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issue with getting special day [duplicate]

I have an date, i need to add no. of days to get future date but weekends should be excluded.
i.e
input date = "9-DEC-2011";
No. of days to add = '13';
next date should be "28-Dec-2011"
Here weekends(sat/sun) are not counted.

Try this
var startDate = "9-DEC-2011";
startDate = new Date(startDate.replace(/-/g, "/"));
var endDate = "", noOfDaysToAdd = 13, count = 0;
while(count < noOfDaysToAdd){
endDate = new Date(startDate.setDate(startDate.getDate() + 1));
if(endDate.getDay() != 0 && endDate.getDay() != 6){
//Date.getDay() gives weekday starting from 0(Sunday) to 6(Saturday)
count++;
}
}
alert(endDate);//You can format this date as per your requirement
Working Demo

#ShankarSangoli
Here's a newer version which avoid recreating a Date object on each loop, note that it's wrapped in a function now.
function calcWorkingDays(fromDate, days) {
var count = 0;
while (count < days) {
fromDate.setDate(fromDate.getDate() + 1);
if (fromDate.getDay() != 0 && fromDate.getDay() != 6) // Skip weekends
count++;
}
return fromDate;
}
alert(calcWorkingDays(new Date("9/DEC/2011"), 13));

Here is an elegant solution without any looping or external library:
function addBusinessDaysToDate(date, days) {
var day = date.getDay();
date = new Date(date.getTime());
date.setDate(date.getDate() + days + (day === 6 ? 2 : +!day) + (Math.floor((days - 1 + (day % 6 || 1)) / 5) * 2));
return date;
}
var date = "9-DEC-2011";
var newDate = addBusinessDaysToDate(new Date(date.replace(/-/g, "/")), 13);
console.log(newDate.toString().replace(/\S+\s(\S+)\s(\d+)\s(\d+)\s.*/, '$2-$1-$3')); // alerts "28-Dec-2011"

or you can be like this
function addWeekdays(date, weekdays) {
var newDate = new Date(date.getTime());
var i = 0;
while (i < weekdays) {
newDate.setDate(newDate.getDate() + 1);
var day = newDate.getDay();
if (day > 1 && day < 7) {
i++;
}
}
return newDate;
}
var currentDate = new Date('10/31/2014');
var targetDate = addWeekdays(currentDate, 45);
alert(targetDate);

Using moment.js:
const DATE_FORMAT = 'D-MMM-YYYY';
const SUNDAY = 0; // moment day index
const SATURDAY = 6; // moment day index
const WEEKENDS = [SATURDAY, SUNDAY];
function addBusinessDays(stringDate, numberOfDays, dateFormat = DATE_FORMAT) {
const date = moment(stringDate, dateFormat);
let count = 0;
while (count < numberOfDays) {
date.add(1, 'day');
// Skip weekends
if (WEEKENDS.includes(date.day())) {
continue;
}
// Increment count
count++;
}
return date.format(dateFormat);
}
// Test cases
console.log(addBusinessDays('3-Mar-2021', 1)); // 4-Mar-2021
console.log(addBusinessDays('3-Mar-2021', 2)); // 5-Mar-2021
console.log(addBusinessDays('3-Mar-2021', 3)); // 8-Mar-2021
console.log(addBusinessDays('3-Mar-2021', 4)); // 9-Mar-2021
console.log(addBusinessDays('3-Mar-2021', 5)); // 10-Mar-2021
console.log(addBusinessDays('9-Dec-2011', 13)); // 28-Dec-2011
console.log(addBusinessDays('10-Dec-2011', 13)); // 28-Dec-2011 (Saturday, so remain on Friday)
console.log(addBusinessDays('11-Dec-2011', 13)); // 28-Dec-2011 (Sunday, so remain on Friday)
console.log(addBusinessDays('12-Dec-2011', 13)); // 29-Dec-2011
console.log(addBusinessDays('13-Dec-2011', 13)); // 30-Dec-2011
console.log(addBusinessDays('14-Dec-2011', 13)); // 2-Jan-2012
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.24.0/moment.min.js"></script>

This question is quite old, but all the previous answers are iterating over the days one by one. That could be inefficient for a large number of days. This works for me, assuming days is a positive int and the startDate is a working day:
function addWorkingDates(startDate, days) {
const current_day = startDate.getDay() - 1; // Week day, starting on Monday
const weekend_days = 2 * parseInt((current_day + days) / 5);
startDate.setDate(changed_to.getDate() + days + weekend_days);
}
addWorkingDates(new Date(),5)

For some reason it was more intuitive to me to try this recursively. This version doesn't account for holidays, but you could change the isValid function to check whatever.
function addWeekdaysToDate(date, numberToAdd) {
var isValid = function(d) { return d.getDay() !== 0 && d.getDay() !== 6 }
if(Math.abs(numberToAdd) > 1) {
return addWeekdaysToDate(
addWeekdaysToDate(date, Math.sign(numberToAdd)),
numberToAdd - Math.sign(numberToAdd)
)
} else if(Math.abs(numberToAdd) === 1) {
var result = new Date(date)
result.setDate(result.getDate() + Math.sign(numberToAdd))
if(isValid(result)) {
return result
} else {
return addWeekdaysToDate(result, Math.sign(numberToAdd))
}
} else if(numberToAdd === 0) {
return date
}
return false
}
console.log(addWeekdaysToDate(new Date(), 1))
console.log(addWeekdaysToDate(new Date(), 5))
console.log(addWeekdaysToDate(new Date(), -7))
In certain browsers you may need a polyfill for Math.sign:
Math.sign = Math.sign || function(x) {
x = +x; // convert to a number
if (x === 0 || isNaN(x)) {
return Number(x);
}
return x > 0 ? 1 : -1;
}

try this solution
<script language="javascript">
function getDateExcludeWeekends(startDay, startMonth, startYear, daysToAdd) {
var sdate = new Date();
var edate = new Date();
var dayMilliseconds = 1000 * 60 * 60 * 24;
sdate.setFullYear(startYear,startMonth,startDay);
edate.setFullYear(startYear,startMonth,startDay+daysToAdd);
var weekendDays = 0;
while (sdate <= edate) {
var day = sdate.getDay()
if (day == 0 || day == 6) {
weekendDays++;
}
sdate = new Date(+sdate + dayMilliseconds);
}
sdate.setFullYear(startYear,startMonth,startDay + weekendDays+daysToAdd);
return sdate;
}
</script>

If you want to get the next working day, from a specific date, use the following code...
function getNextWorkingDay(originalDate) {
var nextWorkingDayFound = false;
var nextWorkingDate = new Date();
var dateCounter = 1;
while (!nextWorkingDayFound) {
nextWorkingDate.setDate(originalDate.getDate() + dateCounter);
dateCounter++;
if (!isDateOnWeekend(nextWorkingDate)) {
nextWorkingDayFound = true;
}
}
return nextWorkingDate;
}
function isDateOnWeekend(date) {
if (date.getDay() === 6 || date.getDay() === 0)
return true;
else
return false;
}

Try this
function calculate() {
var noOfDaysToAdd = 13;
var startDate = "9-DEC-2011";
startDate = new Date(startDate.replace(/-/g, "/"));
var endDate = "";
count = 0;
while (count < noOfDaysToAdd) {
endDate = new Date(startDate.setDate(startDate.getDate() + 1));
if (endDate.getDay() != 0 && endDate.getDay() != 6) {
count++;
}
}
document.getElementById("result").innerHTML = endDate;
}
<div>
Date of book delivery: <span id="result"></span><br /><br />
<input type="button" onclick="calculate();" value="Calculate" />
<br>
<br>
</div>

I want to input date calculated Year and month show two text box like txtYear and txtMonth [duplicate]

How to get the difference between two dates in years, months, and days in JavaScript, like: 10th of April 2010 was 3 years, x month and y days ago?
There are lots of solutions, but they only offer the difference in the format of either days OR months OR years, or they are not correct (meaning not taking care of actual number of days in a month or leap years, etc). Is it really that difficult to do that?
I've had a look at:
http://momentjs.com/ -> can only output the difference in either years, months, OR days
http://www.javascriptkit.com/javatutors/datedifference.shtml
http://www.javascriptkit.com/jsref/date.shtml
http://timeago.yarp.com/
www.stackoverflow.com -> Search function
In php it is easy, but unfortunately I can only use client-side script on that project. Any library or framework that can do it would be fine, too.
Here are a list of expected outputs for date differences:
//Expected output should be: "1 year, 5 months".
diffDate(new Date('2014-05-10'), new Date('2015-10-10'));
//Expected output should be: "1 year, 4 months, 29 days".
diffDate(new Date('2014-05-10'), new Date('2015-10-09'));
//Expected output should be: "1 year, 3 months, 30 days".
diffDate(new Date('2014-05-10'), new Date('2015-09-09'));
//Expected output should be: "9 months, 27 days".
diffDate(new Date('2014-05-10'), new Date('2015-03-09'));
//Expected output should be: "1 year, 9 months, 28 days".
diffDate(new Date('2014-05-10'), new Date('2016-03-09'));
//Expected output should be: "1 year, 10 months, 1 days".
diffDate(new Date('2014-05-10'), new Date('2016-03-11'));

How precise do you need to be? If you do need to take into account common years and leap years, and the exact difference in days between months then you'll have to write something more advanced but for a basic and rough calculation this should do the trick:
today = new Date()
past = new Date(2010,05,01) // remember this is equivalent to 06 01 2010
//dates in js are counted from 0, so 05 is june
function calcDate(date1,date2) {
var diff = Math.floor(date1.getTime() - date2.getTime());
var day = 1000 * 60 * 60 * 24;
var days = Math.floor(diff/day);
var months = Math.floor(days/31);
var years = Math.floor(months/12);
var message = date2.toDateString();
message += " was "
message += days + " days "
message += months + " months "
message += years + " years ago \n"
return message
}
a = calcDate(today,past)
console.log(a) // returns Tue Jun 01 2010 was 1143 days 36 months 3 years ago
Keep in mind that this is imprecise, in order to calculate the date with full precision one would have to have a calendar and know if a year is a leap year or not, also the way I'm calculating the number of months is only approximate.
But you can improve it easily.

Actually, there's a solution with a moment.js plugin and it's very easy.
You might use moment.js
Don't reinvent the wheel again.
Just plug Moment.js Date Range Plugin.
Example:
var starts = moment('2014-02-03 12:53:12');
var ends = moment();
var duration = moment.duration(ends.diff(starts));
// with ###moment precise date range plugin###
// it will tell you the difference in human terms
var diff = moment.preciseDiff(starts, ends, true);
// example: { "years": 2, "months": 7, "days": 0, "hours": 6, "minutes": 29, "seconds": 17, "firstDateWasLater": false }
// or as string:
var diffHuman = moment.preciseDiff(starts, ends);
// example: 2 years 7 months 6 hours 29 minutes 17 seconds
document.getElementById('output1').innerHTML = JSON.stringify(diff)
document.getElementById('output2').innerHTML = diffHuman
<html>
<head>
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.14.1/moment.min.js"></script>
<script src="https://raw.githubusercontent.com/codebox/moment-precise-range/master/moment-precise-range.js"></script>
</head>
<body>
<h2>Difference between "NOW and 2014-02-03 12:53:12"</h2>
<span id="output1"></span>
<br />
<span id="output2"></span>
</body>
</html>

Modified this to be a lot more accurate. It will convert dates to a 'YYYY-MM-DD' format, ignoring HH:MM:SS, and takes an optional endDate or uses the current date, and doesn't care about the order of the values.
function dateDiff(startingDate, endingDate) {
let startDate = new Date(new Date(startingDate).toISOString().substr(0, 10));
if (!endingDate) {
endingDate = new Date().toISOString().substr(0, 10); // need date in YYYY-MM-DD format
}
let endDate = new Date(endingDate);
if (startDate > endDate) {
const swap = startDate;
startDate = endDate;
endDate = swap;
}
const startYear = startDate.getFullYear();
const february = (startYear % 4 === 0 && startYear % 100 !== 0) || startYear % 400 === 0 ? 29 : 28;
const daysInMonth = [31, february, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
let yearDiff = endDate.getFullYear() - startYear;
let monthDiff = endDate.getMonth() - startDate.getMonth();
if (monthDiff < 0) {
yearDiff--;
monthDiff += 12;
}
let dayDiff = endDate.getDate() - startDate.getDate();
if (dayDiff < 0) {
if (monthDiff > 0) {
monthDiff--;
} else {
yearDiff--;
monthDiff = 11;
}
dayDiff += daysInMonth[startDate.getMonth()];
}
return yearDiff + 'Y ' + monthDiff + 'M ' + dayDiff + 'D';
}
// Examples
let dates = [
['2019-05-10','2019-05-10'], // 0Y 0M 0D
['2019-05-09','2019-05-10'], // 0Y 0M 1D
['2018-05-09','2019-05-10'], // 1Y 0M 1D
['2018-05-18','2019-05-10'], // 0Y 11M 23D
['2019-01-09','2019-05-10'], // 0Y 4M 1D
['2019-02-10','2019-05-10'], // 0Y 3M 0D
['2019-02-11','2019-05-10'], // 0Y 2M 27D
['2016-02-11','2019-05-10'], // 3Y 2M 28D - leap year
['1972-11-30','2019-05-10'], // 46Y 5M 10D
['2016-02-11','2017-02-11'], // 1Y 0M 0D
['2016-02-11','2016-03-10'], // 0Y 0M 28D - leap year
['2100-02-11','2100-03-10'], // 0Y 0M 27D - not a leap year
['2017-02-11','2016-02-11'], // 1Y 0M 0D - swapped dates to return correct result
[new Date() - 1000 * 60 * 60 * 24] // 0Y 0M 1D - uses current date
].forEach(([s, e]) => console.log(dateDiff(s, e)));
Older less accurate but much simpler version
#RajeevPNadig's answer was what I was looking for, but his code returns incorrect values as written. This code is not very accurate because it assumes that the sequence of dates from 1 January 1970 is the same as any other sequence of the same number of days. E.g. it calculates the difference from 1 July to 1 September (62 days) as 0Y 2M 3D and not 0Y 2M 0D because 1 Jan 1970 plus 62 days is 3 March.
// startDate must be a date string
function dateAgo(date) {
var startDate = new Date(date);
var diffDate = new Date(new Date() - startDate);
return ((diffDate.toISOString().slice(0, 4) - 1970) + "Y " +
diffDate.getMonth() + "M " + (diffDate.getDate()-1) + "D");
}
Then you can use it like this:
// based on a current date of 2018-03-09
dateAgo('1972-11-30'); // "45Y 3M 9D"
dateAgo('2017-03-09'); // "1Y 0M 0D"
dateAgo('2018-01-09'); // "0Y 2M 0D"
dateAgo('2018-02-09'); // "0Y 0M 28D" -- a little odd, but not wrong
dateAgo('2018-02-01'); // "0Y 1M 5D" -- definitely "feels" wrong
dateAgo('2018-03-09'); // "0Y 0M 0D"
If your use case is just date strings, then this works okay if you just want a quick and dirty 4 liner.

I used this simple code to get difference in Years, Months, days with current date.
var sdt = new Date('1972-11-30');
var difdt = new Date(new Date() - sdt);
alert((difdt.toISOString().slice(0, 4) - 1970) + "Y " + (difdt.getMonth()+1) + "M " + difdt.getDate() + "D");

I think you are looking for the same thing that I wanted. I tried to do this using the difference in milliseconds that javascript provides, but those results do not work in the real world of dates. If you want the difference between Feb 1, 2016 and January 31, 2017 the result I would want is 1 year, 0 months, and 0 days. Exactly one year (assuming you count the last day as a full day, like in a lease for an apartment). However, the millisecond approach would give you 1 year 0 months and 1 day, since the date range includes a leap year. So here is the code I used in javascript for my adobe form (you can name the fields): (edited, there was an error that I corrected)
var f1 = this.getField("LeaseExpiration");
var g1 = this.getField("LeaseStart");
var end = f1.value
var begin = g1.value
var e = new Date(end);
var b = new Date(begin);
var bMonth = b.getMonth();
var bYear = b.getFullYear();
var eYear = e.getFullYear();
var eMonth = e.getMonth();
var bDay = b.getDate();
var eDay = e.getDate() + 1;
if ((eMonth == 0)||(eMonth == 2)||(eMonth == 4)|| (eMonth == 6) || (eMonth == 7) ||(eMonth == 9)||(eMonth == 11))
{
var eDays = 31;
}
if ((eMonth == 3)||(eMonth == 5)||(eMonth == 8)|| (eMonth == 10))
{
var eDays = 30;
}
if (eMonth == 1&&((eYear % 4 == 0) && (eYear % 100 != 0)) || (eYear % 400 == 0))
{
var eDays = 29;
}
if (eMonth == 1&&((eYear % 4 != 0) || (eYear % 100 == 0)))
{
var eDays = 28;
}
if ((bMonth == 0)||(bMonth == 2)||(bMonth == 4)|| (bMonth == 6) || (bMonth == 7) ||(bMonth == 9)||(bMonth == 11))
{
var bDays = 31;
}
if ((bMonth == 3)||(bMonth == 5)||(bMonth == 8)|| (bMonth == 10))
{
var bDays = 30;
}
if (bMonth == 1&&((bYear % 4 == 0) && (bYear % 100 != 0)) || (bYear % 400 == 0))
{
var bDays = 29;
}
if (bMonth == 1&&((bYear % 4 != 0) || (bYear % 100 == 0)))
{
var bDays = 28;
}
var FirstMonthDiff = bDays - bDay + 1;
if (eDay - bDay < 0)
{
eMonth = eMonth - 1;
eDay = eDay + eDays;
}
var daysDiff = eDay - bDay;
if(eMonth - bMonth < 0)
{
eYear = eYear - 1;
eMonth = eMonth + 12;
}
var monthDiff = eMonth - bMonth;
var yearDiff = eYear - bYear;
if (daysDiff == eDays)
{
daysDiff = 0;
monthDiff = monthDiff + 1;
if (monthDiff == 12)
{
monthDiff = 0;
yearDiff = yearDiff + 1;
}
}
if ((FirstMonthDiff != bDays)&&(eDay - 1 == eDays))
{
daysDiff = FirstMonthDiff;
}
event.value = yearDiff + " Year(s)" + " " + monthDiff + " month(s) " + daysDiff + " days(s)"

I have created, yet another one, function for this purpose:
function dateDiff(date) {
date = date.split('-');
var today = new Date();
var year = today.getFullYear();
var month = today.getMonth() + 1;
var day = today.getDate();
var yy = parseInt(date[0]);
var mm = parseInt(date[1]);
var dd = parseInt(date[2]);
var years, months, days;
// months
months = month - mm;
if (day < dd) {
months = months - 1;
}
// years
years = year - yy;
if (month * 100 + day < mm * 100 + dd) {
years = years - 1;
months = months + 12;
}
// days
days = Math.floor((today.getTime() - (new Date(yy + years, mm + months - 1, dd)).getTime()) / (24 * 60 * 60 * 1000));
//
return {years: years, months: months, days: days};
}
Doesn't require any 3rd party libraries. Takes one argument -- date in YYYY-MM-DD format.
https://gist.github.com/lemmon/d27c2d4a783b1cf72d1d1cc243458d56

With dayjs we did it in that way:
export const getAgeDetails = (oldDate: dayjs.Dayjs, newDate: dayjs.Dayjs) => {
const years = newDate.diff(oldDate, 'year');
const months = newDate.diff(oldDate, 'month') - years * 12;
const days = newDate.diff(oldDate.add(years, 'year').add(months, 'month'), 'day');
return {
years,
months,
days,
allDays: newDate.diff(oldDate, 'day'),
};
};
It calculates it perfectly including leap years and different month amount of days.

For quick and easy use I wrote this function some time ago. It returns the diff between two dates in a nice format. Feel free to use it (tested on webkit).
/**
* Function to print date diffs.
*
* #param {Date} fromDate: The valid start date
* #param {Date} toDate: The end date. Can be null (if so the function uses "now").
* #param {Number} levels: The number of details you want to get out (1="in 2 Months",2="in 2 Months, 20 Days",...)
* #param {Boolean} prefix: adds "in" or "ago" to the return string
* #return {String} Diffrence between the two dates.
*/
function getNiceTime(fromDate, toDate, levels, prefix){
var lang = {
"date.past": "{0} ago",
"date.future": "in {0}",
"date.now": "now",
"date.year": "{0} year",
"date.years": "{0} years",
"date.years.prefixed": "{0} years",
"date.month": "{0} month",
"date.months": "{0} months",
"date.months.prefixed": "{0} months",
"date.day": "{0} day",
"date.days": "{0} days",
"date.days.prefixed": "{0} days",
"date.hour": "{0} hour",
"date.hours": "{0} hours",
"date.hours.prefixed": "{0} hours",
"date.minute": "{0} minute",
"date.minutes": "{0} minutes",
"date.minutes.prefixed": "{0} minutes",
"date.second": "{0} second",
"date.seconds": "{0} seconds",
"date.seconds.prefixed": "{0} seconds",
},
langFn = function(id,params){
var returnValue = lang[id] || "";
if(params){
for(var i=0;i<params.length;i++){
returnValue = returnValue.replace("{"+i+"}",params[i]);
}
}
return returnValue;
},
toDate = toDate ? toDate : new Date(),
diff = fromDate - toDate,
past = diff < 0 ? true : false,
diff = diff < 0 ? diff * -1 : diff,
date = new Date(new Date(1970,0,1,0).getTime()+diff),
returnString = '',
count = 0,
years = (date.getFullYear() - 1970);
if(years > 0){
var langSingle = "date.year" + (prefix ? "" : ""),
langMultiple = "date.years" + (prefix ? ".prefixed" : "");
returnString += (count > 0 ? ', ' : '') + (years > 1 ? langFn(langMultiple,[years]) : langFn(langSingle,[years]));
count ++;
}
var months = date.getMonth();
if(count < levels && months > 0){
var langSingle = "date.month" + (prefix ? "" : ""),
langMultiple = "date.months" + (prefix ? ".prefixed" : "");
returnString += (count > 0 ? ', ' : '') + (months > 1 ? langFn(langMultiple,[months]) : langFn(langSingle,[months]));
count ++;
} else {
if(count > 0)
count = 99;
}
var days = date.getDate() - 1;
if(count < levels && days > 0){
var langSingle = "date.day" + (prefix ? "" : ""),
langMultiple = "date.days" + (prefix ? ".prefixed" : "");
returnString += (count > 0 ? ', ' : '') + (days > 1 ? langFn(langMultiple,[days]) : langFn(langSingle,[days]));
count ++;
} else {
if(count > 0)
count = 99;
}
var hours = date.getHours();
if(count < levels && hours > 0){
var langSingle = "date.hour" + (prefix ? "" : ""),
langMultiple = "date.hours" + (prefix ? ".prefixed" : "");
returnString += (count > 0 ? ', ' : '') + (hours > 1 ? langFn(langMultiple,[hours]) : langFn(langSingle,[hours]));
count ++;
} else {
if(count > 0)
count = 99;
}
var minutes = date.getMinutes();
if(count < levels && minutes > 0){
var langSingle = "date.minute" + (prefix ? "" : ""),
langMultiple = "date.minutes" + (prefix ? ".prefixed" : "");
returnString += (count > 0 ? ', ' : '') + (minutes > 1 ? langFn(langMultiple,[minutes]) : langFn(langSingle,[minutes]));
count ++;
} else {
if(count > 0)
count = 99;
}
var seconds = date.getSeconds();
if(count < levels && seconds > 0){
var langSingle = "date.second" + (prefix ? "" : ""),
langMultiple = "date.seconds" + (prefix ? ".prefixed" : "");
returnString += (count > 0 ? ', ' : '') + (seconds > 1 ? langFn(langMultiple,[seconds]) : langFn(langSingle,[seconds]));
count ++;
} else {
if(count > 0)
count = 99;
}
if(prefix){
if(returnString == ""){
returnString = langFn("date.now");
} else if(past)
returnString = langFn("date.past",[returnString]);
else
returnString = langFn("date.future",[returnString]);
}
return returnString;
}

If you are using date-fns and if you dont want to install the Moment.js or the moment-precise-range-plugin. You can use the following date-fns function to get the same result as moment-precise-range-plugin
intervalToDuration({
start: new Date(),
end: new Date("24 Jun 2020")
})
This will give output in a JSON object like below
{
"years": 0,
"months": 0,
"days": 0,
"hours": 19,
"minutes": 35,
"seconds": 24
}
Live Example https://stackblitz.com/edit/react-wvxvql
Link to Documentation https://date-fns.org/v2.14.0/docs/intervalToDuration

Some math is in order.
You can subtract one Date object from another in Javascript, and you'll get the difference between them in milisseconds. From this result you can extract the other parts you want (days, months etc.)
For example:
var a = new Date(2010, 10, 1);
var b = new Date(2010, 9, 1);
var c = a - b; // c equals 2674800000,
// the amount of milisseconds between September 1, 2010
// and August 1, 2010.
Now you can get any part you want. For example, how many days have elapsed between the two dates:
var days = (a - b) / (60 * 60 * 24 * 1000);
// 60 * 60 * 24 * 1000 is the amount of milisseconds in a day.
// the variable days now equals 30.958333333333332.
That's almost 31 days. You can then round down for 30 days, and use whatever remained to get the amounts of hours, minutes etc.

Get the difference between two dates in a human way
This function is capable of returning natural-language-like text. Use it to get responses like:
"4 years, 1 month and 11 days"
"1 year and 2 months"
"11 months and 20 days"
"12 days"
IMPORTANT: date-fns is a dependency
Just copy the code below and plug in a past date into our getElapsedTime function! It will compare the entered date against the present time and return your human-like responses.
import * as dateFns from "https://cdn.skypack.dev/date-fns#2.22.1";
function getElapsedTime(pastDate) {
const duration = dateFns.intervalToDuration({
start: new Date(pastDate),
end: new Date(),
});
let [years, months, days] = ["", "", ""];
if (duration.years > 0) {
years = duration.years === 1 ? "1 year" : `${duration.years} years`;
}
if (duration.months > 0) {
months = duration.months === 1 ? "1 month" : `${duration.months} months`;
}
if (duration.days > 0) {
days = duration.days === 1 ? "1 day" : `${duration.days} days`;
}
let response = [years, months, days].filter(Boolean);
switch (response.length) {
case 3:
response[1] += " and";
response[0] += ",";
break;
case 2:
response[0] += " and";
break;
}
return response.join(" ");
}

Yet another solution, based on some PHP code.
The strtotime function, also based on PHP, can be found here: http://phpjs.org/functions/strtotime/.
Date.dateDiff = function(d1, d2) {
d1 /= 1000;
d2 /= 1000;
if (d1 > d2) d2 = [d1, d1 = d2][0];
var diffs = {
year: 0,
month: 0,
day: 0,
hour: 0,
minute: 0,
second: 0
}
$.each(diffs, function(interval) {
while (d2 >= (d3 = Date.strtotime('+1 '+interval, d1))) {
d1 = d3;
++diffs[interval];
}
});
return diffs;
};
Usage:
> d1 = new Date(2000, 0, 1)
Sat Jan 01 2000 00:00:00 GMT+0100 (CET)
> d2 = new Date(2013, 9, 6)
Sun Oct 06 2013 00:00:00 GMT+0200 (CEST)
> Date.dateDiff(d1, d2)
Object {
day: 5
hour: 0
minute: 0
month: 9
second: 0
year: 13
}

Very old thread, I know, but here's my contribution, as the thread is not solved yet.
It takes leap years into consideration and does not asume any fixed number of days per month or year.
It might be flawed in border cases as I haven't tested it thoroughly, but it works for all the dates provided in the original question, thus I'm confident.
function calculate() {
var fromDate = document.getElementById('fromDate').value;
var toDate = document.getElementById('toDate').value;
try {
document.getElementById('result').innerHTML = '';
var result = getDateDifference(new Date(fromDate), new Date(toDate));
if (result && !isNaN(result.years)) {
document.getElementById('result').innerHTML =
result.years + ' year' + (result.years == 1 ? ' ' : 's ') +
result.months + ' month' + (result.months == 1 ? ' ' : 's ') + 'and ' +
result.days + ' day' + (result.days == 1 ? '' : 's');
}
} catch (e) {
console.error(e);
}
}
function getDateDifference(startDate, endDate) {
if (startDate > endDate) {
console.error('Start date must be before end date');
return null;
}
var startYear = startDate.getFullYear();
var startMonth = startDate.getMonth();
var startDay = startDate.getDate();
var endYear = endDate.getFullYear();
var endMonth = endDate.getMonth();
var endDay = endDate.getDate();
// We calculate February based on end year as it might be a leep year which might influence the number of days.
var february = (endYear % 4 == 0 && endYear % 100 != 0) || endYear % 400 == 0 ? 29 : 28;
var daysOfMonth = [31, february, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
var startDateNotPassedInEndYear = (endMonth < startMonth) || endMonth == startMonth && endDay < startDay;
var years = endYear - startYear - (startDateNotPassedInEndYear ? 1 : 0);
var months = (12 + endMonth - startMonth - (endDay < startDay ? 1 : 0)) % 12;
// (12 + ...) % 12 makes sure index is always between 0 and 11
var days = startDay <= endDay ? endDay - startDay : daysOfMonth[(12 + endMonth - 1) % 12] - startDay + endDay;
return {
years: years,
months: months,
days: days
};
}
<p><input type="text" name="fromDate" id="fromDate" placeholder="yyyy-mm-dd" value="1999-02-28" /></p>
<p><input type="text" name="toDate" id="toDate" placeholder="yyyy-mm-dd" value="2000-03-01" /></p>
<p><input type="button" name="calculate" value="Calculate" onclick="javascript:calculate();" /></p>
<p />
<p id="result"></p>

let startDate = moment(new Date('2017-05-12')); // yyyy-MM-dd
let endDate = moment(new Date('2018-09-14')); // yyyy-MM-dd
let Years = newDate.diff(date, 'years');
let months = newDate.diff(date, 'months');
let days = newDate.diff(date, 'days');
console.log("Year: " + Years, ", Month: " months-(Years*12), ", Days: " days-(Years*365.25)-((365.25*(days- (Years*12)))/12));
Above snippet will print: Year: 1, Month: 4, Days: 2

Using Plane Javascript:
function dateDiffInDays(start, end) {
var MS_PER_DAY = 1000 * 60 * 60 * 24;
var a = new Date(start);
var b = new Date(end);
const diffTime = Math.abs(a - b);
const diffDays = Math.ceil(diffTime / MS_PER_DAY);
console.log("Days: ", diffDays);
// Discard the time and time-zone information.
const utc1 = Date.UTC(a.getFullYear(), a.getMonth(), a.getDate());
const utc2 = Date.UTC(b.getFullYear(), b.getMonth(), b.getDate());
return Math.floor((utc2 - utc1) / MS_PER_DAY);
}
function dateDiffInDays_Months_Years(start, end) {
var m1 = new Date(start);
var m2 = new Date(end);
var yDiff = m2.getFullYear() - m1.getFullYear();
var mDiff = m2.getMonth() - m1.getMonth();
var dDiff = m2.getDate() - m1.getDate();
if (dDiff < 0) {
var daysInLastFullMonth = getDaysInLastFullMonth(start);
if (daysInLastFullMonth < m1.getDate()) {
dDiff = daysInLastFullMonth + dDiff + (m1.getDate() -
daysInLastFullMonth);
} else {
dDiff = daysInLastFullMonth + dDiff;
}
mDiff--;
}
if (mDiff < 0) {
mDiff = 12 + mDiff;
yDiff--;
}
console.log('Y:', yDiff, ', M:', mDiff, ', D:', dDiff);
}
function getDaysInLastFullMonth(day) {
var d = new Date(day);
console.log(d.getDay() );
var lastDayOfMonth = new Date(d.getFullYear(), d.getMonth() + 1, 0);
console.log('last day of month:', lastDayOfMonth.getDate() ); //
return lastDayOfMonth.getDate();
}
Using moment.js:
function dateDiffUsingMoment(start, end) {
var a = moment(start,'M/D/YYYY');
var b = moment(end,'M/D/YYYY');
var diffDaysMoment = b.diff(a, 'days');
console.log('Moments.js : ', diffDaysMoment);
preciseDiffMoments(a,b);
}
function preciseDiffMoments( a, b) {
var m1= a, m2=b;
m1.add(m2.utcOffset() - m1.utcOffset(), 'minutes'); // shift timezone of m1 to m2
var yDiff = m2.year() - m1.year();
var mDiff = m2.month() - m1.month();
var dDiff = m2.date() - m1.date();
if (dDiff < 0) {
var daysInLastFullMonth = moment(m2.year() + '-' + (m2.month() + 1),
"YYYY-MM").subtract(1, 'M').daysInMonth();
if (daysInLastFullMonth < m1.date()) { // 31/01 -> 2/03
dDiff = daysInLastFullMonth + dDiff + (m1.date() -
daysInLastFullMonth);
} else {
dDiff = daysInLastFullMonth + dDiff;
}
mDiff--;
}
if (mDiff < 0) {
mDiff = 12 + mDiff;
yDiff--;
}
console.log('getMomentum() Y:', yDiff, ', M:', mDiff, ', D:', dDiff);
}
Tested the above functions using following samples:
var sample1 = all('2/13/2018', '3/15/2018'); // {'M/D/YYYY'} 30, Y: 0 , M: 1 , D: 2
console.log(sample1);
var sample2 = all('10/09/2019', '7/7/2020'); // 272, Y: 0 , M: 8 , D: 29
console.log(sample2);
function all(start, end) {
dateDiffInDays(start, end);
dateDiffInDays_Months_Years(start, end);
try {
dateDiffUsingMoment(start, end);
} catch (e) {
console.log(e);
}
}

by using Moment library and some custom logic, we can get the exact date difference
var out;
out = diffDate(new Date('2014-05-10'), new Date('2015-10-10'));
display(out);
out = diffDate(new Date('2014-05-10'), new Date('2015-10-09'));
display(out);
out = diffDate(new Date('2014-05-10'), new Date('2015-09-09'));
display(out);
out = diffDate(new Date('2014-05-10'), new Date('2015-03-09'));
display(out);
out = diffDate(new Date('2014-05-10'), new Date('2016-03-09'));
display(out);
out = diffDate(new Date('2014-05-10'), new Date('2016-03-11'));
display(out);
function diffDate(startDate, endDate) {
var b = moment(startDate),
a = moment(endDate),
intervals = ['years', 'months', 'weeks', 'days'],
out = {};
for (var i = 0; i < intervals.length; i++) {
var diff = a.diff(b, intervals[i]);
b.add(diff, intervals[i]);
out[intervals[i]] = diff;
}
return out;
}
function display(obj) {
var str = '';
for (key in obj) {
str = str + obj[key] + ' ' + key + ' '
}
console.log(str);
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.24.0/moment.js"></script>

I did it using a bunch of functions. Pure JavaScript and precise.
This code includes functions that calculate time difference in days, months and years. One of them can be used to get precise time difference for example X years, Y months, Z days. At the end of code I provided some tests.
How it works:
getDaysDiff():
Transforms time difference from milliseconds to days.
getYearsDiff():
There is no worries for effect of months and days of both dates. The function calculates difference in years by moving dates back and forward.
getMonthsDiff() (This one has nothing to do with question, but the concept is used in calExactTimeDiff() and I thought someone may need such a function so I insert it):
This one is a little tricky. The hard work is to deal with month and day of both dates.
If the endDate's month is more than startDate's, this means another year (12 months) is passed. But this is being taken care of in monthsOfFullYears, so the only thing is needed is to add subtraction of month of endDate and startDate.
If the startDate's month is more than endDate's then there is no another year. So we should get the difference between them. Imagine we want to go from month 10 of the current year to 2 of the next year. We can go like this: 11, 12, 1, 2. So we passed 4 months. This is equal to 12 - (10 - 2). We get difference between the months and subtract it from months of a whole year.
Next step is to take care of days of months. If day of endDate is more than or equal to startDate this means another month is passed. So we add 1 to it. But if it's less, then there is nothing to worry about. But in my code I did not do this. Because when I added difference between months I assumed that the days of months are equal. So I already added 1. Thus if day of endDate is less than startDate, I have to decrease months by 1.
There is an exception: if months are equal and endDate's day is less than startDate's, month should be 11.
I used the same concept in calExactTimeDiff().
Hope to be useful :)
// time difference in Days
function getDaysDiff(startDate = new Date(), endDate = new Date()) {
if (startDate > endDate) [startDate, endDate] = [endDate, startDate];
let timeDiff = endDate - startDate;
let timeDiffInDays = Math.floor(timeDiff / (1000 * 3600 * 24));
return timeDiffInDays;
}
// time difference in Months
function getMonthsDiff(startDate = new Date(), endDate = new Date()) {
let monthsOfFullYears = getYearsDiff(startDate, endDate) * 12;
let months = monthsOfFullYears;
// the variable below is not necessary, but I kept it for understanding of code
// we can use "startDate" instead of it
let yearsAfterStart = new Date(
startDate.getFullYear() + getYearsDiff(startDate, endDate),
startDate.getMonth(),
startDate.getDate()
);
let isDayAhead = endDate.getDate() >= yearsAfterStart.getDate();
if (startDate.getMonth() == endDate.getMonth() && !isDayAhead) {
months = 11;
return months;
}
if (endDate.getMonth() >= yearsAfterStart.getMonth()) {
let diff = endDate.getMonth() - yearsAfterStart.getMonth();
months += (isDayAhead) ? diff : diff - 1;
}
else {
months += isDayAhead
? 12 - (startDate.getMonth() - endDate.getMonth())
: 12 - (startDate.getMonth() - endDate.getMonth()) - 1;
}
return months;
}
// time difference in Years
function getYearsDiff(startDate = new Date(), endDate = new Date()) {
if (startDate > endDate) [startDate, endDate] = [endDate, startDate];
let yearB4End = new Date(
endDate.getFullYear() - 1,
endDate.getMonth(),
endDate.getDate()
);
let year = 0;
year = yearB4End > startDate
? yearB4End.getFullYear() - startDate.getFullYear()
: 0;
let yearsAfterStart = new Date(
startDate.getFullYear() + year + 1,
startDate.getMonth(),
startDate.getDate()
);
if (endDate >= yearsAfterStart) year++;
return year;
}
// time difference in format: X years, Y months, Z days
function calExactTimeDiff(firstDate, secondDate) {
if (firstDate > secondDate)
[firstDate, secondDate] = [secondDate, firstDate];
let monthDiff = 0;
let isDayAhead = secondDate.getDate() >= firstDate.getDate();
if (secondDate.getMonth() >= firstDate.getMonth()) {
let diff = secondDate.getMonth() - firstDate.getMonth();
monthDiff += (isDayAhead) ? diff : diff - 1;
}
else {
monthDiff += isDayAhead
? 12 - (firstDate.getMonth() - secondDate.getMonth())
: 12 - (firstDate.getMonth() - secondDate.getMonth()) - 1;
}
let dayDiff = 0;
if (isDayAhead) {
dayDiff = secondDate.getDate() - firstDate.getDate();
}
else {
let b4EndDate = new Date(
secondDate.getFullYear(),
secondDate.getMonth() - 1,
firstDate.getDate()
)
dayDiff = getDaysDiff(b4EndDate, secondDate);
}
if (firstDate.getMonth() == secondDate.getMonth() && !isDayAhead)
monthDiff = 11;
let exactTimeDiffUnits = {
yrs: getYearsDiff(firstDate, secondDate),
mths: monthDiff,
dys: dayDiff,
};
return `${exactTimeDiffUnits.yrs} years, ${exactTimeDiffUnits.mths} months, ${exactTimeDiffUnits.dys} days`
}
let s = new Date(2012, 4, 12);
let e = new Date(2008, 5, 24);
console.log(calExactTimeDiff(s, e));
s = new Date(2001, 7, 4);
e = new Date(2016, 6, 9);
console.log(calExactTimeDiff(s, e));
s = new Date(2011, 11, 28);
e = new Date(2021, 3, 6);
console.log(calExactTimeDiff(s, e));
s = new Date(2020, 8, 7);
e = new Date(2021, 8, 6);
console.log(calExactTimeDiff(s, e));

There a a couple of npm packages that help in doing this. Below is a list gathered from various sources. I find the date-fns version to be the most simplest.
1. date-fns
You can use intervalToDuration, formatDuration from date-fns to humanize a duration in desired format like below:
import { intervalToDuration, formatDuration } from 'date-fns'
let totalDuration = intervalToDuration({
start: new Date(1929, 0, 15, 12, 0, 0),
end: new Date(1968, 3, 4, 19, 5, 0)
});
let textDuration = formatDuration(totalDuration, { format: ['years', 'months'], delimiter: ', ' })
// Output: "39 years, 2 months"
clone the above code from here for trying it yourself: https://runkit.com/embed/diu9o3qe53j4
2. luxon + humanize-duration
you can use luxon to extract the duration between dates and humanize that using humanize-duration like below:
const DateTime = luxon.DateTime;
const Interval = luxon.Interval;
const start = DateTime.fromSQL("2020-06-19 11:14:00");
const finish = DateTime.fromSQL("2020-06-21 13:11:00");
const formatted = Interval
.fromDateTimes(start, finish)
.toDuration()
.valueOf();
console.log(humanizeDuration(formatted))
// output: 2 days, 1 hour, 57 minutes
console.log(humanizeDuration(formatted, { language: 'es' }))
// output: 2 días, 1 hora, 57 minutos
console.log(humanizeDuration(formatted, { language: 'ru' }))
// output: 2 дня, 1 час, 57 минут
<script src="https://cdn.jsdelivr.net/npm/luxon#1.25.0/build/global/luxon.min.js"></script>
<script src="https://cdn.jsdelivr.net/npm/humanize-duration#3.25.1/humanize-duration.min.js"></script>
reference to above code: https://stackoverflow.com/a/65651515/6908282

I would personally use http://www.datejs.com/, really handy. Specifically, look at the time.js file: http://code.google.com/p/datejs/source/browse/trunk/src/time.js

Time span in full Days, Hours, Minutes, Seconds, Milliseconds:
// Extension for Date
Date.difference = function (dateFrom, dateTo) {
var diff = { TotalMs: dateTo - dateFrom };
diff.Days = Math.floor(diff.TotalMs / 86400000);
var remHrs = diff.TotalMs % 86400000;
var remMin = remHrs % 3600000;
var remS = remMin % 60000;
diff.Hours = Math.floor(remHrs / 3600000);
diff.Minutes = Math.floor(remMin / 60000);
diff.Seconds = Math.floor(remS / 1000);
diff.Milliseconds = Math.floor(remS % 1000);
return diff;
};
// Usage
var a = new Date(2014, 05, 12, 00, 5, 45, 30); //a: Thu Jun 12 2014 00:05:45 GMT+0400
var b = new Date(2014, 02, 12, 00, 0, 25, 0); //b: Wed Mar 12 2014 00:00:25 GMT+0400
var diff = Date.difference(b, a);
/* diff: {
Days: 92
Hours: 0
Minutes: 5
Seconds: 20
Milliseconds: 30
TotalMs: 7949120030
} */

Neither of the codes work for me, so I use this instead for months and days:
function monthDiff(d2, d1) {
var months;
months = (d2.getFullYear() - d1.getFullYear()) * 12;
months -= d1.getMonth() + 1;
months += d2.getMonth() + 1;
return months <= 0 ? 0 : months;
}
function daysInMonth(date) {
return new Date(date.getYear(), date.getMonth() + 1, 0).getDate();
}
function diffDate(date1, date2) {
if (date2 && date2.getTime() && !isNaN(date2.getTime())) {
var months = monthDiff(date1, date2);
var days = 0;
if (date1.getUTCDate() >= date2.getUTCDate()) {
days = date1.getUTCDate() - date2.getUTCDate();
}
else {
months--;
days = date1.getUTCDate() - date2.getUTCDate() + daysInMonth(date2);
}
// Use the variables months and days how you need them.
}
}

The following is an algorithm which gives correct but not totally precise since it does not take into account leap year. It also assumes 30 days in a month. A good usage for example is if someone lives in an address from 12/11/2010 to 11/10/2011, it can quickly tells that the person lives there for 10 months and 29 days. From 12/11/2010 to 11/12/2011 is 11 months and 1 day. For certain types of applications, that kind of precision is sufficient. This is for those types of applications because it aims for simplicity:
var datediff = function(start, end) {
var diff = { years: 0, months: 0, days: 0 };
var timeDiff = end - start;
if (timeDiff > 0) {
diff.years = end.getFullYear() - start.getFullYear();
diff.months = end.getMonth() - start.getMonth();
diff.days = end.getDate() - start.getDate();
if (diff.months < 0) {
diff.years--;
diff.months += 12;
}
if (diff.days < 0) {
diff.months = Math.max(0, diff.months - 1);
diff.days += 30;
}
}
return diff;
};
Unit tests

To calculate the difference between two dates in Years, Months, Days, Minutes, Seconds, Milliseconds using TypeScript/ JavaScript
dateDifference(actualDate) {
// Calculate time between two dates:
const date1 = actualDate; // the date you already commented/ posted
const date2: any = new Date(); // today
let r = {}; // object for clarity
let message: string;
const diffInSeconds = Math.abs(date2 - date1) / 1000;
const days = Math.floor(diffInSeconds / 60 / 60 / 24);
const hours = Math.floor(diffInSeconds / 60 / 60 % 24);
const minutes = Math.floor(diffInSeconds / 60 % 60);
const seconds = Math.floor(diffInSeconds % 60);
const milliseconds =
Math.round((diffInSeconds - Math.floor(diffInSeconds)) * 1000);
const months = Math.floor(days / 31);
const years = Math.floor(months / 12);
// the below object is just optional
// if you want to return an object instead of a message
r = {
years: years,
months: months,
days: days,
hours: hours,
minutes: minutes,
seconds: seconds,
milliseconds: milliseconds
};
// check if difference is in years or months
if (years === 0 && months === 0) {
// show in days if no years / months
if (days > 0) {
if (days === 1) {
message = days + ' day';
} else { message = days + ' days'; }
} else if (hours > 0) {
if (hours === 1) {
message = hours + ' hour';
} else {
message = hours + ' hours';
}
} else {
// show in minutes if no years / months / days
if (minutes === 1) {
message = minutes + ' minute';
} else {message = minutes + ' minutes';}
}
} else if (years === 0 && months > 0) {
// show in months if no years
if (months === 1) {
message = months + ' month';
} else {message = months + ' months';}
} else if (years > 0) {
// show in years if years exist
if (years === 1) {
message = years + ' year';
} else {message = years + ' years';}
}
return 'Posted ' + message + ' ago';
// this is the message a user see in the view
}
However, you can update the above logic for the message to show seconds and milliseconds too or else use the object 'r' to format the message whatever way you want.
If you want to directly copy the code, you can view my gist with the above code here

I know it is an old thread, but I'd like to put my 2 cents based on the answer by #Pawel Miech.
It is true that you need to convert the difference into milliseconds, then you need to make some math. But notice that, you need to do the math in backward manner, i.e. you need to calculate years, months, days, hours then minutes.
I used to do some thing like this:
var mins;
var hours;
var days;
var months;
var years;
var diff = new Date() - new Date(yourOldDate);
// yourOldDate may be is coming from DB, for example, but it should be in the correct format ("MM/dd/yyyy hh:mm:ss:fff tt")
years = Math.floor((diff) / (1000 * 60 * 60 * 24 * 365));
diff = Math.floor((diff) % (1000 * 60 * 60 * 24 * 365));
months = Math.floor((diff) / (1000 * 60 * 60 * 24 * 30));
diff = Math.floor((diff) % (1000 * 60 * 60 * 24 * 30));
days = Math.floor((diff) / (1000 * 60 * 60 * 24));
diff = Math.floor((diff) % (1000 * 60 * 60 * 24));
hours = Math.floor((diff) / (1000 * 60 * 60));
diff = Math.floor((diff) % (1000 * 60 * 60));
mins = Math.floor((diff) / (1000 * 60));
But, of course, this is not precise because it assumes that all years have 365 days and all months have 30 days, which is not true in all cases.

Its very simple please use the code below and it will give the difference in that format according to this //3 years 9 months 3 weeks 5 days 15 hours 50 minutes
Date.getFormattedDateDiff = function(date1, date2) {
var b = moment(date1),
a = moment(date2),
intervals = ['years','months','weeks','days'],
out = [];
for(var i=0; i<intervals.length; i++){
var diff = a.diff(b, intervals[i]);
b.add(diff, intervals[i]);
out.push(diff + ' ' + intervals[i]);
}
return out.join(', ');
};
var today = new Date(),
newYear = new Date(today.getFullYear(), 0, 1),
y2k = new Date(2000, 0, 1);
//(AS OF NOV 29, 2016)
//Time since New Year: 0 years, 10 months, 4 weeks, 0 days
console.log( 'Time since New Year: ' + Date.getFormattedDateDiff(newYear, today) );
//Time since Y2K: 16 years, 10 months, 4 weeks, 0 days
console.log( 'Time since Y2K: ' + Date.getFormattedDateDiff(y2k, today) );

This code should give you desired results
//************************** Enter your dates here **********************//
var startDate = "10/05/2014";
var endDate = "11/3/2016"
//******* and press "Run", you will see the result in a popup *********//
var noofdays = 0;
var sdArr = startDate.split("/");
var startDateDay = parseInt(sdArr[0]);
var startDateMonth = parseInt(sdArr[1]);
var startDateYear = parseInt(sdArr[2]);
sdArr = endDate.split("/")
var endDateDay = parseInt(sdArr[0]);
var endDateMonth = parseInt(sdArr[1]);
var endDateYear = parseInt(sdArr[2]);
console.log(startDateDay+' '+startDateMonth+' '+startDateYear);
var yeardays = 365;
var monthArr = [31,,31,30,31,30,31,31,30,31,30,31];
var noofyears = 0
var noofmonths = 0;
if((startDateYear%4)==0) monthArr[1]=29;
else monthArr[1]=28;
if(startDateYear == endDateYear){
noofyears = 0;
noofmonths = getMonthDiff(startDate,endDate);
if(noofmonths < 0) noofmonths = 0;
noofdays = getDayDiff(startDate,endDate);
}else{
if(endDateMonth < startDateMonth){
noofyears = (endDateYear - startDateYear)-1;
if(noofyears < 1) noofyears = 0;
}else{
noofyears = endDateYear - startDateYear;
}
noofmonths = getMonthDiff(startDate,endDate);
if(noofmonths < 0) noofmonths = 0;
noofdays = getDayDiff(startDate,endDate);
}
alert(noofyears+' year, '+ noofmonths+' months, '+ noofdays+' days');
function getDayDiff(startDate,endDate){
if(endDateDay >=startDateDay){
noofdays = 0;
if(endDateDay > startDateDay) {
noofdays = endDateDay - startDateDay;
}
}else{
if((endDateYear%4)==0) {
monthArr[1]=29;
}else{
monthArr[1] = 28;
}
if(endDateMonth != 1)
noofdays = (monthArr[endDateMonth-2]-startDateDay) + endDateDay;
else
noofdays = (monthArr[11]-startDateDay) + endDateDay;
}
return noofdays;
}
function getMonthDiff(startDate,endDate){
if(endDateMonth > startDateMonth){
noofmonths = endDateMonth - startDateMonth;
if(endDateDay < startDateDay){
noofmonths--;
}
}else{
noofmonths = (12-startDateMonth) + endDateMonth;
if(endDateDay < startDateDay){
noofmonths--;
}
}
return noofmonths;
}
https://jsfiddle.net/moremanishk/hk8c419f/

You should try using date-fns. Here's how I did it using intervalToDuration and formatDuration functions from date-fns.
let startDate = Date.parse("2010-10-01 00:00:00 UTC");
let endDate = Date.parse("2020-11-01 00:00:00 UTC");
let duration = intervalToDuration({start: startDate, end: endDate});
let durationInWords = formatDuration(duration, {format: ["years", "months", "days"]}); //output: 10 years 1 month

since I had to use moment-hijri (hijri calendar) and couldn't use moment.diff() method, I came up with this solution. can also be used with moment.js
var momenti = require('moment-hijri')
//calculate hijri
var strt = await momenti(somedateobject)
var until = await momenti()
var years = await 0
var months = await 0
var days = await 0
while(strt.valueOf() < until.valueOf()){
await strt.add(1, 'iYear');
await years++
}
await strt.subtract(1, 'iYear');
await years--
while(strt.valueOf() < until.valueOf()){
await strt.add(1, 'iMonth');
await months++
}
await strt.subtract(1, 'iMonth');
await months--
while(strt.valueOf() < until.valueOf()){
await strt.add(1, 'day');
await days++
}
await strt.subtract(1, 'day');
await days--
await console.log(years)
await console.log(months)
await console.log(days)

A solution with the ECMAScript "Temporal API" which is currently (as of 5th March 2022) in Stage 3 of Active Proposals, which will the method we will do this in the future (soon).
Here is a solution with the current temporal-polyfill
<script type='module'>
import * as TemporalModule from 'https://cdn.jsdelivr.net/npm/#js-temporal/polyfill#0.3.0/dist/index.umd.js'
const Temporal = temporal.Temporal;
//----------------------------------------
function dateDiff(start, end, maxUnit) {
return (Temporal.PlainDate.from(start).until(Temporal.PlainDate.from(end),{largestUnit:maxUnit}).toString()).match(/(\d*Y)|(\d*M)|(\d*D)/g).join(" ");
}
//----------------------------------------
console.log("Diff in (years, months, days): ",dateDiff("1963-02-03","2022-03-06","year"))
console.log("Diff in (months, days) : ",dateDiff("1963-02-03","2022-03-06","month"))
console.log("Diff in (days) : ",dateDiff("1963-02-03","2022-03-06","day"))
</script>

Your expected output is not correct. For example difference between '2014-05-10' and '2015-03-09' is not 9 months, 27 days
the correct answer is
(05-10 to 05-31) = 21 days
(2014-06 to 2015-03) = 9 months
(03-01 to 03-09) = 9 days
total is 9 months and 30 days
WARNING: An ideal function would be aware of leap years and days count in every month, but I found the results of this function accurate enough for my current task, so I shared it with you
function diffDate(date1, date2)
{
var daysDiff = Math.ceil((Math.abs(date1 - date2)) / (1000 * 60 * 60 * 24));
var years = Math.floor(daysDiff / 365.25);
var remainingDays = Math.floor(daysDiff - (years * 365.25));
var months = Math.floor((remainingDays / 365.25) * 12);
var days = Math.ceil(daysDiff - (years * 365.25 + (months / 12 * 365.25)));
return {
daysAll: daysDiff,
years: years,
months: months,
days:days
}
}
console.log(diffDate(new Date('2014-05-10'), new Date('2015-10-10')));
console.log(diffDate(new Date('2014-05-10'), new Date('2015-10-09')));
console.log(diffDate(new Date('2014-05-10'), new Date('2015-09-09')));
console.log(diffDate(new Date('2014-05-10'), new Date('2015-03-09')));
console.log(diffDate(new Date('2014-05-10'), new Date('2016-03-09')));
console.log(diffDate(new Date('2014-05-10'), new Date('2016-03-11')));

Calculating hour difference in Javascript, but only certain hours

Unsure of the best way to describe this, but I need to calculate the difference in hours (rounded down) but only between 8pm and 6am (or rather 20:00 - 06:00 in this case!)
For example:
22:00 - 04:00 (6 hours)
02:40 - 10:20 (4 hours)
20:00 - 06:00 (10 hours)
Unfortunately I need to work on exact dates, because some will span over multiple days - and just to add to the confusion, I also need to exclude certain dates entirely for bank holidays (which I have a list of in an array) but have absolutely no idea how to implement this - any suggestions would be very welcome, thank you

Just going off what the inputs in your sample looks like:
// According to your example, your inputs are strings...
// t1 = "22:00"
// t2 = "04:00";
function hoursDiff(t1, t2){
// Parse out the times, using radix 10 to
// avoid octal edge cases ("08:00" & "09:00")
var time1 = parseInt( t1, 10 );
var time2 = parseInt( t2, 10 );
var hours = 0;
while ( time1 !== time2 ){
time1++;
hours++;
// If we've passed midnight, reset
// to 01:00AM
if ( time1 === 25 ){
time1 = 1;
}
}
return hours;
}

Ok, this was a nice puzzle. It really cost me too much time but it was fun.
Full working code below (jsfiddle here):
function isHoliday(/*Date*/ date) {
for(var i = 0; i < holidays.length; i++) {
if(holidays[i].getTime() == date.getTime()) {
return true;
}
}
return false;
}
function diffHours(/*Date*/ d1, /*Date*/ d2) {
var date1 = new Date(d1.getUTCFullYear()+"-"+(d1.getUTCMonth()+1)+"-"+d1.getUTCDate()+" UTC");
var date2 = new Date(d2.getUTCFullYear()+"-"+(d2.getUTCMonth()+1)+"-"+d2.getUTCDate()+" UTC");
var sum = 0;
var oneday = 24*3600*1000;
var hours, date;
// first day
if(!isHoliday(date1)) {
// decrease by a whole day first (will be added later)
sum -= 10;
// add real hours
hours = d1.getUTCHours() + d1.getUTCMinutes() / 60;
if(hours <= 6) {
sum += 10 - hours;
} else if(hours <= 20) {
sum += 4;
} else {
sum += 24 - hours;
}
}
// last day
if(!isHoliday(date2)) {
// decrease by a whole day first (will be added later)
sum -= 10;
// add real hours
hours = d2.getUTCHours() + d2.getUTCMinutes() / 60;
if(hours <= 6) {
sum += hours;
} else if(hours <= 20) {
sum += 6;
} else {
sum += hours - 14;
}
}
// whole days
while(date1 <= date2) {
if(!isHoliday(date1)) {
sum += 10;
}
// increase date by 1 day
date1.setTime(date1.getTime() + oneday);
}
return Math.floor(sum);
}
// ==============
// examples below
// --------------
// array of Dates (in UTC) to skip
var holidays = [
new Date("2012-01-04 UTC"),
];
for(var i = 0; i < holidays.length; i++) {
console.log('holiday: ', holidays[i].toUTCString());
}
a = new Date("2012-01-01 12:00 UTC");
b = new Date("2012-01-02 12:00 UTC");
c = new Date("2012-01-02 22:00 UTC");
d = new Date("2012-01-03 07:00 UTC");
e = new Date("2012-01-05 12:00 UTC");
console.log({d1: a.toUTCString(), d2: b.toUTCString(), hours: diffHours(a, b)});
console.log({d1: b.toUTCString(), d2: c.toUTCString(), hours: diffHours(b, c)});
console.log({d1: c.toUTCString(), d2: d.toUTCString(), hours: diffHours(c, d)});
console.log({d1: d.toUTCString(), d2: e.toUTCString(), hours: diffHours(d, e)});

Add no. of days in a date to get next date(excluding weekends)

I have an date, i need to add no. of days to get future date but weekends should be excluded.
i.e
input date = "9-DEC-2011";
No. of days to add = '13';
next date should be "28-Dec-2011"
Here weekends(sat/sun) are not counted.

Try this
var startDate = "9-DEC-2011";
startDate = new Date(startDate.replace(/-/g, "/"));
var endDate = "", noOfDaysToAdd = 13, count = 0;
while(count < noOfDaysToAdd){
endDate = new Date(startDate.setDate(startDate.getDate() + 1));
if(endDate.getDay() != 0 && endDate.getDay() != 6){
//Date.getDay() gives weekday starting from 0(Sunday) to 6(Saturday)
count++;
}
}
alert(endDate);//You can format this date as per your requirement
Working Demo

#ShankarSangoli
Here's a newer version which avoid recreating a Date object on each loop, note that it's wrapped in a function now.
function calcWorkingDays(fromDate, days) {
var count = 0;
while (count < days) {
fromDate.setDate(fromDate.getDate() + 1);
if (fromDate.getDay() != 0 && fromDate.getDay() != 6) // Skip weekends
count++;
}
return fromDate;
}
alert(calcWorkingDays(new Date("9/DEC/2011"), 13));

Here is an elegant solution without any looping or external library:
function addBusinessDaysToDate(date, days) {
var day = date.getDay();
date = new Date(date.getTime());
date.setDate(date.getDate() + days + (day === 6 ? 2 : +!day) + (Math.floor((days - 1 + (day % 6 || 1)) / 5) * 2));
return date;
}
var date = "9-DEC-2011";
var newDate = addBusinessDaysToDate(new Date(date.replace(/-/g, "/")), 13);
console.log(newDate.toString().replace(/\S+\s(\S+)\s(\d+)\s(\d+)\s.*/, '$2-$1-$3')); // alerts "28-Dec-2011"

or you can be like this
function addWeekdays(date, weekdays) {
var newDate = new Date(date.getTime());
var i = 0;
while (i < weekdays) {
newDate.setDate(newDate.getDate() + 1);
var day = newDate.getDay();
if (day > 1 && day < 7) {
i++;
}
}
return newDate;
}
var currentDate = new Date('10/31/2014');
var targetDate = addWeekdays(currentDate, 45);
alert(targetDate);

Using moment.js:
const DATE_FORMAT = 'D-MMM-YYYY';
const SUNDAY = 0; // moment day index
const SATURDAY = 6; // moment day index
const WEEKENDS = [SATURDAY, SUNDAY];
function addBusinessDays(stringDate, numberOfDays, dateFormat = DATE_FORMAT) {
const date = moment(stringDate, dateFormat);
let count = 0;
while (count < numberOfDays) {
date.add(1, 'day');
// Skip weekends
if (WEEKENDS.includes(date.day())) {
continue;
}
// Increment count
count++;
}
return date.format(dateFormat);
}
// Test cases
console.log(addBusinessDays('3-Mar-2021', 1)); // 4-Mar-2021
console.log(addBusinessDays('3-Mar-2021', 2)); // 5-Mar-2021
console.log(addBusinessDays('3-Mar-2021', 3)); // 8-Mar-2021
console.log(addBusinessDays('3-Mar-2021', 4)); // 9-Mar-2021
console.log(addBusinessDays('3-Mar-2021', 5)); // 10-Mar-2021
console.log(addBusinessDays('9-Dec-2011', 13)); // 28-Dec-2011
console.log(addBusinessDays('10-Dec-2011', 13)); // 28-Dec-2011 (Saturday, so remain on Friday)
console.log(addBusinessDays('11-Dec-2011', 13)); // 28-Dec-2011 (Sunday, so remain on Friday)
console.log(addBusinessDays('12-Dec-2011', 13)); // 29-Dec-2011
console.log(addBusinessDays('13-Dec-2011', 13)); // 30-Dec-2011
console.log(addBusinessDays('14-Dec-2011', 13)); // 2-Jan-2012
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.24.0/moment.min.js"></script>

This question is quite old, but all the previous answers are iterating over the days one by one. That could be inefficient for a large number of days. This works for me, assuming days is a positive int and the startDate is a working day:
function addWorkingDates(startDate, days) {
const current_day = startDate.getDay() - 1; // Week day, starting on Monday
const weekend_days = 2 * parseInt((current_day + days) / 5);
startDate.setDate(changed_to.getDate() + days + weekend_days);
}
addWorkingDates(new Date(),5)

For some reason it was more intuitive to me to try this recursively. This version doesn't account for holidays, but you could change the isValid function to check whatever.
function addWeekdaysToDate(date, numberToAdd) {
var isValid = function(d) { return d.getDay() !== 0 && d.getDay() !== 6 }
if(Math.abs(numberToAdd) > 1) {
return addWeekdaysToDate(
addWeekdaysToDate(date, Math.sign(numberToAdd)),
numberToAdd - Math.sign(numberToAdd)
)
} else if(Math.abs(numberToAdd) === 1) {
var result = new Date(date)
result.setDate(result.getDate() + Math.sign(numberToAdd))
if(isValid(result)) {
return result
} else {
return addWeekdaysToDate(result, Math.sign(numberToAdd))
}
} else if(numberToAdd === 0) {
return date
}
return false
}
console.log(addWeekdaysToDate(new Date(), 1))
console.log(addWeekdaysToDate(new Date(), 5))
console.log(addWeekdaysToDate(new Date(), -7))
In certain browsers you may need a polyfill for Math.sign:
Math.sign = Math.sign || function(x) {
x = +x; // convert to a number
if (x === 0 || isNaN(x)) {
return Number(x);
}
return x > 0 ? 1 : -1;
}

try this solution
<script language="javascript">
function getDateExcludeWeekends(startDay, startMonth, startYear, daysToAdd) {
var sdate = new Date();
var edate = new Date();
var dayMilliseconds = 1000 * 60 * 60 * 24;
sdate.setFullYear(startYear,startMonth,startDay);
edate.setFullYear(startYear,startMonth,startDay+daysToAdd);
var weekendDays = 0;
while (sdate <= edate) {
var day = sdate.getDay()
if (day == 0 || day == 6) {
weekendDays++;
}
sdate = new Date(+sdate + dayMilliseconds);
}
sdate.setFullYear(startYear,startMonth,startDay + weekendDays+daysToAdd);
return sdate;
}
</script>

If you want to get the next working day, from a specific date, use the following code...
function getNextWorkingDay(originalDate) {
var nextWorkingDayFound = false;
var nextWorkingDate = new Date();
var dateCounter = 1;
while (!nextWorkingDayFound) {
nextWorkingDate.setDate(originalDate.getDate() + dateCounter);
dateCounter++;
if (!isDateOnWeekend(nextWorkingDate)) {
nextWorkingDayFound = true;
}
}
return nextWorkingDate;
}
function isDateOnWeekend(date) {
if (date.getDay() === 6 || date.getDay() === 0)
return true;
else
return false;
}

Try this
function calculate() {
var noOfDaysToAdd = 13;
var startDate = "9-DEC-2011";
startDate = new Date(startDate.replace(/-/g, "/"));
var endDate = "";
count = 0;
while (count < noOfDaysToAdd) {
endDate = new Date(startDate.setDate(startDate.getDate() + 1));
if (endDate.getDay() != 0 && endDate.getDay() != 6) {
count++;
}
}
document.getElementById("result").innerHTML = endDate;
}
<div>
Date of book delivery: <span id="result"></span><br /><br />
<input type="button" onclick="calculate();" value="Calculate" />
<br>
<br>
</div>

add/subtract business days in Javascript

I need a Date.prototype.addBusDays function
that'll take an integer as the number of working days to add to the date.
However, there are two considerations: 1. Weekends, 2. Holidays (which I imagine would be a preset array to compare against. If beginning date and end date contain 3 holidays, then you push out the end date by 3)
I have come across some scripts online, one dilemma I can think of is, lets say you address all the weekends first, then you do the holidays, what if you +1 day (due to holiday), and your end date is pushed into a weekends again...<
Any ideas?
Thanks!
EDIT:
This is a part of a scheduling tool I am developing, which mean the dates will be tied to tasks which are linked together. Adding 1 day to a task, will trigger a recalculation of everything tied to it, potentially all dates in the database.

Datageek's solution helped me but I needed to augment it. This still doesn't do holidays but does do working days with the option of including Sat and/or Sun, and does support adding negative days:-
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;
}

Have a look at the following implementation. Sourced from 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);

(Updated) I've put this algorithm through its paces and it seems stable, though it does use recursion for holiday processing:
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;
}

I expanded on khellendros74's answer for a project of mine that needed to disable Sundays and mailing holidays in the datepicker and return two dates on press of a button: three business days (i.e. non-holiday and non-Sunday) after the date picked in the datepicker (a field with an id of "calendar") and six business days after the date picked in the datepicker and then put those two results into a couple of disabled input fields (handDelivered and mailed). The button press calls the function calculateDates. Here is that code:
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;
}

Simple solution to solve the whole problem; you can just loop through the days to skip weekdays and holidays:
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

This only takes weekends into account and not holidays, but it's a start...
function mod(x, y) {
// https://stackoverflow.com/a/4467559/2173455
return ((x % y) + y) % y;
}
function calculateDateDiff(date, diff) {
let returnDate = new Date(date.getTime());
let daysLeftToAdd = Math.abs(diff);
let weekendDays = 0;
let weekDay = returnDate.getDay();
while(daysLeftToAdd >= 0) {
if(weekDay == 0 || weekDay == 6) {
weekendDays++;
}
else {
daysLeftToAdd--;
}
weekDay = mod(diff > 0 ? weekDay + 1 : weekDay - 1, 7);
}
returnDate.setDate(diff > 0 ?
returnDate.getDate() + diff + weekendDays :
returnDate.getDate() + diff - weekendDays
);
return returnDate;
}