How do I get the number of days between two dates in JavaScript?

Question

How do I get the number of days between two dates in JavaScript? For example, given two dates in input boxes:

<input id="first" value="1/1/2000"/>
<input id="second" value="1/1/2001"/>

<script>
  alert(datediff("day", first, second)); // what goes here?
</script>

Answer 1

Here is a quick and dirty implementation of datediff, as a proof of concept to solve the problem as presented in the question. It relies on the fact that you can get the elapsed milliseconds between two dates by subtracting them, which coerces them into their primitive number value (milliseconds since the start of 1970).

// new Date("dateString") is browser-dependent and discouraged, so we'll write
// a simple parse function for U.S. date format (which does no error checking)
function parseDate(str) {
    var mdy = str.split('/');
    return new Date(mdy[2], mdy[0]-1, mdy[1]);
}

function datediff(first, second) {
    // Take the difference between the dates and divide by milliseconds per day.
    // Round to nearest whole number to deal with DST.
    return Math.round((second-first)/(1000*60*60*24));
}

alert(datediff(parseDate(first.value), parseDate(second.value)));

<input id="first" value="1/1/2000"/>
<input id="second" value="1/1/2001"/>

You should be aware that the "normal" Date APIs (without "UTC" in the name) operate in the local timezone of the user's browser, so in general you could run into issues if your user is in a timezone that you don't expect, and your code will have to deal with Daylight Saving Time transitions. You should carefully read the documentation for the Date object and its methods, and for anything more complicated, strongly consider using a library that offers more safe and powerful APIs for date manipulation.

• Numbers and Dates -- MDN JavaScript Guide
• Date -- MDN JavaScript reference

_{Also, for illustration purposes, the snippet uses named access on the window object for brevity, but in production you should use standardized APIs like getElementById, or more likely, some UI framework.}

Answer 2

As of this writing, only one of the other answers correctly handles DST (daylight saving time) transitions. Here are the results on a system located in California:

                                        1/1/2013- 3/10/2013- 11/3/2013-
User       Formula                      2/1/2013  3/11/2013  11/4/2013  Result
---------  ---------------------------  --------  ---------  ---------  ---------
Miles                   (d2 - d1) / N   31        0.9583333  1.0416666  Incorrect
some         Math.floor((d2 - d1) / N)  31        0          1          Incorrect
fuentesjr    Math.round((d2 - d1) / N)  31        1          1          Correct
toloco     Math.ceiling((d2 - d1) / N)  31        1          2          Incorrect

N = 86400000

Although Math.round returns the correct results, I think it's somewhat clunky. Instead, by explicitly accounting for changes to the UTC offset when DST begins or ends, we can use exact arithmetic:

function treatAsUTC(date) {
    var result = new Date(date);
    result.setMinutes(result.getMinutes() - result.getTimezoneOffset());
    return result;
}

function daysBetween(startDate, endDate) {
    var millisecondsPerDay = 24 * 60 * 60 * 1000;
    return (treatAsUTC(endDate) - treatAsUTC(startDate)) / millisecondsPerDay;
}

alert(daysBetween($('#first').val(), $('#second').val()));

Explanation

JavaScript date calculations are tricky because Date objects store times internally in UTC, not local time. For example, 3/10/2013 12:00 AM Pacific Standard Time (UTC-08:00) is stored as 3/10/2013 8:00 AM UTC, and 3/11/2013 12:00 AM Pacific Daylight Time (UTC-07:00) is stored as 3/11/2013 7:00 AM UTC. On this day, midnight to midnight local time is only 23 hours in UTC!

Although a day in local time can have more or less than 24 hours, a day in UTC is always exactly 24 hours.¹ The daysBetween method shown above takes advantage of this fact by first calling treatAsUTC to adjust both local times to midnight UTC, before subtracting and dividing.

^{1. JavaScript ignores leap seconds.}

Answer 3

The easiest way to get the difference between two dates:

var diff =  Math.floor(( Date.parse(str2) - Date.parse(str1) ) / 86400000); 

You get the difference days (or NaN if one or both could not be parsed). The parse date gived the result in milliseconds and to get it by day you have to divided it by 24 * 60 * 60 * 1000

If you want it divided by days, hours, minutes, seconds and milliseconds:

function dateDiff( str1, str2 ) {
    var diff = Date.parse( str2 ) - Date.parse( str1 ); 
    return isNaN( diff ) ? NaN : {
        diff : diff,
        ms : Math.floor( diff            % 1000 ),
        s  : Math.floor( diff /     1000 %   60 ),
        m  : Math.floor( diff /    60000 %   60 ),
        h  : Math.floor( diff /  3600000 %   24 ),
        d  : Math.floor( diff / 86400000        )
    };
}

