Returns 300
Select [dbo].[udf-Business-Minutes]('2016-09-30 05:00','2016-09-30 19:00:00') --Returns 540
Select [dbo].[udf-Business-Minutes]('2016-09-29 05:00','2016-09-30 19:00:00') --Returns 1080
Select [dbo].[udf-Business-Minutes]('2015-12-31 12:30','2016-01-03 11:30:00') --Returns 300页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/39795557
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