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) {
    let count = 0;
    const curDate = new Date(startDate.getTime());
    while (curDate <= endDate) {
        const dayOfWeek = curDate.getDay();
        if(dayOfWeek !== 0 && dayOfWeek !== 6) count++;
        curDate.setDate(curDate.getDate() + 1);
    }
    alert(count);
    return count;
}

实现它的最简单方法是在开始日期和结束日期之间寻找这些天数。

编辑：我添加了一个额外的验证，以确保只从holidays数组中减去工作日。

$(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>

我对@OscarGarcia 采取了类似的方法，主要是作为练习，因为我的 JS 生rust了。

虽然看起来很相似，但如果假期恰好在周六或周日，请注意不要将一天减去两次。这样，您可以预先加载重复日期列表（例如 12 月 25 日、1 月 1 日、7 月 4 日，这些日期可能是也可能不是工作日 - 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 是一个星期天......使用这个代码，它只会被减去一次（因为是星期天）而不是两次（就像我们只检查它是否是国定假日一样）

你也可以试试这段代码：

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)
  }
}

最佳答案实际上有效，但有一个缺陷。
当圣日在星期六或星期日时，它仍然减少一天。

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--;
      }
  }
}