Here is my refactored version of James version:

function mydiff(date1,date2,interval) {
    var second=1000, minute=second*60, hour=minute*60, day=hour*24, week=day*7;
    date1 = new Date(date1);
    date2 = new Date(date2);
    var timediff = date2 - date1;
    if (isNaN(timediff)) return NaN;
    switch (interval) {
        case "years": return date2.getFullYear() - date1.getFullYear();
        case "months": return (
            ( date2.getFullYear() * 12 + date2.getMonth() )
            -
            ( date1.getFullYear() * 12 + date1.getMonth() )
        );
        case "weeks"  : return Math.floor(timediff / week);
        case "days"   : return Math.floor(timediff / day); 
        case "hours"  : return Math.floor(timediff / hour); 
        case "minutes": return Math.floor(timediff / minute);
        case "seconds": return Math.floor(timediff / second);
        default: return undefined;
    }
}

Answer 4

I recommend using the moment.js library (http://momentjs.com/docs/#/displaying/difference/). It handles daylight savings time correctly and in general is great to work with.

Example:

var start = moment("2013-11-03");
var end = moment("2013-11-04");
end.diff(start, "days")
1

Answer 5

Steps

1. Set start date
2. Set end date
3. Calculate difference
4. Convert milliseconds to days

Native JS

const startDate  = '2020-01-01';
const endDate    = '2020-03-15';

const diffInMs   = new Date(endDate) - new Date(startDate)
const diffInDays = diffInMs / (1000 * 60 * 60 * 24);

---

Comment

I know this is not part of your questions but in general, I would not recommend doing any date calculation or manipulation in vanilla JavaScript and rather use a library like date-fns, Luxon or moment.js for it due to many edge cases.

---

Using a library

Date-fns

https://date-fns.org/v2.16.1/docs/differenceInDays

const differenceInDays = require('date-fns/differenceInDays');

const startDate  = '2020-01-01';
const endDate    = '2020-03-15';

const diffInDays = differenceInDays(new Date(endDate), new Date(startDate));

Luxon

https://moment.github.io/luxon/docs/class/src/datetime.js~DateTime.html#instance-method-diff

const { DateTime } = require('luxon');

const startDate  = '2020-01-01';
const endDate    = '2020-03-15';

const diffInDays = DateTime.fromISO(endDate).diff(DateTime.fromISO(startDate), 'days').toObject().days;

Moment.js

https://momentjs.com/docs/#/displaying/difference/

const moment = require('moment');

const startDate  = '2020-01-01';
const endDate    = '2020-03-15';

const diffInDays = moment(endDate).diff(moment(startDate), 'days');

Examples on RunKit

https://runkit.com/marcobiedermann/5f573e211e85e7001a37ab00

Answer 6

I would go ahead and grab this small utility and in it you will find functions to this for you. Here's a short example:

        <script type="text/javascript" src="date.js"></script>
        <script type="text/javascript">
            var minutes = 1000*60;
            var hours = minutes*60;
            var days = hours*24;

            var foo_date1 = getDateFromFormat("02/10/2009", "M/d/y");
            var foo_date2 = getDateFromFormat("02/12/2009", "M/d/y");

            var diff_date = Math.round((foo_date2 - foo_date1)/days);
            alert("Diff date is: " + diff_date );
        </script>

Answer 7

Using Moment.js

var future = moment('05/02/2015');
var start = moment('04/23/2015');
var d = future.diff(start, 'days'); // 9
console.log(d);

<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.17.1/moment-with-locales.min.js"></script>

Answer 8

Date values in JS are datetime values.

So, direct date computations are inconsistent:

(2013-11-05 00:00:00) - (2013-11-04 10:10:10) < 1 day

for example we need to convert de 2nd date:

(2013-11-05 00:00:00) - (2013-11-04 00:00:00) = 1 day

the method could be truncate the mills in both dates:

var date1 = new Date('2013/11/04 00:00:00');
var date2 = new Date('2013/11/04 10:10:10'); //less than 1
var start = Math.floor(date1.getTime() / (3600 * 24 * 1000)); //days as integer from..
var end = Math.floor(date2.getTime() / (3600 * 24 * 1000)); //days as integer from..
var daysDiff = end - start; // exact dates
console.log(daysDiff);

date2 = new Date('2013/11/05 00:00:00'); //1

var start = Math.floor(date1.getTime() / (3600 * 24 * 1000)); //days as integer from..
var end = Math.floor(date2.getTime() / (3600 * 24 * 1000)); //days as integer from..
var daysDiff = end - start; // exact dates
console.log(daysDiff);

