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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:

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'));
deceze
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Chris
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34 Answers34

39

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.

klenium
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Pawel Miech
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    Thank you for your answer. It is great, but I am afraid exactly what I said I was not looking for. As said, what you wrote is the easy part - considering the exact number of days in a month, leap years, ... that is the difficult thing. – Chris Jul 19 '13 at 22:31
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    Thanks for your answer. but I have modified little bit, which link is given bellow: http://jsfiddle.net/PUSQU/8 – Sarwar Hasan Nov 20 '14 at 09:20
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    This is a nice answer, however I'm surprised no one ever took into consideration that this is off some.. since the months and years math would be based on a year with 372 days in it – chris Jan 02 '22 at 17:45
34

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.

Mordred
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    @RobG I think I clearly stated the limitations in the answer, and will incorporate your comment as an explanation. Not entirely sure why the downvote, given that many of the other answers here have the exact same issue. This is intended to be simple and quick, at the expense of perfect accuracy. – Mordred May 10 '19 at 16:26
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    @RobG Added a much longer, but accurate version for you as well. – Mordred May 10 '19 at 18:27
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    This is exactly what I was looking for, thanks! – Vladimir Jovanović Dec 13 '20 at 10:25
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    This needs a ton more upvotes. it cures the issues experienced by the answers above it. Well done! – Robert Hollon Feb 22 '21 at 03:25
  • How to modify this answer and get weekDiff? – Bhautik Apr 01 '21 at 21:10
  • If you are still around...curious why startDate is wrapped in 2 new Date()s? If I use today as a start date, it changes it to yesterday. In my situation, I am testing to see if the dates are equal so I can return "Today" – Tony Smith May 12 '21 at 13:59
  • @TonySmith I believe the reason was sanitation and trying to make sure we're comparing apples to apples. The function handles both dates in the format `'2021-05-12'`, `'2021-05-12T12:00:00'` and `new Date()`. The first new Date() takes the string, creates a date, and then lops off the hours to make sure we're looking at `'2021-05-12'` only, then does a new Date() on that to use for comparisons. The thought was that it would account for variance in HH:MM:SS in the supplied dates. However, timezones might throw that off. If you know the date is in local you can add back in `getTimezoneOffset()`. – Mordred May 12 '21 at 21:01
  • The problem is that the dates are generated by `new Date(new Date(startingDate).toISOString().substr(0, 10))`, where the result of the *substr* method is a date in YYYY-MM-DD format that is parsed as UTC, so may go back or forward 1 day depending on the host timezone offset and the time for the date. If the times are the same then start and end will have compensating errors, but if the times are different they may not. – RobG Feb 27 '22 at 23:53
  • Consider a user with offset +10. A date for 2022-02-27 09:00 is 2022-02-26 UTC, so the day before. A date for 2022-02-28 11:00 is 2022-02-28 UTC, or same day. So the difference appears to be 2 days when it likely should be 1. This may also manifest itself over daylight saving boundaries. The issue is just about parsing of the strings, if they're treated as UTC always then the problem goes away. – RobG Feb 27 '22 at 23:59
32

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>
Inanc Gumus
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    you pay a big penalty of over 100k when all you need is a 2 line function – Charles Okwuagwu Oct 01 '17 at 09:33
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    Minified version is only 51KB. ~20KB if server hosting has gzip compression enable (cloudflare hosting does). Not to mention that the browser may have already cached from another site you've visited that's using the same CDN reference. Considering how this library also does many other useful stuff with dates, I would consider this to be a pretty good trade off considering how difficult dates are to deal with in JavaScript. – Adrian Sanguineti Nov 20 '17 at 02:18
  • Don't just optimize for the computer, optimize for the programmer. Your 2 line func can become 20-30 in time. Using a tested and a consistent library is better in my book. – Inanc Gumus Nov 20 '17 at 08:58
  • you can't always use a plugin. and cannot always afford even 20KB – user151496 Mar 04 '18 at 22:10
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    then don't use it for those cases. – Inanc Gumus Mar 05 '18 at 00:49
  • I would also like to recommend the use of date-fns https://date-fns.org/ you can get the things that you only need + it has more nicer api :) – Alleo Indong Sep 14 '18 at 09:53
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    the answer is unfairly at very bottom. I've already found the library - but I'd save a couple hours if the answer was on the top – feech Nov 21 '19 at 00:39
13

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");
Rajeev P Nadig
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    This code as written returns incorrect results. Assuming today's date is 2018-03-09: `sdt = new Date("2017-03-09")` will alert "1Y 1M 1D", `sdt = new Date("2018-02-09")` will alert "0Y 1M 29D", `sdt = new Date("2018-03-09")` will alert "0Y 1M 1D". My edit for this answer was rejected, so I added the corrected code as a separate answer. – Mordred Mar 09 '18 at 20:48
9

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)"
Murray
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7

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

Jakub Pelák
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7

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.

Patryk Janik
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6

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;
}
Chris
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  • Hey Chris I tried this answer and it did not work. `getNiceTime(new Date('2014-05-10'), new Date('2015-10-10'), 4, true);` should return `1 year, 5 months, 1 hour ago` but did return `1 year, 5 months, 2 days, 1 hour ago` – Whitecat Nov 16 '15 at 18:58
5

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

teja463
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4

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.

Geeky Guy
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  • Thank you Renan for taking time to answer. Very nice, however, the same Pawelmhm - it's simple to get the number of miliseconds or days between two dates - but considering days in a month and leap years etc, for a precise calculation is tricky... – Chris Jul 19 '13 at 22:33
4

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(" ");
}

Felipe Chernicharo
  • 3,619
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3

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
}
webtweakers
  • 715
  • 7
  • 19
2

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>
David SN
  • 31
  • 2
2
   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

Nilesh Patel
  • 557
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  • an added note, if you get longer dates from server ie. `2018-10-31T00:58:08.041Z` `dateVar.substring(0,9)` will work a treat – mewc Oct 31 '18 at 00:57
  • @mewc the lib has sophisticated parser, check out the documentation https://momentjs.com/docs/#/parsing/ – feech Nov 21 '19 at 00:42
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); 
    }
}
Yash
  • 9,250
  • 2
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  • 74
1

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>
Sachin Gaur
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  • 6
1

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));
MaHdi
  • 53
  • 6
  • this does not work if it's say 1 day less than a year between the 2 days so for instance "2021-07-15" to "2022-07-14" will return "0 years, -1 months, 30 days" – Touchpad Jul 01 '21 at 09:40
  • 1
    Thank you, @Touchpad. I fixed it. – MaHdi Jul 01 '21 at 19:11
1

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>
Mohsen Alyafei
  • 4,765
  • 3
  • 30
  • 42
1

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

Gangula
  • 5,193
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0

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

Joe Minichino
  • 2,793
  • 20
  • 20
  • Thank you for sharing Joe. However, I could not find that datejs is able to tell an exact time difference between two dates like "4 years, 2 month and 11 days ago"? – Chris Jul 19 '13 at 22:37
0

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
} */
Oleg
  • 1,291
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  • 16
0

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.
    }
}
Filgaia
  • 1,046
  • 10
  • 11
0

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

Kevin Le - Khnle
  • 10,579
  • 11
  • 54
  • 80
0

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

Sai Nikhil
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0

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.

Dr. MAF
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0

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) );
Hammad ul Hasan
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0

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/

0

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
Rinzin
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0

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)
mdehghani
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0

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')));
Accountant م
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0

Pure javascript, no "wild" difference calculations

I know this is an old and long thread and I'm sure there's a plenty of great answers already, but I'd like to add my two cents - a simple and transparent solution, that is independent from any library and works precisely. Simply leaving all the magic on javascript Date object.

The function expects two dates in ISO format (their order doesn't matter), returning an object with years, months and days (and a label for it).

