Question

我正在试图弄清楚如何计算TSQL中两个日期之间“星期二”的数量？

“星期二”可以是任何价值。

Answer 1

谢谢t-clausen.dk，保存了几天。没有每天的实例：

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

Answer 2

declare @from datetime= '9/20/2011' 
declare @to datetime  = '9/28/2011' 

select datediff(day, -6, @to)/7-datediff(day, -5, @from)/7

在@from找到星期二之前的第一个星期一的一周。
找到@to
减去周数

Answer 3

@ t-clausen.dk＆amp; Andriy M回复t-clausen.dks response and comments

查询使用的事实是1900-01-01是星期一。并且1900-01-01是日期0.

select dateadd(day,0,0)

datediff - 函数的第二个参数是startdate。

所以你要比较'1899-12-26'和你的@ to-date和'1899-12-26'是星期二

select datename(dw,dateadd(day, 0, -6)), datename(dw, '1899-12-26')

使用相同事实的第二个日期也是如此。

事实上，你可以与任何已知的星期二和相应的星期三进行比较（这不是你正在研究的日期间隔）。

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]

基本上算法是“所有星期二减去所有星期三”。

Answer 4

查看此问题：Count work days between two dates

有几种方法可以为你的问题利用这个问题的答案。

如何计算SQL中两个日期之间“星期二”的数量？

4 个答案: