http://topic.csdn.net/u/20070322/12/2a8d5fcf-2d5b-49e1-b58a-c7277e15ee95.html
方法1
if exists (select * from dbo.sysobjects where id = object_id(N '[dbo].[f_WorkDay] ') and xtype in (N 'FN ', N 'IF ', N 'TF '))
drop function [dbo].[f_WorkDay]
GO
--计算两个日期相差的工作天数
CREATE FUNCTION f_WorkDay(
@dt_begin datetime, --计算的开始日期
@dt_end datetime --计算的结束日期
)RETURNS int
AS
BEGIN
DECLARE @workday int,@i int,@bz bit,@dt datetime
IF @dt_begin> @dt_end
SELECT @bz=1,@dt=@dt_begin,@dt_begin=@dt_end,@dt_end=@dt
ELSE
SET @bz=0
SELECT @i=DATEDIFF(Day,@dt_begin,@dt_end)+1,
@workday=@i/7*5,
@dt_begin=DATEADD(Day,@i/7*7,@dt_begin)
WHILE @dt_begin <=@dt_end
BEGIN
SELECT @workday=CASE
WHEN (@@DATEFIRST+DATEPART(Weekday,@dt_begin)-1)%7 BETWEEN 1 AND 5
THEN @workday+1 ELSE @workday END,
@dt_begin=@dt_begin+1
END
RETURN(CASE WHEN @bz=1 THEN -@workday ELSE @workday END)
END
GO
方法2
create function fn_test(@begin datetime,@end datetime)
returns int
As
BEGIN
declare @i int,@j int
set @i=0
set @j=0
if @end> @begin
begin
while dateadd(d,@i,@begin) <=@end
begin
if datepart(weekday,dateadd(d,@i,@begin)) not in(1,7)
set @j=@j+1
set @i=@i+1
end
end
return @j
END
