Question

如何获取开始日期td { cursor: pointer; }（7月1日，周五）和下周二之间的日期差异（以天为单位）？

谢谢！

Answer 1

嗯。怎么样？

select (case when datename(dow, date) = 'Monday' then 1
             when datename(dow, date) = 'Tuesday' then 7
             when datename(dow, date) = 'Wednesday' then 6
             when datename(dow, date) = 'Thursday' then 5
             when datename(dow, date) = 'Friday' then 4
             when datename(dow, date) = 'Saturday' then 3
             when datename(dow, date) = 'Sunday' then 2
        end)

Answer 2

你可以试试这个：

SELECT DATEDIFF(day,'2014-06-05','2014-08-05') AS DiffDate

参考：http://www.w3schools.com/sql/func_datediff.asp

Answer 3

递归cte怎么样？

;with cte as
(SELECT 0 as d
, DATENAME(dw, DATEADD(day, 0, CONVERT(date, '20160701'))) dname
union all
select d + 1
, DATENAME(dw, DATEADD(day, d + 1, CONVERT(date, '20160701'))) dname
from cte
where d < 6
)
select d as NumberOfDaysUntil
, dname
from cte
where dname = 'Tuesday'

从当天开始，它获得一周的工作日并找到第一个名称匹配。如果您想跳过开始的日期，请将0更改为1，将< 6更改为< 7

SQL Server：如何找到给定日期和下一个相对日期之间的差异？

3 个答案: