2017-06-13 90 views
-2

我已经找遍了互联网之间,我终于这段代码在 Weekend Exclusion排除节假日2个日期

OR

function calcBusinessDays(dDate1, dDate2) { // input given as Date objects 
    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 
    } 

我不知道如何添加代码,以排除假期。我知道我必须创建一个数组来存储假期,但之后我坚持了一点。我需要你的帮助,非常感谢。

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你想排除周末假期吗? –

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我想排除周末和假期。 –

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在这种情况下,您必须知道假期的日期。你有这些信息吗? –

回答

0

通过假期阵列只是循环和检查日期落在范围内,也检查节日是在这种情况下,它已经算周六或周日。下面是修改的例子。

function calcBusinessDays(dDate1, dDate2, holidays) { // input given as Date objects 
    var iWeeks, iDateDiff, iAdjust = 0, i; 
    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 

    for(i = 0; i < holidays.length; i++) { 
     if(holidays[i] >= dDate1 && holidays[i] <= dDate2 && holidays[i].getDay() != 0 && holidays[i].getDay() != 6) { 
     iDateDiff--; 
     } 
    } 

    return (iDateDiff + 1); // add 1 because dates are inclusive 
    } 

var holidays = [ new Date(2017, 5, 2), new Date(2017, 5, 3), new Date(2017, 5, 4), new Date(2017, 5, 5) ]; 

calcBusinessDays(new Date(2017,1,1), new Date(), holidays); 
// returns 93 
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呀呀,我发现它的工作,谢谢!Q –