Answer 9

Better to get rid of DST, Math.ceil, Math.floor etc. by using UTC times:

var firstDate = Date.UTC(2015,01,2);
var secondDate = Date.UTC(2015,04,22);
var diff = Math.abs((firstDate.valueOf() 
    - secondDate.valueOf())/(24*60*60*1000));

This example gives difference 109 days. 24*60*60*1000 is one day in milliseconds.

Answer 10

To Calculate days between 2 given dates you can use the following code.Dates I use here are Jan 01 2016 and Dec 31 2016

var day_start = new Date("Jan 01 2016");
var day_end = new Date("Dec 31 2016");
var total_days = (day_end - day_start) / (1000 * 60 * 60 * 24);
document.getElementById("demo").innerHTML = Math.round(total_days);

<h3>DAYS BETWEEN GIVEN DATES</h3>
<p id="demo"></p>

Answer 11

It is possible to calculate a full proof days difference between two dates resting across different TZs using the following formula:

var start = new Date('10/3/2015');
var end = new Date('11/2/2015');
var days = (end - start) / 1000 / 60 / 60 / 24;
console.log(days);
// actually its 30 ; but due to daylight savings will show 31.0xxx
// which you need to offset as below
days = days - (end.getTimezoneOffset() - start.getTimezoneOffset()) / (60 * 24);
console.log(days);

Answer 12

I found this question when I want do some calculate on two date, but the date have hours and minutes value, I modified @michael-liu 's answer to fit my requirement, and it passed my test.

diff days 2012-12-31 23:00 and 2013-01-01 01:00 should equal 1. (2 hour) diff days 2012-12-31 01:00 and 2013-01-01 23:00 should equal 1. (46 hour)

function treatAsUTC(date) {
    var result = new Date(date);
    result.setMinutes(result.getMinutes() - result.getTimezoneOffset());
    return result;
}

var millisecondsPerDay = 24 * 60 * 60 * 1000;
function diffDays(startDate, endDate) {
    return Math.floor(treatAsUTC(endDate) / millisecondsPerDay) - Math.floor(treatAsUTC(startDate) / millisecondsPerDay);
}

Answer 13

var start= $("#firstDate").datepicker("getDate");
var end= $("#SecondDate").datepicker("getDate");
var days = (end- start) / (1000 * 60 * 60 * 24);
 alert(Math.round(days));

jsfiddle example :)

Answer 14

This may not be the most elegant solution, but it seems to answer the question with a relatively simple bit of code, I think. Can't you use something like this:

function dayDiff(startdate, enddate) {
  var dayCount = 0;

  while(enddate >= startdate) {
    dayCount++;
    startdate.setDate(startdate.getDate() + 1);
  }

return dayCount; 
}

This is assuming you are passing date objects as parameters.

Answer 15

I think the solutions aren't correct 100% I would use ceil instead of floor, round will work but it isn't the right operation.

function dateDiff(str1, str2){
    var diff = Date.parse(str2) - Date.parse(str1); 
    return isNaN(diff) ? NaN : {
        diff: diff,
        ms: Math.ceil(diff % 1000),
        s: Math.ceil(diff / 1000 % 60),
        m: Math.ceil(diff / 60000 % 60),
        h: Math.ceil(diff / 3600000 % 24),
        d: Math.ceil(diff / 86400000)
    };
}

Answer 16

What about using formatDate from DatePicker widget? You could use it to convert the dates in timestamp format (milliseconds since 01/01/1970) and then do a simple subtraction.

Answer 17

function timeDifference(date1, date2) {
  var oneDay = 24 * 60 * 60; // hours*minutes*seconds
  var oneHour = 60 * 60; // minutes*seconds
  var oneMinute = 60; // 60 seconds
  var firstDate = date1.getTime(); // convert to milliseconds
  var secondDate = date2.getTime(); // convert to milliseconds
  var seconds = Math.round(Math.abs(firstDate - secondDate) / 1000); //calculate the diffrence in seconds
  // the difference object
  var difference = {
    "days": 0,
    "hours": 0,
    "minutes": 0,
    "seconds": 0,
  }
  //calculate all the days and substract it from the total
  while (seconds >= oneDay) {
    difference.days++;
    seconds -= oneDay;
  }
  //calculate all the remaining hours then substract it from the total
  while (seconds >= oneHour) {
    difference.hours++;
    seconds -= oneHour;
  }
  //calculate all the remaining minutes then substract it from the total 
  while (seconds >= oneMinute) {
    difference.minutes++;
    seconds -= oneMinute;
  }
  //the remaining seconds :
  difference.seconds = seconds;
  //return the difference object
  return difference;
}
console.log(timeDifference(new Date(2017,0,1,0,0,0),new Date()));

