[javascript] How to calculate date difference in JavaScript?

I want to calculate date difference in days, hours, minutes, seconds, milliseconds, nanoseconds. How can I do it?

This question is related to javascript datetime

The answer is


based on javascript runtime prototype implementation you can use simple arithmetic to subtract dates as in bellow

var sep = new Date(2020, 07, 31, 23, 59, 59);
var today = new Date();
var diffD = Math.floor((sep - today) / (1000 * 60 * 60 * 24));
console.log('Day Diff: '+diffD);

the difference return answer as milliseconds, then you have to convert it by division:

  • by 1000 to convert to second
  • by 1000×60 convert to minute
  • by 1000×60×60 convert to hour
  • by 1000×60×60×24 convert to day

Another solution is convert difference to a new Date object and get that date's year(diff from 1970), month, day etc.

var date1 = new Date(2010, 6, 17);
var date2 = new Date(2013, 12, 18);
var diff = new Date(date2.getTime() - date1.getTime());
// diff is: Thu Jul 05 1973 04:00:00 GMT+0300 (EEST)

console.log(diff.getUTCFullYear() - 1970); // Gives difference as year
// 3

console.log(diff.getUTCMonth()); // Gives month count of difference
// 6

console.log(diff.getUTCDate() - 1); // Gives day count of difference
// 4

So difference is like "3 years and 6 months and 4 days". If you want to take difference in a human readable style, that can help you.


this should work just fine if you just need to show what time left, since JavaScript uses frames for its time you'll have get your End Time - The Time RN after that we can divide it by 1000 since apparently 1000 frames = 1 seconds, after that you can use the basic math of time, but there's still a problem to this code, since the calculation is static, it can't compensate for the different day total in a year (360/365/366), the bunch of IF after the calculation is to make it null if the time is lower than 0, hope this helps even though it's not exactly what you're asking :)

var now = new Date();
var end = new Date("End Time");
var total = (end - now) ;
var totalD =  Math.abs(Math.floor(total/1000));

var years = Math.floor(totalD / (365*60*60*24));
var months = Math.floor((totalD - years*365*60*60*24) / (30*60*60*24));
var days = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24)/ (60*60*24));
var hours = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24)/ (60*60));
var minutes = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24 - hours*60*60)/ (60));
var seconds = Math.floor(totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24 - hours*60*60 - minutes*60);

var Y = years < 1 ? "" : years + " Years ";
var M = months < 1 ? "" : months + " Months ";
var D = days < 1 ? "" : days + " Days ";
var H = hours < 1 ? "" : hours + " Hours ";
var I = minutes < 1 ? "" : minutes + " Minutes ";
var S = seconds < 1 ? "" : seconds + " Seconds ";
var A = years == 0 && months == 0 && days == 0 && hours == 0 && minutes == 0 && seconds == 0 ? "Sending" : " Remaining";

document.getElementById('txt').innerHTML = Y + M + D + H + I + S + A;

var DateDiff = {

    inDays: function(d1, d2) {
        var t2 = d2.getTime();
        var t1 = d1.getTime();

        return parseInt((t2-t1)/(24*3600*1000));
    },

    inWeeks: function(d1, d2) {
        var t2 = d2.getTime();
        var t1 = d1.getTime();

        return parseInt((t2-t1)/(24*3600*1000*7));
    },

    inMonths: function(d1, d2) {
        var d1Y = d1.getFullYear();
        var d2Y = d2.getFullYear();
        var d1M = d1.getMonth();
        var d2M = d2.getMonth();

        return (d2M+12*d2Y)-(d1M+12*d1Y);
    },

    inYears: function(d1, d2) {
        return d2.getFullYear()-d1.getFullYear();
    }
}

var dString = "May, 20, 1984";

var d1 = new Date(dString);
var d2 = new Date();

document.write("<br />Number of <b>days</b> since "+dString+": "+DateDiff.inDays(d1, d2));
document.write("<br />Number of <b>weeks</b> since "+dString+": "+DateDiff.inWeeks(d1, d2));
document.write("<br />Number of <b>months</b> since "+dString+": "+DateDiff.inMonths(d1, d2));
document.write("<br />Number of <b>years</b> since "+dString+": "+DateDiff.inYears(d1, d2));

Code sample taken from here.


_x000D_
_x000D_
function daysInMonth (month, year) {_x000D_
    return new Date(year, month, 0).getDate();_x000D_
}_x000D_
function getduration(){_x000D_
_x000D_
let A= document.getElementById("date1_id").value_x000D_
let B= document.getElementById("date2_id").value_x000D_
_x000D_
let C=Number(A.substring(3,5))_x000D_
let D=Number(B.substring(3,5))_x000D_
let dif=D-C_x000D_
let arr=[];_x000D_
let sum=0;_x000D_
for (let i=0;i<dif+1;i++){_x000D_
  sum+=Number(daysInMonth(i+C,2019))_x000D_
}_x000D_
let sum_alter=0;_x000D_
for (let i=0;i<dif;i++){_x000D_
  sum_alter+=Number(daysInMonth(i+C,2019))_x000D_
}_x000D_
let no_of_month=(Number(B.substring(3,5)) - Number(A.substring(3,5)))_x000D_
let days=[];_x000D_
if ((Number(B.substring(3,5)) - Number(A.substring(3,5)))>0||Number(B.substring(0,2)) - Number(A.substring(0,2))<0){_x000D_
days=Number(B.substring(0,2)) - Number(A.substring(0,2)) + sum_alter_x000D_
}_x000D_
_x000D_
if ((Number(B.substring(3,5)) == Number(A.substring(3,5)))){_x000D_
console.log(Number(B.substring(0,2)) - Number(A.substring(0,2)) + sum_alter)_x000D_
}_x000D_
_x000D_
time_1=[]; time_2=[]; let hour=[];_x000D_
 time_1=document.getElementById("time1_id").value_x000D_
 time_2=document.getElementById("time2_id").value_x000D_
  if (time_1.substring(0,2)=="12"){_x000D_
     time_1="00:00:00 PM"_x000D_
  }_x000D_
if (time_1.substring(9,11)==time_2.substring(9,11)){_x000D_
hour=Math.abs(Number(time_2.substring(0,2)) - Number(time_1.substring(0,2)))_x000D_
}_x000D_
if (time_1.substring(9,11)!=time_2.substring(9,11)){_x000D_
hour=Math.abs(Number(time_2.substring(0,2)) - Number(time_1.substring(0,2)))+12_x000D_
}_x000D_
let min=Math.abs(Number(time_1.substring(3,5))-Number(time_2.substring(3,5)))_x000D_
document.getElementById("duration_id").value=days +" days "+ hour+"  hour " + min+"  min " _x000D_
}
_x000D_
<input type="text" id="date1_id" placeholder="28/05/2019">_x000D_
<input type="text" id="date2_id" placeholder="29/06/2019">_x000D_
<br><br>_x000D_
<input type="text" id="time1_id" placeholder="08:01:00 AM">_x000D_
<input type="text" id="time2_id" placeholder="00:00:00 PM">_x000D_
<br><br>_x000D_
<button class="text" onClick="getduration()">Submit </button>_x000D_
<br><br>_x000D_
<input type="text" id="duration_id" placeholder="days hour min">
_x000D_
_x000D_
_x000D_


Sorry but flat millisecond calculation is not reliable Thanks for all the responses, but few of the functions I tried are failing either on 1. A date near today's date 2. A date in 1970 or 3. A date in a leap year.

Approach that best worked for me and covers all scenario e.g. leap year, near date in 1970, feb 29 etc.

var someday = new Date("8/1/1985");
var today = new Date();
var years = today.getFullYear() - someday.getFullYear();

// Reset someday to the current year.
someday.setFullYear(today.getFullYear());

// Depending on when that day falls for this year, subtract 1.
if (today < someday)
{
    years--;
}
document.write("Its been " + years + " full years.");

If you are using moment.js then it is pretty simple to find date difference.

var now  = "04/09/2013 15:00:00";
var then = "04/09/2013 14:20:30";

moment.utc(moment(now,"DD/MM/YYYY HH:mm:ss").diff(moment(then,"DD/MM/YYYY HH:mm:ss"))).format("HH:mm:ss")

var d1=new Date(2011,0,1); // jan,1 2011
var d2=new Date(); // now

var diff=d2-d1,sign=diff<0?-1:1,milliseconds,seconds,minutes,hours,days;
diff/=sign; // or diff=Math.abs(diff);
diff=(diff-(milliseconds=diff%1000))/1000;
diff=(diff-(seconds=diff%60))/60;
diff=(diff-(minutes=diff%60))/60;
days=(diff-(hours=diff%24))/24;

console.info(sign===1?"Elapsed: ":"Remains: ",
             days+" days, ",
             hours+" hours, ",
             minutes+" minutes, ",
             seconds+" seconds, ",
             milliseconds+" milliseconds.");

