Difference between two dates in years, months, day

Question

I've been searching for 4 hours now, and have not found a solution to get the difference between two dates in years, months, and days in JavaScript, like: 10th of April 2010 was 3 years, x month and y days ago.

There are lots of solutions, but they only offer the difference in the format of either days OR months OR years, or they are not correct (meaning not taking care of actual number of days in a month or leap years, etc). Is it really that difficult to do that?

I've had a look at:

• http://momentjs.com/ -> can only output the difference in either years, months, OR days
• http://www.javascriptkit.com/javatutors/datedifference.shtml
• http://www.javascriptkit.com/jsref/date.shtml
• http://timeago.yarp.com/
• www.stackoverflow.com -> Search function

In php it is easy, but unfortunately I can only use client-side script on that project. Any library or framework that can do it would be fine, too.

Here are a list of expected outputs for date differences:

//Expected output should be: "1 year, 5 months".
diffDate(new Date('2014-05-10'), new Date('2015-10-10'));

//Expected output should be: "1 year, 4 months, 29 days".
diffDate(new Date('2014-05-10'), new Date('2015-10-09'));

//Expected output should be: "1 year, 3 months, 30 days".
diffDate(new Date('2014-05-10'), new Date('2015-09-09'));

//Expected output should be: "9 months, 27 days".
diffDate(new Date('2014-05-10'), new Date('2015-03-09'));

//Expected output should be: "1 year, 9 months, 28 days".
diffDate(new Date('2014-05-10'), new Date('2016-03-09'));

//Expected output should be: "1 year, 10 months, 1 days".
diffDate(new Date('2014-05-10'), new Date('2016-03-11'));

Answer 1

I think you are looking for the same thing that I wanted. I tried to do this using the difference in milliseconds that javascript provides, but those results do not work in the real world of dates. If you want the difference between Feb 1, 2016 and January 31, 2017 the result I would want is 1 year, 0 months, and 0 days. Exactly one year (assuming you count the last day as a full day, like in a lease for an apartment). However, the millisecond approach would give you 1 year 0 months and 1 day, since the date range includes a leap year. So here is the code I used in javascript for my adobe form (you can name the fields): (edited, there was an error that I corrected)

var f1 = this.getField("LeaseExpiration");
var g1 = this.getField("LeaseStart");


var end = f1.value
var begin = g1.value
var e = new Date(end);
var b = new Date(begin);
var bMonth = b.getMonth();
var bYear = b.getFullYear();
var eYear = e.getFullYear();
var eMonth = e.getMonth();
var bDay = b.getDate();
var eDay = e.getDate() + 1;

if ((eMonth == 0)||(eMonth == 2)||(eMonth == 4)|| (eMonth == 6) || (eMonth == 7) ||(eMonth == 9)||(eMonth == 11))

{
var eDays =  31;
}

if ((eMonth == 3)||(eMonth == 5)||(eMonth == 8)|| (eMonth == 10))

{
var eDays = 30;
}

if (eMonth == 1&&((eYear % 4 == 0) && (eYear % 100 != 0)) || (eYear % 400 == 0))
{
var eDays = 29;
}

if (eMonth == 1&&((eYear % 4 != 0) || (eYear % 100 == 0)))
{
var eDays = 28;
}


if ((bMonth == 0)||(bMonth == 2)||(bMonth == 4)|| (bMonth == 6) || (bMonth == 7) ||(bMonth == 9)||(bMonth == 11))

{
var bDays =  31;
}

if ((bMonth == 3)||(bMonth == 5)||(bMonth == 8)|| (bMonth == 10))

{
var bDays = 30;
}

if (bMonth == 1&&((bYear % 4 == 0) && (bYear % 100 != 0)) || (bYear % 400 == 0))
{
var bDays = 29;
}

if (bMonth == 1&&((bYear % 4 != 0) || (bYear % 100 == 0)))
{
var bDays = 28;
}


var FirstMonthDiff = bDays - bDay + 1;


if (eDay - bDay < 0)
{

eMonth = eMonth - 1;
eDay = eDay + eDays;

}

var daysDiff = eDay - bDay;

if(eMonth - bMonth < 0)
{
eYear = eYear - 1;
eMonth = eMonth + 12;
}

var monthDiff = eMonth - bMonth;

var yearDiff = eYear - bYear;

if (daysDiff == eDays)
{
daysDiff = 0;
monthDiff = monthDiff + 1;

if (monthDiff == 12)
{
monthDiff = 0;
yearDiff = yearDiff + 1;
}

}

if ((FirstMonthDiff != bDays)&&(eDay - 1 == eDays))

{
daysDiff = FirstMonthDiff;

