Difference in Months between two dates in JavaScri

Question

How would I work out the difference for two Date() objects in JavaScript, while only return the number of months in the difference?

Any help would be great :)

Answer 1

Here's a function that accurately provides the number of months between 2 dates.
The default behavior only counts whole months, e.g. 3 months and 1 day will result in a difference of 3 months. You can prevent this by setting the roundUpFractionalMonths param as true, so a 3 month and 1 day difference will be returned as 4 months.

The accepted answer above (T.J. Crowder's answer) isn't accurate, it returns wrong values sometimes.

For example, monthDiff(new Date('Jul 01, 2015'), new Date('Aug 05, 2015')) returns 0 which is obviously wrong. The correct difference is either 1 whole month or 2 months rounded-up.

Here's the function I wrote:

function getMonthsBetween(date1,date2,roundUpFractionalMonths)
{
    //Months will be calculated between start and end dates.
    //Make sure start date is less than end date.
    //But remember if the difference should be negative.
    var startDate=date1;
    var endDate=date2;
    var inverse=false;
    if(date1>date2)
    {
        startDate=date2;
        endDate=date1;
        inverse=true;
    }

    //Calculate the differences between the start and end dates
    var yearsDifference=endDate.getFullYear()-startDate.getFullYear();
    var monthsDifference=endDate.getMonth()-startDate.getMonth();
    var daysDifference=endDate.getDate()-startDate.getDate();

    var monthCorrection=0;
    //If roundUpFractionalMonths is true, check if an extra month needs to be added from rounding up.
    //The difference is done by ceiling (round up), e.g. 3 months and 1 day will be 4 months.
    if(roundUpFractionalMonths===true && daysDifference>0)
    {
        monthCorrection=1;
    }
    //If the day difference between the 2 months is negative, the last month is not a whole month.
    else if(roundUpFractionalMonths!==true && daysDifference<0)
    {
        monthCorrection=-1;
    }

    return (inverse?-1:1)*(yearsDifference*12+monthsDifference+monthCorrection);
};

Answer 2

The definition of "the number of months in the difference" is subject to a lot of interpretation. :-)

You can get the year, month, and day of month from a JavaScript date object. Depending on what information you're looking for, you can use those to figure out how many months are between two points in time.

For instance, off-the-cuff, this finds out how many full months lie between two dates, not counting partial months (e.g., excluding the month each date is in):

function monthDiff(d1, d2) {
    var months;
    months = (d2.getFullYear() - d1.getFullYear()) * 12;
    months -= d1.getMonth() + 1;
    months += d2.getMonth();
    return months <= 0 ? 0 : months;
}

monthDiff(
    new Date(2008, 10, 4), // November 4th, 2008
    new Date(2010, 2, 12)  // March 12th, 2010
);
// Result: 15: December 2008, all of 2009, and Jan & Feb 2010

monthDiff(
    new Date(2010, 0, 1),  // January 1st, 2010
    new Date(2010, 2, 12)  // March 12th, 2010
);
// Result: 1: February 2010 is the only full month between them

monthDiff(
    new Date(2010, 1, 1),  // February 1st, 2010
    new Date(2010, 2, 12)  // March 12th, 2010
);
// Result: 0: There are no *full* months between them

(Note that month values in JavaScript start with 0 = January.)

Including fractional months in the above is much more complicated, because three days in a typical February is a larger fraction of that month (~10.714%) than three days in August (~9.677%), and of course even February is a moving target depending on whether it's a leap year.

There are also some date and time libraries available for JavaScript that probably make this sort of thing easier.

Answer 3

If you need to count full months, regardless of the month being 28, 29, 30 or 31 days. Below should work.

var months = to.getMonth() - from.getMonth() 
    + (12 * (to.getFullYear() - from.getFullYear()));

if(to.getDate() < from.getDate()){
    months--;
}
return months;

This is an extended version of the answer https://stackoverflow.com/a/4312956/1987208 but fixes the case where it calculates 1 month for the case from 31st of January to 1st of February (1day).

This will cover the following;

• 1st Jan to 31st Jan ---> 30days ---> will result in 0 (logical since it is not a full month)
• 1st Feb to 1st Mar ---> 28 or 29 days ---> will result in 1 (logical since it is a full month)
• 15th Feb to 15th Mar ---> 28 or 29 days ---> will result in 1 (logical since a month passed)
• 31st Jan to 1st Feb ---> 1 day ---> will result in 0 (obvious but the mentioned answer in the post results in 1 month)

Answer 4

To expand on @T.J.'s answer, if you're looking for simple months, rather than full calendar months, you could just check if d2's date is greater than or equal to than d1's. That is, if d2 is later in its month than d1 is in its month, then there is 1 more month. So you should be able to just do this:

function monthDiff(d1, d2) {
    var months;
    months = (d2.getFullYear() - d1.getFullYear()) * 12;
    months -= d1.getMonth() + 1;
    months += d2.getMonth();
    // edit: increment months if d2 comes later in its month than d1 in its month
    if (d2.getDate() >= d1.getDate())
        months++
    // end edit
    return months <= 0 ? 0 : months;
}

monthDiff(
    new Date(2008, 10, 4), // November 4th, 2008
    new Date(2010, 2, 12)  // March 12th, 2010
);
// Result: 16; 4 Nov – 4 Dec '08, 4 Dec '08 – 4 Dec '09, 4 Dec '09 – 4 March '10

This doesn't totally account for time issues (e.g. 3 March at 4:00pm and 3 April at 3:00pm), but it's more accurate and for just a couple lines of code.