How to calculate date difference in JavaScript?

Question

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

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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:

/*  Return number of days between d0 and d1.
**  Returns positive if d0 < d1, otherwise negative.
**
**  e.g. between 2000-02-28 and 2001-02-28 there are 366 days
**       between 2015-12-28 and 2015-12-29 there is 1 day
**       between 2015-12-28 23:59:59 and 2015-12-29 00:00:01 there is 1 day
**       between 2015-12-28 00:00:01 and 2015-12-28 23:59:59 there are 0 days
**        
**  @param {Date} d0  - start date
**  @param {Date} d1  - end date
**  @returns {number} - whole number of days between d0 and d1
**
*/
function daysDifference(d0, d1) {
  var diff = new Date(+d1).setHours(12) - new Date(+d0).setHours(12);
  return Math.round(diff/8.64e7);
}

// Simple formatter
function formatDate(date){
  return [date.getFullYear(),('0'+(date.getMonth()+1)).slice(-2),('0'+date.getDate()).slice(-2)].join('-');
}

// Examples
[[new Date(2000,1,28), new Date(2001,1,28)],  // Leap year
 [new Date(2001,1,28), new Date(2002,1,28)],  // Not leap year
 [new Date(2017,0,1),  new Date(2017,1,1)] 
].forEach(function(dates) {
  document.write('From ' + formatDate(dates[0]) + ' to ' + formatDate(dates[1]) +
                 ' is ' + daysDifference(dates[0],dates[1]) + ' days<br>');
});

Answer 5

<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>

Answer 6

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

Answer 7

adding to @paresh mayani 's answer, to work like Facebook - showing how much time has passed in sec/min/hours/weeks/months/years

var DateDiff = {

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

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


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

        return parseInt((t2-t1)/60000);
    },

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

        return parseInt((t2-t1)/3600000);
    },

    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"; //will also get (Y-m-d H:i:s)
    
    var d1 = new Date(dString);
    var d2 = new Date();
    
    var timeLaps = DateDiff.inSeconds(d1, d2);
    var dateOutput = "";
    
    
    if (timeLaps<60)
    {
      dateOutput = timeLaps+" seconds";
    }
    else  
    {
      timeLaps = DateDiff.inMinutes(d1, d2);
      if (timeLaps<60)
      {
        dateOutput = timeLaps+" minutes";
      }
      else
      {
        timeLaps = DateDiff.inHours(d1, d2);
        if (timeLaps<24)
        {
          dateOutput = timeLaps+" hours";
        }
        else
        {
            timeLaps = DateDiff.inDays(d1, d2);
            if (timeLaps<7)
            {
              dateOutput = timeLaps+" days";
            }
            else
            {
                timeLaps = DateDiff.inWeeks(d1, d2);
                if (timeLaps<4)
                {
                  dateOutput = timeLaps+" weeks";
                }
                else
                {
                    timeLaps = DateDiff.inMonths(d1, d2);
                    if (timeLaps<12)
                    {
                      dateOutput = timeLaps+" months";
                    }
                    else
                    {
                      timeLaps = DateDiff.inYears(d1, d2);
                      dateOutput = timeLaps+" years";
                    }
                }
            }
        }
      }
    }
    
    alert (dateOutput);

Answer 8

With momentjs it's simple:

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

Answer 9

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      
}

Answer 10

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.");

Answer 11

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))

Answer 12

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

Answer 13

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

Answer 14

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.");

Answer 15

This code will return the difference between two dates in days:

const previous_date = new Date("2019-12-23");
const current_date = new Date();

const current_year = current_date.getFullYear();
const previous_date_year = 
previous_date.getFullYear();

const difference_in_years = current_year - 
previous_date_year;

let months = current_date.getMonth();
months = months + 1; // for making the indexing 
// of months from 1

for(let i = 0; i < difference_in_years; i++){
months = months + 12;
}

let days = current_date.getDate();

days = days + (months * 30.417);

console.log(`The days between ${current_date} and 
${previous_date} are : ${days} (approximately)`);

Answer 16

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")

Answer 17

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"));

