On Jul 6, 2005, at 2:07 PM, Ryan A wrote:

Hi,

I'm confused, this should give me the age as 17 instead of 16...but it does
not...any ideas why?

<?php print date("Y:m:d");
$age="1988-07-06";

$day1=strtotime($age);
$day2 = strtotime(date("Y-m-d"));
$dif_s = ($day2-$day1);
$dif_d = ($dif_s/60/60/24);
$age = floor(($dif_d/365.24));

echo $age;
?>

Thanks,
Ryan

--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php



If I change $day2 to =time(); instead of strtotime() I get 17. Could be some weirdness going on with using date()?.


Edward Vermillion
[EMAIL PROTECTED]

--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php

Reply via email to