I am calculating the number of days between the 'from' and 'to' date. For example, if the from date is 13/04/2010 and the to date is 15/04/2010 the result should be
How do I get the result using JavaScript?

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7 Answers
const oneDay = 24 * 60 * 60 * 1000; // hours*minutes*seconds*milliseconds
const firstDate = new Date(2008, 1, 12);
const secondDate = new Date(2008, 1, 22);
const diffDays = Math.round(Math.abs((firstDate - secondDate) / oneDay));

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84Warning: not all days are 24 hours long. If your date range spans a daylight saving change, you'll lose or gain an hour (typically). Use Math.round() on the result (avoid floor or ceil). – Mark Sep 20 '11 at 02:33
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12In fact i'd preffer Math.ceil here since even if 2.01 days are left saying 3 days left makes more sense that sayin 2 days left. – 5hahiL Nov 17 '12 at 10:20
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20@Mark 's comment about using Math.round was added to code of the answer. Don't round the result again, like someone I know... (ok it was me) – Aardvark May 14 '14 at 20:38
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Why doesn't it count it as 5 days when it's between June 1st to June 5th? – adamj Jun 09 '14 at 09:42
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What about leap years? – Kumait Jul 10 '15 at 09:09
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@Kumait What about them? – MaxVT Jul 10 '15 at 13:28
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My fault, this must be handled internally in the Date object – Kumait Jul 13 '15 at 02:13
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7I tested this with 2015-02-1 to 2015-03-01 and got 31 days. There's something wrong here... EDIT: Okay yeah.. javascript months are from 0-12 so if you're getting the input from a jquery datepicker or a regular normal date, subtract 1 form the month – Robert Mennell Oct 13 '15 at 23:07
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19Note that this code still takes into account the TIME of the given dates when counting. i.e. checking the number of days between 3PM on the 1st and 00:00:00 on the 2nd will yield zero days. To fix this, set both dates to midnight before comparison, i.e. add: firstDate.setHours(0, 0, 0); secondDate.setHours(0, 0, 0); This way you can say daysBetween(new Date(2016,1,10,15), new Date(2016,1,11))); and still come up with 1 day difference. – Howard Feb 10 '16 at 13:37
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2Nice, it threw me for a moment that JavaScript has January as 0, so I had put a 1 thinking that was January, but it was actually February, throwing off my calculation. Thanks for this result. – Chris Hawkes Mar 28 '16 at 00:28
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When I copied this code to my console and after hitting return, I get undefined in my console. Any reason? – Ilyas karim Oct 05 '17 at 15:45
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@Ilyaskarim if you're using a console, you probably want to see the resulting difference. typing `diffDays` produces the expected result. – MaxVT Oct 19 '17 at 20:55
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I was having an off by one error depending on the time of day, @Howard 's suggestion fixed my problem! – reggaeguitar Feb 16 '18 at 17:03
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To prevent red squiggly lines on the last line of the above code, replace the last line with the following code: `const diffDays = Math.round(Math.abs((firstDate.valueOf() - secondDate.valueOf()) / oneDay));` – Devner Jan 16 '22 at 21:27
Here is a function that does this:
function days_between(date1, date2) {
// The number of milliseconds in one day
const ONE_DAY = 1000 * 60 * 60 * 24;
// Calculate the difference in milliseconds
const differenceMs = Math.abs(date1 - date2);
// Convert back to days and return
return Math.round(differenceMs / ONE_DAY);
}

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12Avoid floor because it will give the wrong result if a day in the range is shorter due to daylight saving. – Mark Sep 20 '11 at 02:38
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why alway return Thu Jan 01 1970 08:00:00 GMT+0800 (Indochina Time) ?? – huykon225 Jun 20 '19 at 08:54
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Better make it `Date.UTC(date1.getFullYear(), date1.getMonth(), date1.getDate())` for both dates and use `Math.floor`? – naveen Apr 12 '22 at 15:16
Here's what I use. If you just subtract the dates, it won't work across the Daylight Savings Time Boundary (eg April 1 to April 30 or Oct 1 to Oct 31). This drops all the hours to make sure you get a day and eliminates any DST problem by using UTC.
var nDays = ( Date.UTC(EndDate.getFullYear(), EndDate.getMonth(), EndDate.getDate()) -
Date.UTC(StartDate.getFullYear(), StartDate.getMonth(), StartDate.getDate())) / 86400000;
as a function:
function DaysBetween(StartDate, EndDate) {
// The number of milliseconds in all UTC days (no DST)
const oneDay = 1000 * 60 * 60 * 24;
// A day in UTC always lasts 24 hours (unlike in other time formats)
const start = Date.UTC(EndDate.getFullYear(), EndDate.getMonth(), EndDate.getDate());
const end = Date.UTC(StartDate.getFullYear(), StartDate.getMonth(), StartDate.getDate());
// so it's safe to divide by 24 hours
return (start - end) / oneDay;
}
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2@RuudLenders because day in UTC always lasts 24 hours, unlike in other time formats, so it's safe do divide by 24 hours – Dan Apr 26 '17 at 18:56
Here is my implementation:
function daysBetween(one, another) {
return Math.round(Math.abs((+one) - (+another))/8.64e7);
}
+<date>
does the type coercion to the integer representation and has the same effect as <date>.getTime()
and 8.64e7
is the number of milliseconds in a day.

