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[p] 思路是找出該月1號是星期幾，推出該周的周末是幾號，把兩個值放入一個二維數(shù)組，依此推出下面幾周的起始號和結(jié)束號，然后把所在取的號比較得出所在周 [br] <-% [br] '/////////////////////////////////////////// [br] 'function:getweek(date) [br] 'description:取得該日期所在月份的當(dāng)前周 [br] 'para:date要計(jì)算的日期 [br] 'return:所在周數(shù) [br] 'author:愚人(yurensky@hotmail.com) [br] '////////////////////////////////////////// [br] [br] function getweek(strdate) [br] dim strmon,stryear,nweek [br] [br] strdate = formatdatetime(strdate,2) [br] strmon = month(strdate) [br] stryear = year(strdate) [br] strday = day(strdate) [br] nweek = weekday(formatdatetime(stryear&"-"&strmon&"-1",2)) - 1 [br] [br] dim arrrange(5,2) [br] [br] for i = 0 to 4 [br] if i = 0 then [br] arrrange(i,0) = 1 [br] if nweek = 6 then [br] arrrange(i,1) = arrrange(i,0) [br] else [br] arrrange(i,1) = arrrange(i,0) + 6 - nweek [br] end if [br] else [br] arrrange(i,0) = arrrange(i-1,1)+1 [br] arrrange(i,1) = arrrange(i,0) + 6 [br] end if [br] next [br] [br] dim currweek [br] [br] for i = 0 to 4 [br] if arrrange(i,0) <= strday and strday <= arrrange(i,1) then [br] currweek = i + 1 [br] end if [br] next [br] [br] if currweek = "" then [br] currweek = 0 [br] end if [br] [br] getweek = currweek [br] end function [br] [br] [br] 'test [br] dim nweek,cdate [br] cdate = now() [br] nweek = getweek(cdate) [br] [br] response.write cdate & "是"&year(cdate) &"年"&month(cdate)&"月的第"&nweek&"周" [br] %->[/p]
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[p]另外一個思路：[br]<-script type="text/vbs"-> [br] month_week=int(datepart("ww",now)-((datepart("y",now)-datepart("d",now))/7)) [br] alert(month_week) [br] 'datepart("ww",now) 現(xiàn)在是一年第多少周 [br] '(datepart("y",now) 現(xiàn)在是一年第多少天 [br] 'datepart("d",now) 現(xiàn)在是這個月第多少天 [br] '年的天數(shù)減這個月的天數(shù) 除上7就是這個月之前的周數(shù) [br] '總的周數(shù)－上個月的周數(shù),表示今天是這個月的多少周 [br] '如果有問題，請高手指點(diǎn) [br] <-/script->[/p]
[p] [/p]
[p]另一個方法：[br]<-script type="text/vbs"-> [br] dim month_week [br] month_week = datediff("ww", dateadd("d",-datepart("d",now), now), now) [br] 'dateadd("d",-datepart("d",now), now) '上個月的月未 與今天對比 [br] alert(month_week) [br] <-/script->[/p][/align]