SQL取出 所有周六 周日的日期

SQL取出 所有周六 周日的日期

create table SatSun([id] int identity(1,1),[date] datetime,[weekday] char(6))

go

declare @datetime datetime,@weekday char(6)

set @datetime='2007-1-1'

while @datetime<='2007-12-31'

begin

select @weekday=datename(weekday,@datetime)

if @weekday= '星期六' insert SatSun([date],[weekday]) values(@datetime,@weekday)

if @weekday='星期日' insert SatSun([date],[weekday]) values(@datetime,'星期天')

select @datetime=dateadd(day,1,@datetime)

end

go

select * from satsun

drop table satsun

/*

id          date                                                   weekday

----------- ------------------------------------------------------ -------

1           2007-01-06 00:00:00.000                                星期六

2           2007-01-07 00:00:00.000                                星期天

3           2007-01-13 00:00:00.000                                星期六

4           2007-01-14 00:00:00.000                                星期天

5           2007-01-20 00:00:00.000                                星期六

6           2007-01-21 00:00:00.000                                星期天

7           2007-01-27 00:00:00.000                                星期六

8           2007-01-28 00:00:00.000                                星期天

9           2007-02-03 00:00:00.000                                星期六

……

100         2007-12-16 00:00:00.000                                星期天

101         2007-12-22 00:00:00.000                                星期六

102         2007-12-23 00:00:00.000                                星期天

103         2007-12-29 00:00:00.000                                星期六

104         2007-12-30 00:00:00.000                                星期天

*/（所影响的行数为 104 行）

****************************************************************************

declare @t table(dd datetime)

declare @d1 datetime

select @d1='2007-1-1'

while(year(@d1) <2008)

begin

insert into @t select @d1

set @d1=dateadd(dd,1,@d1)

end

select * from @t where datepart(weekday,dd)=6 or datepart(weekday,dd)=7

****************************************************************************

declare @i int

declare @d datetime

set @i=1

set @d='2007-1-1'

Create table #a (Dat datetime)

while year(@d)=2007

begin

insert into #a values (@d )

set @i=@i+1

set @d=dateadd(day,1,@d)

end

select *,datepart(dw,Dat) as a from #a where  datepart(dw,Dat) =1 or datepart(dw,Dat) =7

****************************************************************************

/*

功能: 计算在某一段时间内某周几（如星期一）的所有日期

设计：OK_008

时间：2006-10

*/

DECLARE @Date datetime

DECLARE @StartDate datetime

DECLARE @EndDate datetime

DECLARE @WeekDay int

DECLARE @i int

SET DATEFIRST 7 --设置每周的第一天

SET @StartDate='2006-01-01' --统计的开始日期

SET @EndDate='2006-12-31' --统计的结束日期

SET @WeekDay=1 --根据实际的@@DATEFIRST而定，一般默认是7，如 @StartDate='2006-01-01'时候， @WeekDay=3表示星期二

SET @i=DATEPART(weekday,@StartDate)

PRINT '每周的第1天设置@@DATEFIRST: '+CAST(@@DATEFIRST AS nvarchar(1))

PRINT '开始日期对应一周的第几天: '+CAST(@i AS nvarchar(1))

IF(@i<=@WeekDay AND @i<7)

SET @i=@WeekDay-@i

ELSE IF(@i<=@WeekDay AND @i=7)

SET @i=@i-@WeekDay

ELSE

SET @i=@@DATEFIRST-@i+@WeekDay

SET @Date=DATEADD(day,@i,@StartDate)

WHILE @Date<=@EndDate

BEGIN

IF(@StartDate<=@Date) PRINT CONVERT(nvarchar(10),@Date,121)

SET @Date=DATEADD(Week,1,@Date)

END

GO

/* ==============运行结果================*/

/*

每周的第1天设置@@DATEFIRST: 7

开始日期对应一周的第几天: 1

2006-01-01

2006-01-08

2006-01-15

2006-01-22

2006-01-29

..........

*/

select dateadd(day,x,col),'星期二' from

(

select cast('2006-1-1' as datetime) as col

)a cross join

(

SELECT top 365 b8.i+b7.i + b6.i + b5.i + b4.i +b3.i +b2.i + b1.i + b0.i x

FROM(SELECT 0 i UNION ALL SELECT 1) b0

CROSS JOIN(SELECT 0 i UNION ALL SELECT 2) b1

CROSS JOIN(SELECT 0 i UNION ALL SELECT 4) b2

CROSS JOIN(SELECT 0 i UNION ALL SELECT 8) b3

CROSS JOIN(SELECT 0 i UNION ALL SELECT 16) b4

CROSS JOIN(SELECT 0 i UNION ALL SELECT 32) b5

CROSS JOIN(SELECT 0 i UNION ALL SELECT 64) b6

CROSS JOIN(SELECT 0 i UNION ALL SELECT 128) b7

CROSS JOIN(SELECT 0 i UNION ALL SELECT 256) b8

order by 1

)b

where datepart(dw,dateadd(day,x,col))=3 (这个地方改值换取其他日期)

2006-01-03 00:00:00.000                                星期二

2006-01-10 00:00:00.000                                星期二

2006-01-17 00:00:00.000                                星期二

2006-01-24 00:00:00.000                                星期二

2006-01-31 00:00:00.000                                星期二

****************************************************************************

DECLARE @t TABLE(date0 DATETIME)

DECLARE @st DATETIME,@et DATETIME

SET @st='2007-01-01'

SET @et='2008-01-01'

WHILE @st <@et

BEGIN

INSERT INTO @t VALUES(@st)

SELECT @st=DATEADD(DAY,1,@st)

END

SELECT date0,DATENAME(weekday,date0) FROM @t WHERE DATEPART(weekday,date0+@@DATEFIRST-1) IN (6,7)

****************************************************************************

由于工作需要，在SQL Server 2005 下面写了一个计算两个日期之间相差工作日的函数。函数是以一个星期5天工作日计算，没有剔除五一国庆等假期。代码如下：

SET ANSI_NULLS ON

GO

SET QUOTED_IDENTIFIER ON

GO

-- =============================================

-- Author:  Sinmen

-- Create date: 2007-11-01

-- Description: 计算两个日期之间相差的工作日

-- =============================================

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)

--判断传入的开始日期是否比结束日期大

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