Question

Without using PHP 5.3’s date_diff function (I’m using PHP 5.2.17), is there a simple and accurate way to do this? I am thinking of something like the code below, but I don’t know how to account for leap years:

$days = ceil(abs( strtotime('2000-01-25') - strtotime('2010-02-20') ) / 86400);
$months = ???;

I’m trying to work out the number of months old a person is.

Answer 1

$date1 = '2000-01-25';
$date2 = '2010-02-20';

$ts1 = strtotime($date1);
$ts2 = strtotime($date2);

$year1 = date('Y', $ts1);
$year2 = date('Y', $ts2);

$month1 = date('m', $ts1);
$month2 = date('m', $ts2);

$diff = (($year2 - $year1) * 12) + ($month2 - $month1);

You may want to include the days somewhere too, depending on whether you mean whole months or not. Hope you get the idea though.