在Javascript中添加/减去工作日

Question

我需要一个Date.prototype.addBusDays函数，它将整数作为要添加到日期的工作日数。

但是，有两个注意事项：1。周末，2。假期（我想这将是一个预设数组来比较。如果开始日期和结束日期包含3个假期，那么你将结束日期推出3）

我在网上遇到过一些剧本，我可以想到的一个难题是，让我们说你先解决所有周末，然后你做假期，如果你+1天（由于假期），你的结束日期会被推入再过一个周末... <

有任何想法吗？ 谢谢！

编辑：

这是我正在开发的调度工具的一部分，这意味着日期将与链接在一起的任务相关联。 向任务添加1天将触发重新计算与其关联的所有内容，可能是数据库中的所有日期。

Answer 1

Datageek的解决方案对我有所帮助，但我需要对其进行补充。 这仍然不是假期，但确实在工作日可以选择包括周六和/或太阳，并支持增加负面日期： -

function AddWorkingDays(datStartDate, lngNumberOfWorkingDays, blnIncSat, blnIncSun) {
    var intWorkingDays = 5;
    var intNonWorkingDays = 2;
    var intStartDay = datStartDate.getDay(); // 0=Sunday ... 6=Saturday
    var intOffset;
    var intModifier = 0;

    if (blnIncSat) { intWorkingDays++; intNonWorkingDays--; }
    if (blnIncSun) { intWorkingDays++; intNonWorkingDays--; }
    var newDate = new Date(datStartDate)
    if (lngNumberOfWorkingDays >= 0) {
        // Moving Forward
        if (!blnIncSat && blnIncSun) {
            intOffset = intStartDay;
        } else {
            intOffset = intStartDay - 1;
        }
        // Special start Saturday rule for 5 day week
        if (intStartDay == 6 && !blnIncSat && !blnIncSun) {
            intOffset -= 6;
            intModifier = 1;
        }
    } else {
        // Moving Backward
        if (blnIncSat && !blnIncSun) {
            intOffset = intStartDay - 6;
        } else {
            intOffset = intStartDay - 5;
        }
        // Special start Sunday rule for 5 day week
        if (intStartDay == 0 && !blnIncSat && !blnIncSun) {
            intOffset++;
            intModifier = 1;
        }
    }
    // ~~ is used to achieve integer division for both positive and negative numbers
    newDate.setTime(datStartDate.getTime() + (new Number((~~((lngNumberOfWorkingDays + intOffset) / intWorkingDays) * intNonWorkingDays) + lngNumberOfWorkingDays + intModifier)*86400000));
    return newDate;
}

Answer 2

看看下面的实现。 来自about.com

addWeekdays = function(date, dd) {
  var wks = Math.floor(dd/5);
  var dys = dd.mod(5);
  var dy = this.getDay();
  if (dy === 6 && dys > -1) {
     if (dys === 0) {dys-=2; dy+=2;}
     dys++; dy -= 6;
  }
  if (dy === 0 && dys < 1) {
    if (dys === 0) {dys+=2; dy-=2;}
    dys--; dy += 6;
  }
  if (dy + dys > 5) dys += 2;
  if (dy + dys < 1) dys -= 2;
  date.setDate(date.getDate()+wks*7+dys);
}

var date = new Date();
addWeekdays(date, 9);

Answer 3

（更新）虽然它确实使用递归进行假日处理，但我已经将这个算法放在它的步伐中并且看起来很稳定。

 holidays = [new Date("2/13/2019"), new Date("2/19/2019")]; function addWorkdays(workdays, startDate) { //Make adjustments if the start date is on a weekend let dayOfWeek = startDate.getDay(); let adjustedWorkdays = Math.abs(workdays); if (0 == dayOfWeek || 6 == dayOfWeek) { adjustedWorkdays += (Math.abs((dayOfWeek % 5) + Math.sign(workdays)) % 2) + 1; dayOfWeek = (dayOfWeek - 6) * -1; } let endDate = new Date(startDate); endDate.setDate(endDate.getDate() + (((Math.floor(((workdays >= 0 ? dayOfWeek - 1 : 6 - dayOfWeek) + adjustedWorkdays) / 5) * 2) + adjustedWorkdays) * (workdays < 0 ? -1 : 1))); //If we cross holidays, recompute our end date accordingly let numHolidays = holidays.reduce(function(total, holiday) { return (holiday >= Math.min(startDate, endDate) && holiday <= Math.max(startDate, endDate)) ? total + 1 : total; }, 0); if (numHolidays > 0) { endDate.setDate(endDate.getDate() + Math.sign(workdays)); return addWorkdays((numHolidays - 1) * Math.sign(workdays), endDate); } else return endDate; } 

Answer 4

我会做一个循环。 继续添加天数，直到你达到有效的一天。

Answer 5

我扩展了khellendros74对我的项目的回答，该项目需要在日期选择器中禁用星期日和邮寄假期，并在按下按钮时返回两个日期：在选择日期之后的三个工作日（即非假日和非星期日） datepicker（id为“calendar”的字段）和在datepicker中选择的日期后六个工作日，然后将这两个结果放入几个禁用的输入字段（handDelivered和mailed）。 按下按钮调用函数calculateDates。 这是代码：

var disabledDates = ['11/11/2015', '11/26/2015', '12/25/2015', '01/01/2016','01/18/2016', '02/15/2016','05/30/2016', '07/04/2016','09/05/2016','10/10/2016','11/11/2016','11/24/2016', '12/26/2016','01/02/2017','01/16/2017', '02/20/2017','05/29/2017', '07/04/2017','09/04/2017','10/09/2017','11/10/2017','11/23/2017', '12/25/2017','01/01/2018','01/15/2018', '02/19/2018','05/28/2018', '07/04/2018','09/03/2018','10/08/2018','11/12/2018','11/22/2018', '12/25/2018','01/01/2019','01/21/2019', '02/18/2019','05/27/2019', '07/04/2019','09/02/2019','10/14/2019','11/11/2019','11/28/2019', '12/25/2019','01/01/2020','01/20/2020', '02/17/2020','05/25/2020', '07/03/2020','09/07/2020','10/11/2020','11/26/2020','11/26/2020', '12/25/2020'];

$(function(){

    $('#calendar').datepicker({
        dateFormat: 'mm/dd/yy',
        beforeShowDay: editDays
    });

    function editDays(date) {
        for (var i = 0; i < disabledDates.length; i++) {
            if (new Date(disabledDates[i]).toString() == date.toString() || date.getDay() == 0) {             
                 return [false];
            }
        }
        return [true];
     }   

});

function calculateDates()
{
    if( !$('#calendar').val()){
        alert("Please enter a date.");
        document.getElementById('calendar').focus();
        return false;
    }

    var dayThreeAdd = 0;
    var daySixAdd = 0;

    for (var i = 0; i < disabledDates.length; i++) {
        var oneDays = AddWorkingDays($('#calendar').val(),1,true,false);
        var twoDays = AddWorkingDays($('#calendar').val(),2,true,false);
        var threeDays = AddWorkingDays($('#calendar').val(),3,true,false);
        var fourDays = AddWorkingDays($('#calendar').val(),4,true,false);
        var fiveDays = AddWorkingDays($('#calendar').val(),5,true,false);
        var sixDays = AddWorkingDays($('#calendar').val(),6,true,false);

        if (new Date(disabledDates[i]).toString() == oneDays.toString()) {
             dayThreeAdd++;
             daySixAdd++;
        }
        if (new Date(disabledDates[i]).toString() == twoDays.toString()) {             
             dayThreeAdd++;
             daySixAdd++;
        }
        if (new Date(disabledDates[i]).toString() == threeDays.toString()) {             
             dayThreeAdd++;
             daySixAdd++;
        }
        if (new Date(disabledDates[i]).toString() == fourDays.toString()) {
            daySixAdd++;
        }
        if (new Date(disabledDates[i]).toString() == fiveDays.toString()) {
            daySixAdd++;
        }
        if (new Date(disabledDates[i]).toString() == sixDays.toString()) {
            daySixAdd++;
        }

    }

    var threeDays = AddWorkingDays($('#calendar').val(),(3 + dayThreeAdd),true,false);
    var sixDays = AddWorkingDays($('#calendar').val(),(6 + daySixAdd),true,false);

    $('#handDelivered').val((threeDays.getMonth()+1) + '/' + threeDays.getDate() + '/' + (threeDays.getYear()+1900));
    $('#mailed').val((sixDays.getMonth()+1) + '/' + sixDays.getDate() + '/' + (sixDays.getYear()+1900));



}

function AddWorkingDays(datStartDate, lngNumberOfWorkingDays, blnIncSat, blnIncSun) {
    datStartDate = new Date(datStartDate);
    var intWorkingDays = 5;
    var intNonWorkingDays = 2;
    var intStartDay = datStartDate.getDay(); // 0=Sunday ... 6=Saturday
    var intOffset;
    var intModifier = 0;

    if (blnIncSat) { intWorkingDays++; intNonWorkingDays--; }
    if (blnIncSun) { intWorkingDays++; intNonWorkingDays--; }
    var newDate = new Date(datStartDate)
    if (lngNumberOfWorkingDays >= 0) {
        // Moving Forward
        if (!blnIncSat && blnIncSun) {
            intOffset = intStartDay;
        } else {
            intOffset = intStartDay - 1;
        }
        // Special start Saturday rule for 5 day week
        if (intStartDay == 6 && !blnIncSat && !blnIncSun) {
            intOffset -= 6;
            intModifier = 1;
        }
    } else {
        // Moving Backward
        if (blnIncSat && !blnIncSun) {
            intOffset = intStartDay - 6;
        } else {
            intOffset = intStartDay - 5;
        }
        // Special start Sunday rule for 5 day week
        if (intStartDay == 0 && !blnIncSat && !blnIncSun) {
            intOffset++;
            intModifier = 1;
        }
    }
    // ~~ is used to achieve integer division for both positive and negative numbers
    newDate.setTime(datStartDate.getTime() + (new Number((~~((lngNumberOfWorkingDays + intOffset) / intWorkingDays) * intNonWorkingDays) + lngNumberOfWorkingDays + intModifier)*86400000));
    return newDate;
}

Answer 6

解决整个问题的简单解决方案; 你可以循环过去，以跳过工作日和假期：

 Date.prototype.holidays = { // fill in common holidays all: [ '0101', // Jan 01 '1225' // Dec 25 ], 2016: [ // add year specific holidays '0104' // Jan 04 2016 ], 2017: [ // And so on for other years. ] }; Date.prototype.addWorkingDays = function(days) { while (days > 0) { this.setDate(this.getDate() + 1); if (!this.isHoliday()) days--; } return this; }; Date.prototype.substractWorkingDays = function(days) { while (days > 0) { this.setDate(this.getDate() - 1); if (!this.isHoliday()) days--; } return this; }; Date.prototype.isHoliday = function() { function zeroPad(n) { n |= 0; return (n < 10 ? '0' : '') + n; } // if weekend return true from here it self; if (this.getDay() == 0 || this.getDay() == 6) { return true; } var day = zeroPad(this.getMonth() + 1) + zeroPad(this.getDate()); // if date is present in the holiday list return true; return !!~this.holidays.all.indexOf(day) || (this.holidays[this.getFullYear()] ? !!~this.holidays[this.getFullYear()].indexOf(day) : false); }; // Uasage var date = new Date('2015-12-31'); date.addWorkingDays(10); alert(date.toDateString()); // Mon Jan 18 2016 date.substractWorkingDays(10); alert(date.toDateString()) // Thu Dec 31 2015