Calculating number of years between 2 dates in PHP

Question

I need to get the number of years from 2 dates provided. Here's my code:

function daysDifference($endDate, $beginDate)
{
   $date_parts1=explode("-", $beginDate);
   $date_parts2=explode("-", $endDate);

   $start_date=gregoriantojd($date_parts1[1], $date_parts1[2], $date_parts1[0]);
   $end_date=gregoriantojd($date_parts2[1], $date_parts2[2], $date_parts2[0]);
   $diff = $end_date - $start_date;
   echo $diff;
   $years = floor($diff / (365.25*60*60*24));
   return $years;
}

echo daysDifference('2011-03-12','2008-03-09');

The $diff gives a number output. When I return $years, am getting 0. What have I done wrong?

Please search before posting. [How to calculate the difference between two dates using PHP?](http://stackoverflow.com/questions/676824/how-to-calculate-the-difference-between-two-dates-using-php) — Jon Gauthier, Mar 22 '11 at 05:23
You may want to clarify what you are looking for. Your function is named `daysDifference` and returns `$years` and uses `gregoriantojd`. — Matthew, Mar 22 '11 at 05:32

score 159 · Answer 1 · answered Mar 22 '11 at 05:29

159

$d1 = new DateTime('2011-03-12');
$d2 = new DateTime('2008-03-09');

$diff = $d2->diff($d1);

echo $diff->y;

answered Mar 22 '11 at 05:29

Matthew

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4

Yeah, this one should be the recommended one. – Aistis Jul 27 '17 at 08:08

score 11 · Answer 2 · answered Jun 22 '16 at 14:15

11

On a PHP 5.2 box (yeah really, they still exist) so without DateTime::diff() support I ended up using this:

$dateString='10-05-1975';
$years = round((time()-strtotime($dateString))/(3600*24*365.25))

answered Jun 22 '16 at 14:15

Nick de Kruijk

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score -1 · Answer 3 · answered Mar 22 '11 at 05:35

the $start_date and $end_date' value is the number of days, not seconds. so you should not divide the $diff with 365.25*60*60*24.

function daysDifference($endDate, $beginDate)
{

   $date_parts1 = explode("-", $beginDate);
   $date_parts2 = explode("-", $endDate);
   $start_date = gregoriantojd($date_parts1[1], $date_parts1[2], $date_parts1[0]);
   $end_date = gregoriantojd($date_parts2[1], $date_parts2[2], $date_parts2[0]);
   $diff = abs($end_date - $start_date);
   $years = floor($diff / 365.25);
   return $years;
}

echo daysDifference('2011-03-12','2008-03-09');

score -2 · Answer 4 · edited May 08 '14 at 06:30

Guys heres a good code i used to use.....its not my idea ....i just got it some where from the net...

It also have a variety of options hope this helps.........

    function datediff($interval, $datefrom, $dateto, $using_timestamps = false) {
    /*
    $interval can be:
    yyyy - Number of full years
    q - Number of full quarters
    m - Number of full months
    y - Difference between day numbers
        (eg 1st Jan 2004 is "1", the first day. 2nd Feb 2003 is "33". The datediff is "-32".)
    d - Number of full days
    w - Number of full weekdays
    ww - Number of full weeks
    h - Number of full hours
    n - Number of full minutes
    s - Number of full seconds (default)
    */

    if (!$using_timestamps) {
        $datefrom = strtotime($datefrom, 0);
        $dateto = strtotime($dateto, 0);
    }
    $difference = $dateto - $datefrom; // Difference in seconds

    switch($interval) {

    case 'yyyy': // Number of full years

        $years_difference = floor($difference / 31536000);
        if (mktime(date("H", $datefrom), date("i", $datefrom), date("s", $datefrom), date("n", $datefrom), date("j", $datefrom), date("Y", $datefrom)+$years_difference) > $dateto) {
            $years_difference--;
        }
        if (mktime(date("H", $dateto), date("i", $dateto), date("s", $dateto), date("n", $dateto), date("j", $dateto), date("Y", $dateto)-($years_difference+1)) > $datefrom) {
            $years_difference++;
        }
        $datediff = $years_difference;
        break;

    case "q": // Number of full quarters

        $quarters_difference = floor($difference / 8035200);
        while (mktime(date("H", $datefrom), date("i", $datefrom), date("s", $datefrom), date("n", $datefrom)+($quarters_difference*3), date("j", $dateto), date("Y", $datefrom)) < $dateto) {
            $months_difference++;
        }
        $quarters_difference--;
        $datediff = $quarters_difference;
        break;

    case "m": // Number of full months

        $months_difference = floor($difference / 2678400);
        while (mktime(date("H", $datefrom), date("i", $datefrom), date("s", $datefrom), date("n", $datefrom)+($months_difference), date("j", $dateto), date("Y", $datefrom)) < $dateto) {
            $months_difference++;
        }
        $months_difference--;
        $datediff = $months_difference;
        break;

    case 'y': // Difference between day numbers

        $datediff = date("z", $dateto) - date("z", $datefrom);
        break;

    case "d": // Number of full days

        $datediff = floor($difference / 86400);
        break;

    case "w": // Number of full weekdays

        $days_difference = floor($difference / 86400);
        $weeks_difference = floor($days_difference / 7); // Complete weeks
        $first_day = date("w", $datefrom);
        $days_remainder = floor($days_difference % 7);
        $odd_days = $first_day + $days_remainder; // Do we have a Saturday or Sunday in the remainder?
        if ($odd_days > 7) { // Sunday
            $days_remainder--;
        }
        if ($odd_days > 6) { // Saturday
            $days_remainder--;
        }
        $datediff = ($weeks_difference * 5) + $days_remainder;
        break;

    case "ww": // Number of full weeks

        $datediff = floor($difference / 604800);
        break;

    case "h": // Number of full hours

        $datediff = floor($difference / 3600);
        break;

    case "n": // Number of full minutes

        $datediff = floor($difference / 60);
        break;

    default: // Number of full seconds (default)

        $datediff = $difference;
        break;
    }    

    return $datediff;

} 

echo datediff('yyyy','2009-01-16','2011-03-16');

score -4 · Answer 5 · answered Mar 22 '11 at 05:24

-4

If you want the number of years between two dates, why not just use the following :

function yearsDifference($endDate, $beginDate)
{
   $date_parts1=explode("-", $beginDate);
   $date_parts2=explode("-", $endDate);
   return $date_parts2[0] - $date_parts1[0];
}

echo yearsDifference('2011-03-12','2008-03-09');

Which, in this case, will get you :

answered Mar 22 '11 at 05:24

Pascal MARTIN

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he wants the difference in Julian Day – xkeshav Mar 22 '11 at 05:27
This may not give the intended result. What about Dec 31 vs Jan 1? I don't think that should be a full year, but it depends on context. – Matthew Mar 22 '11 at 05:34
This one will return only the difference between the year numbers. It is misleading. – Aistis Jul 27 '17 at 08:11
1

this doesnt consider if you need the difference between 01-12-2017 to 01-01-2018 OR 01-12-2017 to 02-12-2018. The result would be the same which is very wrong. – Izac Mac Dec 08 '17 at 02:17

Calculating number of years between 2 dates in PHP

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