Question

这是我对以下SQL的标准：

自2013年1月7日至2013年3月18日

1 - 2013年1月7日后的第一个星期日是2013年1月13日 - 完成（FirstSunday） 2 - 第一个星期日2013年3月18日之前是2013年3月17日 - 完成（LastSunday） 3 - 2013年1月13日至2013年3月17日期间的天数= 63，除以63 = 7 = 9，+ 1 = 10

所以我需要我的约会做的是使用上面第1和第2行（第一个和上个星期日）的结果，所以我得到第3节的正确结果。你能帮忙吗？

Declare @From Datetime
Declare @To Datetime

Set @From = '07 Jan 2013'
Set @To = '18 Mar 2013'

Select dateadd(dd, CASE WHEN datepart(weekday, @From) = 1 THEN 0 ELSE 8 - datepart(weekday, @From) END,@From) as FirstSunday, 
dateadd(dd, CASE WHEN datepart(weekday, @To) = 1 THEN 0 ELSE 1 - datepart(weekday, @To) END,@To) as LastSunday

,datediff(d,@From, @To)

Answer 1

我希望它有所帮助：

Declare @From  Datetime 
Declare @To  Datetime 
Declare @temp  Datetime 
Declare @firstSunday DateTime
Declare @lastSunday DateTime

Set @From = '7 Jan 2013' 
Set @To = '18 Mar 2013'


set @temp = @From
WHILE ( @temp <= @To )
    BEGIN 
        IF DATEPART(dw, @temp) = 1
        BEGIN
            SET @firstSunday = @temp
            BREAK
        END

        set @temp = dateadd(dd,1,@temp)

    END


set @temp = @To
WHILE ( @temp >= @From )
    BEGIN

    IF DATEPART(dw, @temp) = 1
        BEGIN
            SET @lastSunday = @temp
            BREAK
        END

        set @temp = dateadd(dd,-1,@temp)

    END

SELECT @firstSunday, @lastSunday, DATEDIFF(dd,@firstSunday, @lastSunday)/7 +1

Answer 2

Declare @From Datetime
Declare @To Datetime

Set @From = '01 Mar 2013'
Set @To = '31 Mar 2013'


SELECT Datediff(Week, dateadd(dd, CASE WHEN datepart(weekday, @From) = 1 THEN 0 ELSE 8 - datepart(weekday, @From) END,@From), 
dateadd(dd, CASE WHEN datepart(weekday, @To) = 1 THEN 0 ELSE 1 - datepart(weekday, @To) END,@To))+1

Answer 3

我认为这比my previous answer

更优雅但更少可读

Fiddle demo here

declare @From date = '20130107' --'7 Jan 2013' 
declare @To date = '20130318'  --'18 Mar 2013' 

select datediff(day,dateadd(day,wkdays,@from),@to)/7 + 
       case when wkdays = 0 and 
        datename(weekday,dateadd(day,wkdays,@from)) <> 'sunday' then 0 else 1 end 
from (values (0),(1),(2),(3),(4),(5),(6)) t(wkdays)
where dateadd(day,wkdays,@from) between @from and @to
      and datename(weekday,dateadd(day,wkdays,@from)) = 'sunday'

使用修改后的@From和@To日期来获得新结果

3 个答案: