计算Javascript中两个日期之间的工作日除外

Question

我有一个javascript函数，它计算两个日期之间的工作日，它可以工作，但问题是它不考虑假期。如何修改此函数，例如在异常数组中添加假日？

在互联网上搜索了这个问题，但没有找到关于假期的例外情况。

例如假日数组：

var holidays = ['2016-05-03','2016-05-05'];

我有计算这个的功能：

function workingDaysBetweenDates(d0, d1) {
    var startDate = parseDate(d0);
    var endDate = parseDate(d1);  
    // Validate input
    if (endDate < startDate)
        return 0;

    // Calculate days between dates
    var millisecondsPerDay = 86400 * 1000; // Day in milliseconds
    startDate.setHours(0,0,0,1);  // Start just after midnight
    endDate.setHours(23,59,59,999);  // End just before midnight
    var diff = endDate - startDate;  // Milliseconds between datetime objects    
    var days = Math.ceil(diff / millisecondsPerDay);

    // Subtract two weekend days for every week in between
    var weeks = Math.floor(days / 7);
    days = days - (weeks * 2);

    // Handle special cases
    var startDay = startDate.getDay();
    var endDay = endDate.getDay();

    // Remove weekend not previously removed.   
    if (startDay - endDay > 1)         
        days = days - 2;      

    // Remove start day if span starts on Sunday but ends before Saturday
    if (startDay == 0 && endDay != 6)
        days = days - 1  

    // Remove end day if span ends on Saturday but starts after Sunday
    if (endDay == 6 && startDay != 0)
        days = days - 1  

    return days;
}
function parseDate(input) {
    // Transform date from text to date
  var parts = input.match(/(\d+)/g);
  // new Date(year, month [, date [, hours[, minutes[, seconds[, ms]]]]])
  return new Date(parts[0], parts[1]-1, parts[2]); // months are 0-based
}

在jsfiddle中做了一个例子：

JSFiddle example

也许还有一些其他功能可以在Jquery中轻松使用？

Answer 1

实现目标的最简单方法是在开始和结束日期之间寻找这些天。

$(document).ready(function(){
	$('#calc').click(function(){
  var d1 = $('#d1').val();
  var d2 = $('#d2').val();
		$('#dif').text(workingDaysBetweenDates(d1,d2));
	});
});

function workingDaysBetweenDates(d0, d1) {
	var holidays = ['2016-05-03','2016-05-05'];
    var startDate = parseDate(d0);
    var endDate = parseDate(d1);  
    // Validate input
    if (endDate < startDate) {
        return 0;
    }
    // Calculate days between dates
    var millisecondsPerDay = 86400 * 1000; // Day in milliseconds
    startDate.setHours(0,0,0,1);  // Start just after midnight
    endDate.setHours(23,59,59,999);  // End just before midnight
    var diff = endDate - startDate;  // Milliseconds between datetime objects    
    var days = Math.ceil(diff / millisecondsPerDay);
    
    // Subtract two weekend days for every week in between
    var weeks = Math.floor(days / 7);
    days -= weeks * 2;

    // Handle special cases
    var startDay = startDate.getDay();
    var endDay = endDate.getDay();
    
    // Remove weekend not previously removed.   
    if (startDay - endDay > 1) {
        days -= 2;
    }
    // Remove start day if span starts on Sunday but ends before Saturday
    if (startDay == 0 && endDay != 6) {
        days--;  
    }
    // Remove end day if span ends on Saturday but starts after Sunday
    if (endDay == 6 && startDay != 0) {
        days--;
    }
    /* Here is the code */
    for (var i in holidays) {
      if ((holidays[i] >= d0) && (holidays[i] <= d1)) {
      	days--;
      }
    }
    return days;
}
           
function parseDate(input) {
	// Transform date from text to date
  var parts = input.match(/(\d+)/g);
  // new Date(year, month [, date [, hours[, minutes[, seconds[, ms]]]]])
  return new Date(parts[0], parts[1]-1, parts[2]); // months are 0-based
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="text" id="d1" value="2016-05-02"><br>
<input type="text" id="d2" value="2016-05-08">

<p>Working days count: <span id="dif"></span></p>
<button id="calc">Calc</button>

<p>
Now it shows 5 days, but I need for example add holidays 
3 and 5 May (2016-05-03 and 2016-05-05) so the result will be 3 working days
</p>

Answer 2

尝试：

var startDate = new Date('05/03/2016');
var endDate = new Date('05/10/2016');
var numOfDates = getBusinessDatesCount(startDate,endDate);

function getBusinessDatesCount(startDate, endDate) {
    var count = 0;
    var curDate = startDate;
    while (curDate <= endDate) {
        var dayOfWeek = curDate.getDay();
        if(!((dayOfWeek == 6) || (dayOfWeek == 0)))
           count++;
        curDate.setDate(curDate.getDate() + 1);
    }
    alert(count)
    return count;
}

Answer 3

最高答案实际上是有效的，但有缺陷。
当斋月在星期六或星期日时，它仍会减少一天。

将此添加到现有代码中：

.... /* Here is the code */
for (var i in holidays) {
  if ((holidays[i] >= d0) && (holidays[i] <= d1)) {

    // Check if specific holyday is Saturday or Sunday
      var yourDate = new Date(holidays[i]);
      if(yourDate.getDay() === 6 || yourDate.getDay() === 0){

          // If it is.. do nothing

      } else {

          // if it is not, reduce a day..
          days--;
      }
  }
}

Answer 4

简单地从你得到的值（在你的小提琴中）减少数组的长度

.force

您还需要从假日过滤掉周末

var numberofdayswithoutHolidays= 5;
var holidays = ['2016-05-03','2016-05-05'];
alert( numberofdayswithoutHolidays - holidays.length );

Answer 5

您也可以尝试以下代码：

const moment = require('moment-business-days');
/**
 *
 * @param {String} date - iso Date
 * @returns {Number} difference between now and @param date
 */
const calculateDaysLeft = date => {
  try {
     return moment(date).businessDiff(moment(new Date()))
  } catch (err) {
     throw new Error(err)
  }
}

Answer 6

由于我的JS生锈，所以我采取了与@OscarGarcia类似的方法，主要是作为锻炼。

虽然看起来很相似，但要注意的是，如果假日恰好在周六或周日，则每天不减两次。这样，您可以预加载一个重复日期列表（例如12月25日，1月1日，7月4日，该日期可能在或可能不在其他工作日-周一至周五-）

$(document).ready(function(){
    $('#calc').click(function(){
  var d1 = $('#d1').val();
  var d2 = $('#d2').val();
        $('#dif').text(workingDaysBetweenDates(d1,d2));
    });
});
function workingDaysBetweenDates(d0, d1) {
    var startDate = parseDate(d0);
    var endDate = parseDate(d1);
    // populate the holidays array with all required dates without first taking care of what day of the week they happen
    var holidays = ['2018-12-09', '2018-12-10', '2018-12-24', '2018-12-31'];
    // Validate input
    if (endDate < startDate)
        return 0;

    var z = 0; // number of days to substract at the very end
    for (i = 0; i < holidays.length; i++)
    {
        var cand = parseDate(holidays[i]);
        var candDay = cand.getDay();

      if (cand >= startDate && cand <= endDate && candDay != 0 && candDay != 6)
      {
        // we'll only substract the date if it is between the start or end dates AND it isn't already a saturday or sunday
        z++;
      }

    }
    // Calculate days between dates
    var millisecondsPerDay = 86400 * 1000; // Day in milliseconds
    startDate.setHours(0,0,0,1);  // Start just after midnight
    endDate.setHours(23,59,59,999);  // End just before midnight
    var diff = endDate - startDate;  // Milliseconds between datetime objects    
    var days = Math.ceil(diff / millisecondsPerDay);

    // Subtract two weekend days for every week in between
    var weeks = Math.floor(days / 7);
    days = days - (weeks * 2);

    // Handle special cases
    var startDay = startDate.getDay();
    var endDay = endDate.getDay();

    // Remove weekend not previously removed.   
    if (startDay - endDay > 1)         
        days = days - 2;      

    // Remove start day if span starts on Sunday but ends before Saturday
    if (startDay == 0 && endDay != 6)
        days = days - 1  

    // Remove end day if span ends on Saturday but starts after Sunday
    if (endDay == 6 && startDay != 0)
        days = days - 1  

