sql 工作日处理函数(标准节假日)

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