Answer 18

Simple, easy, and sophisticated. This function will be called in every 1 sec to update time.

const year = (new Date().getFullYear());
const bdayDate = new Date("04,11,2019").getTime(); //mmddyyyy

// countdown
let timer = setInterval(function () {

    // get today's date
    const today = new Date().getTime();

    // get the difference
    const diff = bdayDate - today;

    // math
    let days = Math.floor(diff / (1000 * 60 * 60 * 24));
    let hours = Math.floor((diff % (1000 * 60 * 60 * 24)) / (1000 * 60 * 60));
    let minutes = Math.floor((diff % (1000 * 60 * 60)) / (1000 * 60));
    let seconds = Math.floor((diff % (1000 * 60)) / 1000);

}, 1000);

Answer 19

If you have two unix timestamps, you can use this function (made a little more verbose for the sake of clarity):

// Calculate number of days between two unix timestamps
// ------------------------------------------------------------
var daysBetween = function(timeStampA, timeStampB) {
    var oneDay = 24 * 60 * 60 * 1000; // hours * minutes * seconds * milliseconds
    var firstDate = new Date(timeStampA * 1000);
    var secondDate = new Date(timeStampB * 1000);
    var diffDays = Math.round(Math.abs((firstDate.getTime() - secondDate.getTime())/(oneDay)));
    return diffDays;
};

Example:

daysBetween(1096580303, 1308713220); // 2455

Answer 20

Be careful when using milliseconds.

The date.getTime() returns milliseconds and doing math operation with milliseconds requires to include

• Daylight Saving Time (DST)
• checking if both dates have the same time (hours, minutes, seconds, milliseconds)
• make sure what behavior of days diff is required: 19 September 2016 - 29 September 2016 = 1 or 2 days difference?

The example from comment above is the best solution I found so far https://stackoverflow.com/a/11252167/2091095 . But use +1 to its result if you want the to count all days involved.

function treatAsUTC(date) {
    var result = new Date(date);
    result.setMinutes(result.getMinutes() - result.getTimezoneOffset());
    return result;
}

function daysBetween(startDate, endDate) {
    var millisecondsPerDay = 24 * 60 * 60 * 1000;
    return (treatAsUTC(endDate) - treatAsUTC(startDate)) / millisecondsPerDay;
}

var diff = daysBetween($('#first').val(), $('#second').val()) + 1;

Answer 21

Date.prototype.days = function(to) {
  return Math.abs(Math.floor(to.getTime() / (3600 * 24 * 1000)) - Math.floor(this.getTime() / (3600 * 24 * 1000)))
}


console.log(new Date('2014/05/20').days(new Date('2014/05/23'))); // 3 days

console.log(new Date('2014/05/23').days(new Date('2014/05/20'))); // 3 days

Answer 22

I had the same issue in Angular. I do the copy because else he will overwrite the first date. Both dates must have time 00:00:00 (obviously)

 /*
* Deze functie gebruiken we om het aantal dagen te bereken van een booking.
* */
$scope.berekenDagen = function ()
{
    $scope.booking.aantalDagen=0;

    /*De loper is gelijk aan de startdag van je reservatie.
     * De copy is nodig anders overschijft angular de booking.van.
     * */
    var loper = angular.copy($scope.booking.van);

    /*Zolang de reservatie beschikbaar is, doorloop de weekdagen van je start tot einddatum.*/
    while (loper < $scope.booking.tot) {
        /*Tel een dag op bij je loper.*/
        loper.setDate(loper.getDate() + 1);
        $scope.booking.aantalDagen++;
    }

    /*Start datum telt natuurlijk ook mee*/
    $scope.booking.aantalDagen++;
    $scope.infomsg +=" aantal dagen: "+$scope.booking.aantalDagen;
};

Answer 23

I used below code to experiment the posting date functionality for a news post.I calculate the minute or hour or day or year based on the posting date and current date.