Answer 18

function daysInMonth (month, year) {
    return new Date(year, month, 0).getDate();
}
function getduration(){

let A= document.getElementById("date1_id").value
let B= document.getElementById("date2_id").value

let C=Number(A.substring(3,5))
let D=Number(B.substring(3,5))
let dif=D-C
let arr=[];
let sum=0;
for (let i=0;i<dif+1;i++){
  sum+=Number(daysInMonth(i+C,2019))
}
let sum_alter=0;
for (let i=0;i<dif;i++){
  sum_alter+=Number(daysInMonth(i+C,2019))
}
let no_of_month=(Number(B.substring(3,5)) - Number(A.substring(3,5)))
let days=[];
if ((Number(B.substring(3,5)) - Number(A.substring(3,5)))>0||Number(B.substring(0,2)) - Number(A.substring(0,2))<0){
days=Number(B.substring(0,2)) - Number(A.substring(0,2)) + sum_alter
}

if ((Number(B.substring(3,5)) == Number(A.substring(3,5)))){
console.log(Number(B.substring(0,2)) - Number(A.substring(0,2)) + sum_alter)
}

time_1=[]; time_2=[]; let hour=[];
 time_1=document.getElementById("time1_id").value
 time_2=document.getElementById("time2_id").value
  if (time_1.substring(0,2)=="12"){
     time_1="00:00:00 PM"
  }
if (time_1.substring(9,11)==time_2.substring(9,11)){
hour=Math.abs(Number(time_2.substring(0,2)) - Number(time_1.substring(0,2)))
}
if (time_1.substring(9,11)!=time_2.substring(9,11)){
hour=Math.abs(Number(time_2.substring(0,2)) - Number(time_1.substring(0,2)))+12
}
let min=Math.abs(Number(time_1.substring(3,5))-Number(time_2.substring(3,5)))
document.getElementById("duration_id").value=days +" days "+ hour+"  hour " + min+"  min " 
}

<input type="text" id="date1_id" placeholder="28/05/2019">
<input type="text" id="date2_id" placeholder="29/06/2019">
<br><br>
<input type="text" id="time1_id" placeholder="08:01:00 AM">
<input type="text" id="time2_id" placeholder="00:00:00 PM">
<br><br>
<button class="text" onClick="getduration()">Submit </button>
<br><br>
<input type="text" id="duration_id" placeholder="days hour min">

Answer 19

var date1 = new Date("06/30/2019");
var date2 = new Date("07/30/2019");
  
// To calculate the time difference of two dates
var Difference_In_Time = date2.getTime() - date1.getTime();
  
// To calculate the no. of days between two dates
var Difference_In_Days = Difference_In_Time / (1000 * 3600 * 24);
  
//To display the final no. of days (result)
document.write("Total number of days between dates  <br>"
               + date1 + "<br> and <br>" 
               + date2 + " is: <br> " 
               + Difference_In_Days);

Answer 20

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;

Answer 21

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.

Answer 22

I made a below function to get the difference between now and "2021-02-26T21:50:42.123".

The difference return answer as milliseconds, so I convert it by using this formula:

(1000 * 3600 * 24).

function getDiff(dateAcquired) {
      let calDiff = Math.floor(
        (new Date() - new Date(dateAcquired)) / (1000 * 3600 * 24)
      );
      return calDiff;
    }
    console.log(getDiff("2021-02-26T21:50:42.123"));

Answer 23

Can be useful :

const date_diff = (date1, date2) => Math.ceil(Math.abs(date1 - date2)/24 * 60 * 60 * 1000)

or

const date_diff = (date1, date2) => Math.ceil(Math.abs(date1 - date2)/86400000)

where 24 * 60 * 60 * 1000 is (day * minutes * seconds * milliseconds) = 86400000 milliseconds in one day

Thank you

Answer 24

            // the idea is to get time left for new year.
           // Not considering milliseconds as of now, but that 
           //  can be done
           
            var newYear = '1 Jan 2023';
            const secondsInAMin = 60;
            const secondsInAnHour = 60 * secondsInAMin;
            const secondsInADay = 24 * secondsInAnHour;

            function DateDiffJs() {
                var newYearDate = new Date(newYear);
                var currDate = new Date();

                var remainingSecondsInDateDiff = (newYearDate - currDate) / 1000;
                var days = Math.floor(remainingSecondsInDateDiff / secondsInADay);

                var remainingSecondsAfterDays = remainingSecondsInDateDiff - (days * secondsInADay);
                var hours = Math.floor(remainingSecondsAfterDays / secondsInAnHour);

                var remainingSecondsAfterhours = remainingSecondsAfterDays - (hours * secondsInAnHour);
                var mins = Math.floor(remainingSecondsAfterhours / secondsInAMin);

                var seconds = Math.floor(remainingSecondsAfterhours - (mins * secondsInAMin));


                console.log(`days :: ${days}`)
                console.log(`hours :: ${hours}`)
                console.log(`mins :: ${mins}`)
                console.log(`seconds :: ${seconds}`)

            }

            DateDiffJs();