Difference between two dates in years, months, days in JavaScript

JavascriptDateDatediff

Javascript Problem Overview


I've been searching for 4 hours now, and have not found a solution 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:

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'));

Javascript Solutions


Solution 1 - Javascript

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.

Solution 2 - Javascript

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>

Solution 3 - Javascript

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) {
  var 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
  }
  var endDate = new Date(endingDate);
  if (startDate > endDate) {
    var swap = startDate;
    startDate = endDate;
    endDate = swap;
  }
  var startYear = startDate.getFullYear();
  var february = (startYear % 4 === 0 && startYear % 100 !== 0) || startYear % 400 === 0 ? 29 : 28;
  var daysInMonth = [31, february, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];

  var yearDiff = endDate.getFullYear() - startYear;
  var monthDiff = endDate.getMonth() - startDate.getMonth();
  if (monthDiff < 0) {
    yearDiff--;
    monthDiff += 12;
  }
  var 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,'2019-05-10'] // 0Y 0M 1D
].forEach(([s, e]) => console.log(dateDiff(s, e)));


Older less accurate but simpler version

@RajeevPNadig's answer was what I was looking for, but his code returns incorrect values as written. This 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.

Solution 4 - Javascript

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");

Solution 5 - Javascript

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)"

Solution 6 - Javascript

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

Solution 7 - Javascript

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;
}

Solution 8 - Javascript

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.

Solution 9 - Javascript

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.

Solution 10 - Javascript

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

Solution 11 - Javascript

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); 
	}
}

Solution 12 - Javascript

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
}

Solution 13 - Javascript

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>

Solution 14 - Javascript

   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

Solution 15 - Javascript

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/[email protected]";

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(" ");
}

Solution 16 - Javascript

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));

Solution 17 - Javascript

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
} */

Solution 18 - Javascript

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.
	}
}

Solution 19 - Javascript

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

Solution 20 - Javascript

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

Solution 21 - Javascript

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.

Solution 22 - Javascript

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) );

Solution 23 - Javascript

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>

Solution 24 - Javascript

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/

Solution 25 - Javascript

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





Solution 26 - Javascript

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)

Solution 27 - Javascript

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/[email protected]/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(/(\dY)|(\dM)|(\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>

Solution 28 - Javascript

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

Solution 29 - Javascript

I do it this way. Precise? Maybe or maybe not. Try it

<html>
  <head>
    <title> Age Calculator</title>
  </head>

  <input type="date" id="startDate" value="2000-01-01">
  <input type="date" id="endDate"  value="2020-01-01">
  <button onclick="getAge(new Date(document.getElementById('startDate').value), new Date(document.getElementById('endDate').value))">Check Age</button>
  <script>
    function getAge (startDate, endDate) {
      var diff = endDate-startDate
      var age = new Date(new Date("0000-01-01").getTime()+diff)
      var years = age.getFullYear()
      var months = age.getMonth()
      var days = age.getDate()
      console.log(years,"years",months,"months",days-1,"days")
      return (years+"years "+ months+ "months"+ days,"days")
    }
  </script>
</html>

Solution 30 - Javascript

I've stumbled upon this while having the same problem. Here is my code. It totally relies on the JS date function, so leap years are handled, and does not compare days based on hours, so it avoids daylight saving issues.

function dateDiff(start, end) {
    let years = 0, months = 0, days = 0;
    // Day diffence. Trick is to use setDate(0) to get the amount of days
    // from the previous month if the end day less than the start day.
    if (end.getDate() < start.getDate()) {
        months = -1;
        let datePtr = new Date(end);
        datePtr.setDate(0);
        days = end.getDate() + (datePtr.getDate() - start.getDate());
    } else {
        days = end.getDate() - start.getDate();
    }

    if (end.getMonth() < start.getMonth() ||
       (end.getMonth() === start.getMonth() && end.getDate() < start.getDate())) {
        years = -1;
        months += end.getMonth() + (12 - start.getMonth());
    } else {
        months += end.getMonth() - start.getMonth();
    }

    years += end.getFullYear() - start.getFullYear();
    console.log(`${years}y ${months}m ${days}d`);
    return [years, months, days];
}

let a = new Date(2019,6,31);  // 31 Jul 2019
let b = new Date(2022,2, 1);  //  1 Mar 2022

console.log(dateDiff(a, b));  // [2, 7, -2]

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