It can be quite easily extended for hours, minutes etc. Hope it helps out there.

dateRange(firstDate = "1979-07-07", secondDate = "2022-11-19") {
  const dateA = new Date(firstDate);
  const dateB = new Date(secondDate);
  if (dateA.getTime() === dateB.getTime()) {
    return {
      years: 0,
      months: 0,
      days: 0,
      label: "The dates are the same.",
    };
  }

  let startDate = null;
  let endDate = null;

  //the date range should be computed in absolute values
  //it doesn't matter, in which order the dates are supplied
  //so, simply said, startDate = min(dateA, dateB) and endDate = max(dateA, dateB)
  if (dateA < dateB) {
    startDate = dateA;
    //subtracting 1 day from the end date in order to get correct values
    endDate = new Date(
      dateB.getFullYear(),
      dateB.getMonth(),
      dateB.getDate() - 1,
    );
  } else if (dateA > dateB) {
    startDate = dateB;
    endDate = new Date(
      dateA.getFullYear(),
      dateA.getMonth(),
      dateA.getDate() - 1,
    );
  }

  //setting controlDate which will be incremented until reaching endDate
  let controlDate = startDate;
  //setting differences in years, months and days
  let yearsDiff = 0;
  let monthsDiff = 0;
  let daysDiff = 0;

  //incrementing by years until exceeding end date
  do {
    yearsDiff++;
    controlDate = new Date(
      startDate.getFullYear() + yearsDiff,
      startDate.getMonth(),
      startDate.getDate(),
    );
  } while (controlDate.getTime() <= endDate.getTime());

  //end date exceeded, so decrement yearsDiff
  yearsDiff--;

  //incrementing by months until exceeding end date
  do {
    monthsDiff++;
    controlDate = new Date(
      startDate.getFullYear() + yearsDiff,
      startDate.getMonth() + monthsDiff,
      startDate.getDate(),
    );
  } while (controlDate.getTime() <= endDate.getTime());

  //end date exceeded, so decrement monthsDiff
  monthsDiff--;

  //incrementing by days until exceeding end date
  do {
    daysDiff++;
    controlDate = new Date(
      startDate.getFullYear() + yearsDiff,
      startDate.getMonth() + monthsDiff,
      startDate.getDate() + daysDiff,
    );
  } while (controlDate.getTime() <= endDate.getTime());

  //end date exceeded - daysDiff keeps final value

  return {
    years: yearsDiff,
    months: monthsDiff,
    days: daysDiff,
    label:
      "" +
      yearsDiff +
      " year(s), " +
      monthsDiff +
      " month(s), " +
      daysDiff +
      " day(s).",
  };
}
Lojza3D
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-1

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>
-1

This worked for me.

function GetDateDifference(date, otherDate) {
    var dateOnly = date.setHours(0, 0, 0, 0);
    var otherDateOnly = otherDate.setHours(0, 0, 0, 0);
    var milliseconds = Math.abs(dateOnly - otherDateOnly);
    var millisecondsPerDay = 1000 * 60 * 60 * 24;
    return Math.floor(milliseconds / millisecondsPerDay);
}
-2

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]
RobG
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  • The question requires Javascript, not Typescript. I'll vote useful if you fix it. – Paul-Sebastian Manole May 12 '19 at 13:19
  • Just remove the type identifier after the dots. I've shared the best algorithm here. I don't care for your vote. – Alexander Simon Aug 12 '19 at 19:45
  • It's nothing personal. The author needs a Javascript solution. Not a Typescript solution. I'm asking you to do the fix. It's only right. – Paul-Sebastian Manole Aug 12 '19 at 21:03
  • This doesn't handle different length months as shown in the added example. 31 Jul 2019 to 1 Mar 2022 should be 2 years, 6 months and 1 day, but the function returns `[2, 7, -2]`, i.e. 2 years, 7 months and -2 days. – RobG Jan 20 '22 at 03:10