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.
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Answer
$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.