--************
--在两个日期范围里所跨越那几周返回如：1，2表是第一周和第二周，

declare @aa table (date datetime,weekdays int )
declare @i int
set @i=datediff(day,@bdate,@edate)
while(@i>=0)
begin
insert @aa
values (dateadd(day,@i,@bdate),datepart(week,dateadd(day,@i,@bdate)))
set @i=@i-1
end
select weekdays
into #week
from @AA group by weekdays

--************
--在日期范围里减去周六、周日的天数
create function a (@Sdate datetime ,@Edate datetime)
returns int
as
begin
declare @aa table (date datetime)
declare @i int
set @i=datediff(day,@Sdate,@Edate)
while(@i>=0)
begin
insert @aa
values (dateadd(day,@i,@Sdate))
set @i=@i-1
end
select @i= count(*) from @aa where datepart(weekday,date) not in (1,7)
return @i

end

--如：select dbo.A(\'2004-10-01\',\'2004-10-11\')
--返回结果为7

--***********
--输入第几周得到此周的开始、结束日期
declare @FirstDayOfYear datetime--年頭
declare @FirstDayWeekOfYear datetime --第一周的第一天
declare @BDate datetime
declare @EDate datetime

select @FirstDayOfYear= dateadd(yy,datediff(yy,0,getdate()),0)
select @FirstDayWeekOfYear=@FirstDayOfYear - datepart(dw, @FirstDayOfYear)+1

select @EDate=dateadd(ww,@week,@FirstDayWeekOfYear-1 )
select @BDate= dateadd(ww,-1,dateadd(dd,1,@EDate) )

SET @BDate =convert(datetime, convert(char(10),@BDate,101))
SET @EDate =convert(datetime, convert(char(10),@EDate,101))

如果本文没有解决您的问题，请进老妖怪开发者社区提问