请问如何用时间函数实现某两年2月份总共有多少天？

2007-12-16 07:53:48 345 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

js 判断2月份多少天 function backDay() { //返回2月份的天数 var year = prompt('请输入年份'); if (isRunyear(year)) { alert('这一年是闰年，2月份有29天'); } else { alert('这一年是平年，2月份有28天'); } } ... 阅读详情

请问如何用时间函数实现某两年间2月份总共有多少天？

例如实现1980年到2005年间2月份总共多少天？

...全文

Java——Calendar类练习，如何获取任意年份的2月份有多少天需求：键盘录入任意一个年份，获取任意一年的二月有多少天 a: 键盘录入一个年份 b: 创建一个Calendar对象 c: 把这个Calendar的时间设置为a中录入的年的3月1号注意2表示3月 d: 向前推算一天 e: 获取月中的天 f: 输出 package org.westos.demo; import java.util.Calendar; import java.util.Scanner; public class MyT 阅读详情

tired_bird

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切换为时间正序

dobear_0922 2007-12-23

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全部按28天算，再加上中间闰年次数，，，

dawugui 2007-12-23

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CREATE PROCEDURE my_proc(@syear int , @eyear int)

as 

begin

  declare @i as int

  set @i = 0

  while @syear <= @eyear

  begin

      set @i = @i + datediff(day , cast(@syear as varchar) + '-02-01' , cast(@syear as varchar) + '-03-01')

      set @syear = @syear + 1

  end

  print 'result = ' + cast(@i as varchar)

end

go



exec my_proc 1980 , 2005

drop procedure my_proc



/*

result = 735

*/

dawugui 2007-12-23

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CREATE PROCEDURE my_proc(@syear int , @eyear int)

as 

begin

  declare @i as int

  set @i = 0

  while @syear <= @eyear

  begin

      set @i = @i + datediff(day , cast(@syear as varchar) + '-02-01' , cast(@syear as varchar) + '-03-01')

      set @syear = @syear + 1

  end

  print 'result = ' + cast(@i as varchar)

end

go



exec my_proc 1980 , 2005

drop procedure my_proc



/*

result = 735

*/

中国风 2007-12-23

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没遇到楼主说的情况，这样试试..





create   function   F_year(@startDate   nvarchar(4),@EndDate   nvarchar(4)) 

returns   int 

as 

begin 

	declare   @MonthDay   int 

	set   @MonthDay=0 

	while   @startDate!> @EndDate 

	        select   @MonthDay=@MonthDay+

				datediff(dd,

						cast(@startDate+'0201' as datetime),

						cast(Dateadd(month,1,@startDate+'0201') as datetime)

						) , 

	        @startDate=@startDate+1 

	return   @MonthDay 

end 



go

select   dbo.F_year(2006,2007) 



--drop function F_year

tired_bird 2007-12-23

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在运行roy_88的F＿year函数即
create function F_year(@startDate int,@EndDate int)
returns int
as
begin
declare @MonthDay int
set @MonthDay=0
while @startDate!>@EndDate
select @MonthDay=@MonthDay+datediff(dd,rtrim(@startDate)+'0201',dateadd(month,1,rtrim(@startDate)+'0201')),
@startDate=@startDate+1
return @MonthDay
end

能成功，但在执行　select dbo.F_year(2006,2007)
语句时，却处于一直查询状态，不出结果，为什么，我可能哪步出错了

JL99000 2007-12-23

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方法多多
不再写了
主要是个遍历查询的过程

liangCK 2007-12-16

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BS技术帖接分..

-狙击手- 2007-12-16

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接分

liangCK 2007-12-16

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那给分结帖吧.

tired_bird 2007-12-16

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谢谢，刚才没看到roy_88的函数！

liangCK 2007-12-16

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那你使用roy_88的函数就可以了啊.

tired_bird 2007-12-16

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先谢谢大家的帮忙！

但如果把1980　和　2005　换成变量＠yyear1 和＠yyear2　呢？
即实现＠yyear1到yyear2年间2月份总共多少天？能实现吗？感谢！　

liangCK 2007-12-16

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楼上..你累了吧.

