由於工作需要，在SQL Server 2005 下面寫了一個計算兩個日期之間相差工作日的函式。函式是以一個星期5天工作日計算，沒有剔除五一國慶等假期。程式碼如下： 
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
-- =============================================
-- Author:  Sinmen
-- Create date: 2007-11-01
-- Description: 計算兩個日期之間相差的工作日
-- =============================================
ALTER FUNCTION [dbo].[WorkDatediff]
(
@begin_date datetime,
@end_date datetime
)
RETURNS int
AS
BEGIN
DECLARE @return_date_quantity int
DECLARE @temp datetime
DECLARE @week_quantity int
DECLARE @day_quantity int
DECLARE @begin_day_of_week int
DECLARE @end_day_of_week int
DECLARE @add_begin_day_quantity int
DECLARE @add_end_day_quantity int
set @day_quantity = Datediff(d,@begin_date,@end_date)
--判斷傳入的開始日期是否比結束日期大
if @day_quantity < 0
  begin
   set @temp = @begin_date
   set @begin_date = @end_date
   set @end_date = @temp
  end
set @week_quantity = Abs(Datediff(ww,@begin_date,@end_date)) - 1
if @week_quantity < 0
  set @week_quantity = 0
set @begin_day_of_week = Datepart(dw,@begin_date) - 1
set @end_day_of_week = Datepart(dw,@end_date) - 1

set @add_begin_day_quantity = case 
         when @begin_day_of_week > 5 then 0
         else 6 - @begin_day_of_week --(5 - @begin_day_of_week + 1)
        end
set @add_end_day_quantity = case
         when abs(@day_quantity) < 8  then 0 
         when @end_day_of_week > 5 then 5
         else @end_day_of_week
        end
if @day_quantity = 0
  set @return_date_quantity = 0
else
  set @return_date_quantity = @week_quantity * 5 + @add_begin_day_quantity + @add_end_day_quantity
if @day_quantity < 0
  set @return_date_quantity = @return_date_quantity * -1

RETURN @return_date_quantity
END
GO
SET ANSI_NULLS OFF
GO
SET QUOTED_IDENTIFIER OFF
GO

Original text:http://www.cnblogs.com/wayne-ivan/archive/2008/04/10/1146308.html