Re: date() problem
| From: | Edward Vermillion | Date: | Wed, 06 Jul 2005 20:00:11 +0000 |
| Subject: | Re: date() problem | ||
| References: | 1 2 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-218230@lists.php.net to get a copy of this message | ||
On Jul 6, 2005, at 2:35 PM, Edward Vermillion wrote:
On Jul 6, 2005, at 2:07 PM, Ryan A wrote:Actually it's in the math... With the numbers you have here $dif_d would need to be 6209.24/25 to get up to 17, so the fault lies in the 365.24/25 assumption for a day. Don't ask me how to get around it... it makes my head hurt. :D Edward Vermillion evermillion@doggydoo.netHi, 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.phpIf I change $day2 to =time(); instead of strtotime() I get 17. Could be some weirdness going on with using date()?.