var startDate= new Date("Mon Jan 01 2007 11:00:00");
var endDate  =new Date("Tue Jan 02 2007 12:50:00");
var timeStart = startDate.getTime();
var timeEnd = endDate.getTime();
var yearStart = startDate.getFullYear();
var yearEnd   = endDate.getFullYear();
if(yearStart == yearEnd)
 {
  var hourDiff = timeEnd - timeStart; 
  var secDiff = hourDiff / 1000;
  var minDiff = hourDiff / 60 / 1000; 
  var hDiff = hourDiff / 3600 / 1000; 
  var myObj = {};
  myObj.hours = Math.floor(hDiff);
  myObj.minutes = minDiff  
  if(myObj.hours >= 24)
   {
    console.log(Math.floor(myObj.hours/24) + "day(s) ago")
   } 
 else if(myObj.hours>0)
  {
   console.log(myObj.hours +"hour(s) ago")
  }
 else
  {
   console.log(Math.abs(myObj.minutes) +"minute(s) ago")
  }
}
else
{
var yearDiff = yearEnd - yearStart;
console.log( yearDiff +" year(s) ago");
}

Answer 24

if you wanna have an DateArray with dates try this:

<script>
        function getDates(startDate, stopDate) {
        var dateArray = new Array();
        var currentDate = moment(startDate);
        dateArray.push( moment(currentDate).format('L'));

        var stopDate = moment(stopDate);
        while (dateArray[dateArray.length -1] != stopDate._i) {
            dateArray.push( moment(currentDate).format('L'));
            currentDate = moment(currentDate).add(1, 'days');
        }
        return dateArray;
      }
</script>

DebugSnippet

Answer 25

The simple way to calculate days between two dates is to remove both of their time component i.e. setting hours, minutes, seconds and milliseconds to 0 and then subtracting their time and diving it with milliseconds worth of one day.

var firstDate= new Date(firstDate.setHours(0,0,0,0));
var secondDate= new Date(secondDate.setHours(0,0,0,0));
var timeDiff = firstDate.getTime() - secondDate.getTime();
var diffDays =timeDiff / (1000 * 3600 * 24);

Answer 26

function formatDate(seconds, dictionary) {
    var foo = new Date;
    var unixtime_ms = foo.getTime();
    var unixtime = parseInt(unixtime_ms / 1000);
    var diff = unixtime - seconds;
    var display_date;
    if (diff <= 0) {
        display_date = dictionary.now;
    } else if (diff < 60) {
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.second;
        } else {
            display_date = diff + ' ' + dictionary.seconds;
        }
    } else if (diff < 3540) {
        diff = Math.round(diff / 60);
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.minute;
        } else {
            display_date = diff + ' ' + dictionary.minutes;
        }
    } else if (diff < 82800) {
        diff = Math.round(diff / 3600);
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.hour;
        } else {
            display_date = diff + ' ' + dictionary.hours;
        }
    } else {
        diff = Math.round(diff / 86400);
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.day;
        } else {
            display_date = diff + ' ' + dictionary.days;
        }
    }
    return display_date;
}

Answer 27

I took some inspiration from other answers and made the inputs have automatic sanitation. I hope this works well as an improvement over other answers.

//use best practices by labeling your constants.
let MS_PER_SEC = 1000
  , SEC_PER_HR = 60 * 60
  , HR_PER_DAY = 24
  , MS_PER_DAY = MS_PER_SEC * SEC_PER_HR * HR_PER_DAY
;

//let's assume we get Date objects as arguments, otherwise return 0.
function dateDiffInDays(date1, date2) {
    if (!date1 || !date2) {
      return 0;
    }
    return Math.round((date2.getTime() - date1.getTime()) / MS_PER_DAY);
}

// new Date("dateString") is browser-dependent and discouraged, so we'll write
// a simple parse function for U.S. date format. (by @Miles)
function parseDate(str) {
    if (str && str.length > 7 && str.length < 11) {
      let mdy = str.split('/');
      return new Date(mdy[2], mdy[0]-1, mdy[1]);
    }
    return null;
}

function calcInputs() {
  let date1 = document.getElementById("date1")
    , date2 = document.getElementById("date2")
    , resultSpan = document.getElementById("result")
  ;
  if (date1 && date2 && resultSpan) {
    //remove non-date characters
    let date1Val = date1.value.replace(/[^\d\/]/g,'')
      , date2Val = date2.value.replace(/[^\d\/]/g,'')
      , result = dateDiffInDays(parseDate(date1Val), parseDate(date2Val))
    ;
    date1.value = date1Val;
    date2.value = date2Val;
    resultSpan.innerHTML = result + " days";
  }
}
window.onload = function() { calcInputs(); };

