TalkPHP - View Single Post - Most efficient way of playing with dates?

kevthedude · 12-13-2007, 07:57 PM

This is how I modified your code to work:

PHP Code:


			
$first = 1195958278;

$last = mktime(0, 0, 0, date("m", $first)  , date("d", $first)+90, date("Y", $first));

$today = mktime(0, 0, 0, date("m")  , date("d"), date("Y"));

$remaining = floor(($last - $today)/86400);

echo remaining;

$remaining will be your days left. You may have to switch the floor() function to ceil()... I didn't figure out the real difference in the dates but this script shows 71 days remaining. Tell me if this is wrong and I'll fix it up for you.

12-13-2007, 07:57 PM	#3 (permalink)
kevthedude The Wanderer Join Date: Dec 2007 Posts: 7 Thanks: 2	This is how I modified your code to work: PHP Code: `$first = 1195958278; $last = mktime(0, 0, 0, date("m", $first) , date("d", $first)+90, date("Y", $first)); $today = mktime(0, 0, 0, date("m") , date("d"), date("Y")); $remaining = floor(($last - $today)/86400); echo remaining;` $remaining will be your days left. You may have to switch the floor() function to ceil()... I didn't figure out the real difference in the dates but this script shows 71 days remaining. Tell me if this is wrong and I'll fix it up for you.