Re: Adding +7 more days on date() value problem

Why would you do such strange things? It's a lot more simple :P $now = time(); $plus7 = $now+(7*24*60*60); // 7 days, containing 24 hours each, containing 60 minutes each, containing 60 seconds each. echo 'Current Date: '.date('Y-m-d', $now); echo 'Date in 7 days: '.date('Y-m-d', $plus7); now isn't that easier ;) Trevor Gryffyn wrote:

I disagree on the strtotime recommendations that everyone else gave. The mktime() function itself will compensate for leap years, "day = 36" type stuff and all of that. <?php $curdate = date("Y-m-d"); $plus7 = date("Y-m-d",mktime(0,0,0,date("m"),date("d")+7,date("Y"))); Echo "Current Date: $curdate<br>"; Echo "+7 Dats Date: $plus7"; ?> Simple as that! -TG

-----Original Message----- From: Louie Miranda [mailto:lmiranda@gmail.com] Sent: Tuesday, July 27, 2004 11:31 PM To: php-general@lists.php.net Subject: [PHP] Adding +7 more days on date() value problem I can't figure how could i add +7 more days on the DD (Day) value. My problem is, when the day is close on the end of the month like 07/29/2004 when i add +7 on 29 = 36, its obvious we dont have 36 on the calendar. And im creating a program below, to explode it and maybe just add +. But i think its useless if im just going to do + on the DD (Day) value. Hmm :? ## <?php $curdate = date("Y-m-d"); $plus7 = explode("-", $curdate); print (" <p>cut: " .$plus7[0]. " " .$plus7[1]. " " .$plus7[2]. " </p> Current Date: $curdate<br> +7 Dats Date: $plus7 "); ?> ## -- Louie Miranda http://www.axishift.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php