Expressions like "difference in days" are never as simple as they seem. If you have the following dates:

d1: 2011-10-15 23:59:00
d1: 2011-10-16 00:01:00

the difference in time is 2 minutes, should the "difference in days" be 1 or 0? Similar issues arise for any expression of the difference in months, years or whatever since years, months and days are of different lengths and different times (e.g. the day that daylight saving starts is 1 hour shorter than usual and two hours shorter than the day that it ends).

Here is a function for a difference in days that ignores the time, i.e. for the above dates it returns 1.

/*
   Get the number of days between two dates - not inclusive.

   "between" does not include the start date, so days
   between Thursday and Friday is one, Thursday to Saturday
   is two, and so on. Between Friday and the following Friday is 7.

   e.g. getDaysBetweenDates( 22-Jul-2011, 29-jul-2011) => 7.

   If want inclusive dates (e.g. leave from 1/1/2011 to 30/1/2011),
   use date prior to start date (i.e. 31/12/2010 to 30/1/2011).

   Only calculates whole days.

   Assumes d0 <= d1
*/
function getDaysBetweenDates(d0, d1) {

  var msPerDay = 8.64e7;

  // Copy dates so don't mess them up
  var x0 = new Date(d0);
  var x1 = new Date(d1);

  // Set to noon - avoid DST errors
  x0.setHours(12,0,0);
  x1.setHours(12,0,0);

  // Round to remove daylight saving errors
  return Math.round( (x1 - x0) / msPerDay );
}

This can be more concise:

_x000D_
_x000D_
/*  Return number of days between d0 and d1._x000D_
**  Returns positive if d0 < d1, otherwise negative._x000D_
**_x000D_
**  e.g. between 2000-02-28 and 2001-02-28 there are 366 days_x000D_
**       between 2015-12-28 and 2015-12-29 there is 1 day_x000D_
**       between 2015-12-28 23:59:59 and 2015-12-29 00:00:01 there is 1 day_x000D_
**       between 2015-12-28 00:00:01 and 2015-12-28 23:59:59 there are 0 days_x000D_
**        _x000D_
**  @param {Date} d0  - start date_x000D_
**  @param {Date} d1  - end date_x000D_
**  @returns {number} - whole number of days between d0 and d1_x000D_
**_x000D_
*/_x000D_
function daysDifference(d0, d1) {_x000D_
  var diff = new Date(+d1).setHours(12) - new Date(+d0).setHours(12);_x000D_
  return Math.round(diff/8.64e7);_x000D_
}_x000D_
_x000D_
// Simple formatter_x000D_
function formatDate(date){_x000D_
  return [date.getFullYear(),('0'+(date.getMonth()+1)).slice(-2),('0'+date.getDate()).slice(-2)].join('-');_x000D_
}_x000D_
_x000D_
// Examples_x000D_
[[new Date(2000,1,28), new Date(2001,1,28)],  // Leap year_x000D_
 [new Date(2001,1,28), new Date(2002,1,28)],  // Not leap year_x000D_
 [new Date(2017,0,1),  new Date(2017,1,1)] _x000D_
].forEach(function(dates) {_x000D_
  document.write('From ' + formatDate(dates[0]) + ' to ' + formatDate(dates[1]) +_x000D_
                 ' is ' + daysDifference(dates[0],dates[1]) + ' days<br>');_x000D_
});
_x000D_
_x000D_
_x000D_


Ok, there are a bunch of ways you can do that. Yes, you can use plain old JS. Just try:

let dt1 = new Date()
let dt2 = new Date()

Let's emulate passage using Date.prototype.setMinutes and make sure we are in range.

dt1.setMinutes(7)
dt2.setMinutes(42)
console.log('Elapsed seconds:',(dt2-dt1)/1000)

Alternatively you could use some library like js-joda, where you can easily do things like this (directly from docs):

var dt1 = LocalDateTime.parse("2016-02-26T23:55:42.123");
var dt2 = dt1
  .plusYears(6)
  .plusMonths(12)
  .plusHours(2)
  .plusMinutes(42)
  .plusSeconds(12);

// obtain the duration between the two dates
dt1.until(dt2, ChronoUnit.YEARS); // 7
dt1.until(dt2, ChronoUnit.MONTHS); // 84
dt1.until(dt2, ChronoUnit.WEEKS); // 356
dt1.until(dt2, ChronoUnit.DAYS); // 2557
dt1.until(dt2, ChronoUnit.HOURS); // 61370
dt1.until(dt2, ChronoUnit.MINUTES); // 3682242
dt1.until(dt2, ChronoUnit.SECONDS); // 220934532

There are plenty more libraries ofc, but js-joda has an added bonus of being available also in Java, where it has been extensively tested. All those tests have been migrated to js-joda, it's also immutable.


I think this should do it.

let today = new Date();
let form_date=new Date('2019-10-23')
let difference=form_date>today ? form_date-today : today-form_date
let diff_days=Math.floor(difference/(1000*3600*24))

function DateDiff(b, e)
{
    let
        endYear = e.getFullYear(),
        endMonth = e.getMonth(),
        years = endYear - b.getFullYear(),
        months = endMonth - b.getMonth(),
        days = e.getDate() - b.getDate();
    if (months < 0)
    {
        years--;
        months += 12;
    }
    if (days < 0)
    {
        months--;
        days += new Date(endYear, endMonth, 0).getDate();
    }
    return [years, months, days];
}

[years, months, days] = DateDiff(
    new Date("October 21, 1980"),
    new Date("July 11, 2017")); // 36 8 20

With momentjs it's simple:

moment("2016-04-08").fromNow();

Assuming you have two Date objects, you can just subtract them to get the difference in milliseconds:

var difference = date2 - date1;

From there, you can use simple arithmetic to derive the other values.


This is how you can implement difference between dates without a framework.

function getDateDiff(dateOne, dateTwo) {
        if(dateOne.charAt(2)=='-' & dateTwo.charAt(2)=='-'){
            dateOne = new Date(formatDate(dateOne));
            dateTwo = new Date(formatDate(dateTwo));
        }
        else{
            dateOne = new Date(dateOne);
            dateTwo = new Date(dateTwo);            
        }
        let timeDiff = Math.abs(dateOne.getTime() - dateTwo.getTime());
        let diffDays = Math.ceil(timeDiff / (1000 * 3600 * 24));
        let diffMonths = Math.ceil(diffDays/31);
        let diffYears = Math.ceil(diffMonths/12);

        let message = "Difference in Days: " + diffDays + " " +
                      "Difference in Months: " + diffMonths+ " " + 
                      "Difference in Years: " + diffYears;
        return message;
     }

    function formatDate(date) {
         return date.split('-').reverse().join('-');
    }

    console.log(getDateDiff("23-04-2017", "23-04-2018"));

function DateDiff(date1, date2) {
    date1.setHours(0);
    date1.setMinutes(0, 0, 0);
    date2.setHours(0);
    date2.setMinutes(0, 0, 0);
    var datediff = Math.abs(date1.getTime() - date2.getTime()); // difference 
    return parseInt(datediff / (24 * 60 * 60 * 1000), 10); //Convert values days and return value      
}

<html lang="en">
<head>
<script>
function getDateDiff(time1, time2) {
  var str1= time1.split('/');
  var str2= time2.split('/');

  //                yyyy   , mm       , dd
  var t1 = new Date(str1[2], str1[0]-1, str1[1]);
  var t2 = new Date(str2[2], str2[0]-1, str2[1]);

  var diffMS = t1 - t2;    
  console.log(diffMS + ' ms');

  var diffS = diffMS / 1000;    
  console.log(diffS + ' ');

  var diffM = diffS / 60;
  console.log(diffM + ' minutes');

  var diffH = diffM / 60;
  console.log(diffH + ' hours');

  var diffD = diffH / 24;
  console.log(diffD + ' days');
  alert(diffD);
}

//alert(getDateDiff('10/18/2013','10/14/2013'));
</script>
</head>
<body>
  <input type="button" 
       onclick="getDateDiff('10/18/2013','10/14/2013')" 
       value="clickHere()" />

</body>
</html>

use Moment.js for all your JavaScript related date-time calculation

Answer to your question is:

var a = moment([2007, 0, 29]);   
var b = moment([2007, 0, 28]);    
a.diff(b) // 86400000  

Complete details can be found here