    // substract the holiday dates from the original calculation and return to the DOM
    return days - z;
}
function parseDate(input) {
    // Transform date from text to date
  var parts = input.match(/(\d+)/g);
  // new Date(year, month [, date [, hours[, minutes[, seconds[, ms]]]]])
  return new Date(parts[0], parts[1]-1, parts[2]); // months are 0-based
}

2018-12-09是星期天...使用此代码，它将只减去一次（因为是星期日），而不是两次（就像我们只检查是否是国定假日一样）

Answer 7

$(document).ready(() => {
  $('#calc').click(() => {
  var d1 = $('#d1').val();
  var d2 = $('#d2').val();
    $('#dif').text(workingDaysBetweenDates(d1,d2));
  });
});

let workingDaysBetweenDates = (d0, d1) => {
  /* Two working days and an sunday (not working day) */
  var holidays = ['2016-05-03', '2016-05-05', '2016-05-07'];
  var startDate = parseDate(d0);
  var endDate = parseDate(d1);  

// Validate input
  if (endDate < startDate) {
    return 0;
  }

// Calculate days between dates
  var millisecondsPerDay = 86400 * 1000; // Day in milliseconds
  startDate.setHours(0, 0, 0, 1);  // Start just after midnight
  endDate.setHours(23, 59, 59, 999);  // End just before midnight
  var diff = endDate - startDate;  // Milliseconds between datetime objects    
  var days = Math.ceil(diff / millisecondsPerDay);

  // Subtract two weekend days for every week in between
  var weeks = Math.floor(days / 7);
  days -= weeks * 2;

  // Handle special cases
  var startDay = startDate.getDay();
  var endDay = endDate.getDay();
    
  // Remove weekend not previously removed.   
  if (startDay - endDay > 1) {
    days -= 2;
  }
  // Remove start day if span starts on Sunday but ends before Saturday
  if (startDay == 0 && endDay != 6) {
    days--;  
  }
  // Remove end day if span ends on Saturday but starts after Sunday
  if (endDay == 6 && startDay != 0) {
    days--;
  }
  /* Here is the code */
  holidays.forEach(day => {
    if ((day >= d0) && (day <= d1)) {
      /* If it is not saturday (6) or sunday (0), substract it */
      if ((parseDate(day).getDay() % 6) != 0) {
        days--;
      }
    }
  });
  return days;
}
           
function parseDate(input) {
	// Transform date from text to date
  var parts = input.match(/(\d+)/g);
  // new Date(year, month [, date [, hours[, minutes[, seconds[, ms]]]]])
  return new Date(parts[0], parts[1]-1, parts[2]); // months are 0-based
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="text" id="d1" value="2016-05-02"><br>
<input type="text" id="d2" value="2016-05-08">

<p>Working days count: <span id="dif"></span></p>
<button id="calc">Calc</button>

<p>
Now it shows 5 days, but I need for example add holidays 
3 and 5 May (2016-05-03 and 2016-05-05) so the result will be 3 working days
</p>

Answer 8

我认为这种解决方案要简单得多

const numberOfDaysInclusive = (d0, d1) => {
  return 1 + Math.round((d1.getTime()-d0.getTime())/(24*3600*1000));
}

const numberOfWeekends = (d0, d1) => {
    const days = numberOfDaysInclusive(d0, d1); // total number of days
    const sundays = Math.floor((days + (d0.getDay() + 6) % 7) / 7); // number of sundays
    return 2*sundays + (d1.getDay()==6) - (d0.getDay()==0); // multiply sundays by 2 to get both sat and sun, +1 if d1 is saturday, -1 if d0 is sunday
}

const numberOfWeekdays = (d0, d1) => {
    return numberOfDaysInclusive(d0, d1) - numberOfWeekends(d0, d1);
}

Answer 9

获取两个日期之间的所有工作日：

private getCorrectWeekDays(StartDate,EndDate){
 let _weekdays = [0,1,2,3,4];
 var wdArr= [];
 var currentDate = StartDate;
 while (currentDate <= EndDate) {
  if ( _weekdays.includes(currentDate.getDay())){
    wdArr.push(currentDate);
    //if you want to format it to yyyy-mm-dd
    //wdArr.push(currentDate.toISOString().split('T')[0]);
  }
  currentDate.setDate(currentDate.getDate() +1);
}

 return wdArr;
}