xinleile 2007-12-16

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select sum(num) as [1980-2005 二月份天数] from (
select day(dateadd(dd,-1,'1980-03-01')) as num union all

select day(dateadd(dd,-1,'1981-03-01')) as num union all

select day(dateadd(dd,-1,'1982-03-01')) as num union all

select day(dateadd(dd,-1,'1983-03-01')) as num union all

select day(dateadd(dd,-1,'1984-03-01')) as num union all

select day(dateadd(dd,-1,'1985-03-01')) as num union all

select day(dateadd(dd,-1,'1986-03-01')) as num union all

select day(dateadd(dd,-1,'1987-03-01')) as num union all

select day(dateadd(dd,-1,'1988-03-01')) as num union all

select day(dateadd(dd,-1,'1989-03-01')) as num union all

select day(dateadd(dd,-1,'1990-03-01')) as num union all

select day(dateadd(dd,-1,'1991-03-01')) as num union all

select day(dateadd(dd,-1,'1992-03-01')) as num union all

select day(dateadd(dd,-1,'1993-03-01')) as num union all

select day(dateadd(dd,-1,'1994-03-01')) as num union all

select day(dateadd(dd,-1,'1995-03-01')) as num union all

select day(dateadd(dd,-1,'1996-03-01')) as num union all

select day(dateadd(dd,-1,'1997-03-01')) as num union all

select day(dateadd(dd,-1,'1998-03-01')) as num union all

select day(dateadd(dd,-1,'1999-03-01')) as num union all

select day(dateadd(dd,-1,'2000-03-01')) as num union all

select day(dateadd(dd,-1,'2001-03-01')) as num union all

select day(dateadd(dd,-1,'2002-03-01')) as num union all

select day(dateadd(dd,-1,'2003-03-01')) as num union all

select day(dateadd(dd,-1,'2004-03-01')) as num union all

select day(dateadd(dd,-1,'2005-03-01')) as num ) dd

/*
1980-2005 二月份天数
-----------------------------------
735

*/

liangCK 2007-12-16

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我上面错了点.现在正确了.

select top 1000 id=identity(int,0,1) into # from syscolumns a,syscolumns b



select 共有天数=sum(days) from

(

    select days=datepart(day,dateadd(day,-1,convert(char(8),dateadd(month,1,[date]),120)+'1'))

    from

    (

        select [date]=dateadd(year,id,'1980-02-01')

        from #

        where dateadd(year,id,'1980-02-01')<='2005-02-01'

    ) b

) a



drop table #



/*

共有天数        

----------- 

735



（所影响的行数为 1 行）

*/

中国风 2007-12-16

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--只计算二月份的天数和

select dbo.F_year(1980,2007)

            

----------- 

791



（所影响的行数为 1 行）

liangCK 2007-12-16

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select top 1000 id=identity(int,1,1) into # from syscolumns a,syscolumns b



select 共有天数=sum(days) from

(

    select days=datepart(day,dateadd(day,-1,convert(char(8),dateadd(month,1,[date]),120)+'1'))

    from

    (

        select [date]=dateadd(year,id,'1980-02-01')

        from #

        where dateadd(year,id,'1980-02-01')<='2005-02-01'

    ) b

) a



drop table #



/*

共有天数        

----------- 

706



（所影响的行数为 1 行）

*/

中国风 2007-12-16

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create function F_year(@startDate int,@EndDate int)

returns int

as

begin

declare @MonthDay int

set @MonthDay=0

while @startDate!>@EndDate

	select @MonthDay=@MonthDay+datediff(dd,rtrim(@startDate)+'0201',dateadd(month,1,rtrim(@startDate)+'0201')),

	@startDate=@startDate+1

return @MonthDay

end

--

select dbo.F_year(2006,2007)

liangCK 2007-12-16

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select top 1000 id=identity(int,1,1) into # from syscolumns a,syscolumns b



select 共有天数=sum(days) from

(

    select days=datepart(day,dateadd(day,-1,convert(char(8),dateadd(month,1,[date]),120)+'1'))

    from

    (

        select [date]=dateadd(year,id,'1980-02-01')

        from #

        where dateadd(year,id,'1980-02-01')<='2005-02-01'

    ) b

) a



drop table #

请问如何用时间函数实现某两年2月份总共有多少天？

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