//some code examples
console.log(dateDiffInDays(parseDate("1/15/2019"), parseDate("1/30/2019")));
console.log(dateDiffInDays(parseDate("1/15/2019"), parseDate("2/30/2019")));
console.log(dateDiffInDays(parseDate("1/15/2000"), parseDate("1/15/2019")));

<input id="date1" type="text" value="1/1/2000" size="6" onkeyup="calcInputs();" />
<input id="date2" type="text" value="1/1/2019" size="6" onkeyup="calcInputs();"/>
Result: <span id="result"></span>

Answer 28

I recently had the same question, and coming from a Java world, I immediately started to search for a JSR 310 implementation for JavaScript. JSR 310 is a Date and Time API for Java (standard shipped as of Java 8). I think the API is very well designed.

Fortunately, there is a direct port to Javascript, called js-joda.

First, include js-joda in the <head>:

<script
    src="https://cdnjs.cloudflare.com/ajax/libs/js-joda/1.11.0/js-joda.min.js"
    integrity="sha512-piLlO+P2f15QHjUv0DEXBd4HvkL03Orhi30Ur5n1E4Gk2LE4BxiBAP/AD+dxhxpW66DiMY2wZqQWHAuS53RFDg=="
    crossorigin="anonymous"></script>

Then simply do this:

let date1 = JSJoda.LocalDate.of(2020, 12, 1);
let date2 = JSJoda.LocalDate.of(2021, 1, 1);
let daysBetween = JSJoda.ChronoUnit.DAYS.between(date1, date2);

Now daysBetween contains the number of days between. Note that the end date is exclusive.

Answer 29

A Better Solution by

Ignoring time part

it will return 0 if both the dates are same.

function dayDiff(firstDate, secondDate) {
  firstDate = new Date(firstDate);
  secondDate = new Date(secondDate);
  if (!isNaN(firstDate) && !isNaN(secondDate)) {
    firstDate.setHours(0, 0, 0, 0); //ignore time part
    secondDate.setHours(0, 0, 0, 0); //ignore time part
    var dayDiff = secondDate - firstDate;
    dayDiff = dayDiff / 86400000; // divide by milisec in one day
    console.log(dayDiff);
  } else {
    console.log("Enter valid date.");
  }
}

$(document).ready(function() {
  $('input[type=datetime]').datepicker({
    dateFormat: "mm/dd/yy",
    changeMonth: true,
    changeYear: true
  });
  $("#button").click(function() {
    dayDiff($('#first').val(), $('#second').val());
  });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<link rel="stylesheet" href="//code.jquery.com/ui/1.12.1/themes/base/jquery-ui.css">
<script src="//code.jquery.com/ui/1.12.1/jquery-ui.js"></script>

<input type="datetime" id="first" value="12/28/2016" />
<input type="datetime" id="second" value="12/28/2017" />
<input type="button" id="button" value="Calculate">

Answer 30

A contribution, for date before 1970-01-01 and after 2038-01-19

function DateDiff(aDate1, aDate2) {
  let dDay = 0;
  this.isBissexto = (aYear) => {
    return (aYear % 4 == 0 && aYear % 100 != 0) || (aYear % 400 == 0);
  };
  this.getDayOfYear = (aDate) => {
    let count = 0;
    for (let m = 0; m < aDate.getUTCMonth(); m++) {
      count += m == 1 ? this.isBissexto(aDate.getUTCFullYear()) ? 29 : 28 : /(3|5|8|10)/.test(m) ? 30 : 31;
    }
    count += aDate.getUTCDate();
    return count;
  };
  this.toDays = () => {
    return dDay;
  };
  (() => {
    let startDate = aDate1.getTime() <= aDate2.getTime() ? new Date(aDate1.toISOString()) : new Date(aDate2.toISOString());
    let endDate = aDate1.getTime() <= aDate2.getTime() ? new Date(aDate2.toISOString()) : new Date(aDate1.toISOString());
    while (startDate.getUTCFullYear() != endDate.getUTCFullYear()) {
      dDay += (this.isBissexto(startDate.getFullYear())? 366 : 365) - this.getDayOfYear(startDate) + 1;
      startDate = new Date(startDate.getUTCFullYear()+1, 0, 1);
    }
    dDay += this.getDayOfYear(endDate) - this.getDayOfYear(startDate);
  })();
}