2012-03-08 107 views
1

我正在尝试做2件事。在某一个月基于当前日的工作日内最新用Javascript计算几个工作日

  1. 计算数(即今天是2012年3月7日,因此,5个工作日过去了)在给定的去的工作日
  2. 计算数一个月基于当前日(即今天是2012年3月7日,因此,也有这个月剩下17个工作日

这里任何帮助,将不胜感激

编辑:。 这是我”至今尝试过:

function isWeekday(year, month, day) {var day = new Date(year, month, day).getDay();return day !=0 && day !=6;} 
function getWeekdaysInMonth(month, year) {var days = daysInMonth(month, year);var weekdays = 0;for(var i=0; i< days; i++) {if (isWeekday(year, month, i+1)) weekdays++;}return weekdays;} 
function calcBusinessDays(dDate1, dDate2) { 
    var iWeeks, iDateDiff, iAdjust = 0; 
    if (dDate2 < dDate1) return -1;     // error code if dates transposed 
    var iWeekday1 = dDate1.getDay();    // day of week 
    var iWeekday2 = dDate2.getDay(); 
    iWeekday1 = (iWeekday1 == 0) ? 7 : iWeekday1; // change Sunday from 0 to 7 
    iWeekday2 = (iWeekday2 == 0) ? 7 : iWeekday2; 
    if ((iWeekday1 > 5) && (iWeekday2 > 5)) iAdjust = 1; // adjustment if both days on weekend 
    iWeekday1 = (iWeekday1 > 5) ? 5 : iWeekday1; // only count weekdays 
    iWeekday2 = (iWeekday2 > 5) ? 5 : iWeekday2; 
    // calculate differnece in weeks (1000mS * 60sec * 60min * 24hrs * 7 days = 604800000) 
    iWeeks = Math.floor((dDate2.getTime() - dDate1.getTime())/604800000) 
    if (iWeekday1 <= iWeekday2) { 
    iDateDiff = (iWeeks * 5) + (iWeekday2 - iWeekday1) 
    } else { 
    iDateDiff = ((iWeeks + 1) * 5) - (iWeekday1 - iWeekday2) 
    } 
    iDateDiff -= iAdjust       // take into account both days on weekend 
    return (iDateDiff + 1);       // add 1 because dates are inclusive 
} 

不太清楚如何把它放在一起,以获得工作日和工作日。

+1

您到目前为止尝试过了哪些?你试图解决的具体问题是否给你带来麻烦? – maerics 2012-03-08 01:26:50

+0

你是否也想考虑假期呢? – 2012-03-08 01:27:01

+0

Maerics我已经更新了我的要求。雅各布,假期并不重要。 – st4ck0v3rfl0w 2012-03-08 01:34:53

回答

1

下面是一个非常简单的函数,它可以循环使用几天,应该足够快,因为它不应该循环超过31次。如果当天是营业日,则计入过去的天数:

function businessDays(date) { 

    // Copy date 
    var t = new Date(date); 
    // Remember the month number 
    var m = date.getMonth(); 
    var d = date.getDate(); 
    var daysPast = 0, daysToGo = 0; 
    var day; 

    // Count past days 
    while (t.getMonth() == m) { 
    day = t.getDay(); 
    daysPast += (day == 0 || day == 6)? 0 : 1; 
    t.setDate(--d); 
    } 

    // Reset and count days to come 
    t = new Date(date); 
    t.setDate(t.getDate() + 1); 
    d = t.getDate(); 

    while (t.getMonth() == m) { 
    day = t.getDay(); 
    daysToGo += (day == 0 || day == 6)? 0 : 1; 
    t.setDate(++d); 
    } 
    return [daysPast, daysToGo]; 
} 

alert(businessDays(new Date(2012,2,7))); // 7-Mar-2012 => 5, 17