2012-03-08 111 views
1

我正在嘗試做2件事。在某一個月基於當前日的工作日內最新用Javascript計算幾個工作日

  1. 計算數(即今天是2012年3月7日,因此,5個工作日過去了)在給定的去的工作日
  2. 計算數一個月基於當前日(即今天是2012年3月7日,因此,也有這個月剩下17個工作日

這裏任何幫助,將不勝感激

編輯:。 這是我」至今嘗試過:

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,以獲得工作日和工作日。

+1

您到目前爲止嘗試過了哪些?你試圖解決的具體問題是否給你帶來麻煩? – maerics 2012-03-08 01:26:50

+0

你是否也想考慮假期呢? – 2012-03-08 01:27:01

+0

Maerics我已經更新了我的要求。雅各布,假期並不重要。 – st4ck0v3rfl0w 2012-03-08 01:34:53

回答

1

下面是一個非常簡單的函數,它可以循環使用幾天,應該足夠快,因爲它不應該循環超過31次。如果當天是營業日,則計入過去的天數:

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