這個演算法還不夠嚴謹，沒有考慮到周
六、日調換假期等細節的情況。
1、建節假日表
create table [holiday] (
[id] [int] identity (1, 1) not null ,
[name] [nvarchar] (50) collate chinese_prc_ci_as not null ,
[begindate] [datetime] not null ,
[enddate] [datetime] not null ,
[adduser] [nvarchar] (50) collate chinese_prc_ci_as not null ,
[addtime] [datetime] not null ,
constraint [pk_holiday] primary key clustered
([id]
) on [primary]
) on [primary]
go2、計算工作日函式
create function [dbo].[workday]
(@beginday datetime,
@endday datetime
)returns int
asbegin
--set datefirst 1
declare @caldays int
declare @id int
select @caldays=0
while datediff(d, @beginday,@endday)>0
begin
select @beginday=dateadd(day,1,@beginday)
if datepart(dw,@beginday)>1 and datepart(dw,@beginday)<7
begin
select @id=count(*) from holiday where @beginday between begindate and dateadd(s,-1,dateadd(day,1,enddate))
if(@id=0)
select @caldays=@caldays+1
end
end
return @caldays
end3、推算指定日期前n個工作日的日期函式
create function [dbo].[begindate]
(@workday int ,
@endday datetime
)returns datetime
asbegin
declare @beginday datetime
declare @id int
select @beginday=@endday
while datediff(d, @beginday,@endday)<@workday
begin
select @beginday=dateadd(day,-1,@beginday)
if datepart(dw,@beginday)>1 and datepart(dw,@beginday)<7
begin
select @id=count(*) from holiday where @beginday between begindate and dateadd(s,-1,dateadd(day,1,enddate))
if(@id>0)
select @workday=@workday+1
endelse
begin
select @workday=@workday+1
end
endif(datepart(dw,@beginday)=2)
begin
select @beginday=dateadd(day,-2,@beginday)
end
return @beginday
end
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