}
event.value = yearDiff + " Year(s)" + " " + monthDiff + " month(s) " + daysDiff + " days(s)"

Answer 2

The following is an algorithm which gives correct but not totally precise since it does not take into account leap year. It also assumes 30 days in a month. A good usage for example is if someone lives in an address from 12/11/2010 to 11/10/2011, it can quickly tells that the person lives there for 10 months and 29 days. From 12/11/2010 to 11/12/2011 is 11 months and 1 day. For certain types of applications, that kind of precision is sufficient. This is for those types of applications because it aims for simplicity:

var datediff = function(start, end) {
  var diff = { years: 0, months: 0, days: 0 };
  var timeDiff = end - start;

  if (timeDiff > 0) {
    diff.years = end.getFullYear() - start.getFullYear();
    diff.months = end.getMonth() - start.getMonth();
    diff.days = end.getDate() - start.getDate();

    if (diff.months < 0) {
      diff.years--;
      diff.months += 12;
    }

    if (diff.days < 0) {
      diff.months = Math.max(0, diff.months - 1);
      diff.days += 30;
    }
  }

  return diff;
};

Unit tests

Answer 3

Neither of the codes work for me, so I use this instead for months and days:

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

function daysInMonth(date) {
    return new Date(date.getYear(), date.getMonth() + 1, 0).getDate();
}    

function diffDate(date1, date2) {
    if (date2 && date2.getTime() && !isNaN(date2.getTime())) {
        var months = monthDiff(date1, date2);
        var days = 0;

        if (date1.getUTCDate() >= date2.getUTCDate()) {
            days = date1.getUTCDate() - date2.getUTCDate();
        }
        else {
            months--;
            days = date1.getUTCDate() - date2.getUTCDate() + daysInMonth(date2);
        }

        // Use the variables months and days how you need them.
    }
}

Answer 4

I know it is an old thread, but I'd like to put my 2 cents based on the answer by @Pawel Miech.

It is true that you need to convert the difference into milliseconds, then you need to make some math. But notice that, you need to do the math in backward manner, i.e. you need to calculate years, months, days, hours then minutes.

I used to do some thing like this:

    var mins;
    var hours;
    var days;
    var months;
    var years;

    var diff = new Date() - new Date(yourOldDate);  
// yourOldDate may be is coming from DB, for example, but it should be in the correct format ("MM/dd/yyyy hh:mm:ss:fff tt")

    years = Math.floor((diff) / (1000 * 60 * 60 * 24 * 365));
    diff = Math.floor((diff) % (1000 * 60 * 60 * 24 * 365));
    months = Math.floor((diff) / (1000 * 60 * 60 * 24 * 30));
    diff = Math.floor((diff) % (1000 * 60 * 60 * 24 * 30));
    days = Math.floor((diff) / (1000 * 60 * 60 * 24));
    diff = Math.floor((diff) % (1000 * 60 * 60 * 24));
    hours = Math.floor((diff) / (1000 * 60 * 60));
    diff = Math.floor((diff) % (1000 * 60 * 60));
    mins = Math.floor((diff) / (1000 * 60));

But, of course, this is not precise because it assumes that all years have 365 days and all months have 30 days, which is not true in all cases.

Answer 5

This link has the best answer http://forums.asp.net/t/1610039.aspx?How+to+calculate+difference+between+two+dates+in+years

You only need to add a validation for the days, somthing like:

if ( firstDate.getDate() <= now.getDate() )

Answer 6

Time span in full Days, Hours, Minutes, Seconds, Milliseconds:

// Extension for Date
Date.difference = function (dateFrom, dateTo) {
  var diff = { TotalMs: dateTo - dateFrom };
  diff.Days = Math.floor(diff.TotalMs / 86400000);

  var remHrs = diff.TotalMs % 86400000;
  var remMin = remHrs % 3600000;
  var remS   = remMin % 60000;

  diff.Hours        = Math.floor(remHrs / 3600000);
  diff.Minutes      = Math.floor(remMin / 60000);
  diff.Seconds      = Math.floor(remS   / 1000);
  diff.Milliseconds = Math.floor(remS % 1000);
  return diff;
};

// Usage
var a = new Date(2014, 05, 12, 00, 5, 45, 30); //a: Thu Jun 12 2014 00:05:45 GMT+0400 
var b = new Date(2014, 02, 12, 00, 0, 25, 0);  //b: Wed Mar 12 2014 00:00:25 GMT+0400
var diff = Date.difference(b, a);
/* diff: {
  Days: 92
  Hours: 0
  Minutes: 5
  Seconds: 20
  Milliseconds: 30
  TotalMs: 7949120030
} */