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5The type coercion is redundant here because you already use a arithmetic operator to substract the dates (which also results in a type coercion). So you can write it like this: `Math.round(Math.abs(one - another) / 8.64e7);` – A1rPun Aug 25 '15 at 14:23
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4
Adjusted to allow for daylight saving differences. try this:
function daysBetween(date1, date2) {
// adjust diff for for daylight savings
var hoursToAdjust = Math.abs(date1.getTimezoneOffset() /60) - Math.abs(date2.getTimezoneOffset() /60);
// apply the tz offset
date2.addHours(hoursToAdjust);
// The number of milliseconds in one day
var ONE_DAY = 1000 * 60 * 60 * 24
// Convert both dates to milliseconds
var date1_ms = date1.getTime()
var date2_ms = date2.getTime()
// Calculate the difference in milliseconds
var difference_ms = Math.abs(date1_ms - date2_ms)
// Convert back to days and return
return Math.round(difference_ms/ONE_DAY)
}
// you'll want this addHours function too
Date.prototype.addHours= function(h){
this.setHours(this.getHours()+h);
return this;
}

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2There is no need to adjust for the timezone offset, the time value is in UTC. If there is a need for timezone adjustment, it can be applied directly to the minutes using `setMinutes()` rather than converting to hours and using `setHours()`. The arguments to the `set` methods should be integers. If using `setHours` and the offset is not an even multiple of hours, the value will be truncated. – RobG Nov 23 '12 at 22:27
I have written this solution for another post who asked, how to calculate the difference between two dates, so I share what I have prepared:
// Here are the two dates to compare
var date1 = '2011-12-24';
var date2 = '2012-01-01';
// First we split the values to arrays date1[0] is the year, [1] the month and [2] the day
date1 = date1.split('-');
date2 = date2.split('-');
// Now we convert the array to a Date object, which has several helpful methods
date1 = new Date(date1[0], date1[1], date1[2]);
date2 = new Date(date2[0], date2[1], date2[2]);
// We use the getTime() method and get the unixtime (in milliseconds, but we want seconds, therefore we divide it through 1000)
date1_unixtime = parseInt(date1.getTime() / 1000);
date2_unixtime = parseInt(date2.getTime() / 1000);
// This is the calculated difference in seconds
var timeDifference = date2_unixtime - date1_unixtime;
// in Hours
var timeDifferenceInHours = timeDifference / 60 / 60;
// and finaly, in days :)
var timeDifferenceInDays = timeDifferenceInHours / 24;
alert(timeDifferenceInDays);
You can skip some steps in the code, I have written it so to make it easy to understand.
You'll find a running example here: http://jsfiddle.net/matKX/

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3Note that months are zero indexed, so you should have `new Date(date1[0], --date1[1], date1[2]);`. Also, using `parseInt` will truncate the milliseconds. Simpler to just subtract the date objects and convert the difference to days: `Math.round((date1 - date2) / 8.64e7)` or if whole days are required with no rounding, just truncate the decimal part: `(date1 - date2) / 8.64e7 | 0`. – RobG Nov 23 '12 at 22:20
From my little date difference calculator:
var startDate = new Date(2000, 1-1, 1); // 2000-01-01
var endDate = new Date(); // Today
// Calculate the difference of two dates in total days
function diffDays(d1, d2)
{
var ndays;
var tv1 = d1.valueOf(); // msec since 1970
var tv2 = d2.valueOf();
ndays = (tv2 - tv1) / 1000 / 86400;
ndays = Math.round(ndays - 0.5);
return ndays;
}
So you would call:
var nDays = diffDays(startDate, endDate);
(Full source at http://david.tribble.com/src/javascript/jstimespan.html.)
Addendum
The code can be improved by changing these lines:
var tv1 = d1.getTime(); // msec since 1970
var tv2 = d2.getTime();

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I had to take out the - 0.5 from the round function for it to work properly for some reason, I'm thinking it was because of the minus 1 for the month since its 0 indexed – Yohn May 19 '15 at 15:35