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 /*=================================================================*/ if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[f_WorkDayADD]') and xtype in (N'FN', N'IF', N'TF')) drop function [dbo].[f_WorkDayADD] GO --在指定日期上,增加指定工作天数后的日期 CREATE FUNCTION f_WorkDayADD( @date datetime, --基础日期 @workday int --要增加的工作日数 )RETURNS datetime AS BEGIN DECLARE @bz int --增加整周的天数 SELECT @bz=CASE WHEN @workday<0 THEN -1 ELSE 1 END ,@date=DATEADD(Week,@workday/5,@date) ,@workday=@workday%5 --增加不是整周的工作天数 WHILE @workday<>0 SELECT @date=DATEADD(Day,@bz,@date), @workday=CASE WHEN (@@DATEFIRST+DATEPART(Weekday,@date)-1)%7 BETWEEN 1 AND 5 THEN @workday-@bz ELSE @workday END --避免处理后的日期停留在非工作日上 WHILE (@@DATEFIRST+DATEPART(Weekday,@date)-1)%7 in(0,6) SET @date=DATEADD(Day,@bz,@date) RETURN(@date) END
9.工作日处理函数(标准节假日)
最新推荐文章于 2024-05-21 14:23:18 发布