小智 15
谢谢t-clausen.dk,保存了几天.为了得到每天的实例:
declare @from datetime= '3/1/2013'
declare @to datetime = '3/31/2013'
select
datediff(day, -7, @to)/7-datediff(day, -6, @from)/7 AS MON,
datediff(day, -6, @to)/7-datediff(day, -5, @from)/7 AS TUE,
datediff(day, -5, @to)/7-datediff(day, -4, @from)/7 AS WED,
datediff(day, -4, @to)/7-datediff(day, -3, @from)/7 AS THU,
datediff(day, -3, @to)/7-datediff(day, -2, @from)/7 AS FRI,
datediff(day, -2, @to)/7-datediff(day, -1, @from)/7 AS SAT,
datediff(day, -1, @to)/7-datediff(day, 0, @from)/7 AS SUN
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t-c*_*.dk 10
declare @from datetime= '9/20/2011'
declare @to datetime = '9/28/2011'
select datediff(day, -6, @to)/7-datediff(day, -5, @from)/7
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@t-clausen.dk 和 Andriy M 作为对t-clausen.dks 回复和评论的回应
该查询使用 1900-01-01 是星期一这一事实。1900-01-01 是日期 0。
select dateadd(day,0,0)
- 函数的第二个参数datediff是开始日期。
因此,您将“1899-12-26”与您的@to-date 进行比较,“1899-12-26”是星期二
select datename(dw,dateadd(day, 0, -6)), datename(dw, '1899-12-26')
第二次约会也使用相同的事实。
事实上,您可以与任何已知的星期二和相应的星期三(不在您正在调查的日期间隔内)进行比较。
declare @from datetime= '2011-09-19'
declare @to datetime = '2011-10-15'
select datediff(day, '2011-09-13', @to)/7-datediff(day, '2011-09-14', @from)/7 as [works]
,datediff(day, '2011-10-18', @to)/7-datediff(day, '2011-10-19', @from)/7 as [works too]
,datediff(day, '2011-09-27', @to)/7-datediff(day, '2011-09-28', @from)/7 as [dont work]
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基本上,该算法是“所有星期二减去所有星期三”。
查看这个问题:计算两个日期之间的工作日
您也可以通过几种方法利用该问题的答案。