Could someone then help me modify the PHP script so I won't have this timezone issue? I don't understand from looking at the date page on the PHP web site the change(s) I need to make. Thanks, Ron
<? $date1 = strtotime($date1); $date2 = strtotime($date2); $factor = 86400; $difference = (($date1 - $date2) / $factor); On Mon, 2008-03-24 at 07:24 -0700, Jim Lucas wrote: > Ron Piggott wrote: > > I have this math equation this list helped me generate a few weeks ago. > > The purpose is to calculate how many days have passed between 2 dates. > > > > Right now my output ($difference) is 93.9583333333 days. > > > > I am finding this a little weird. Does anyone see anything wrong with > > the way this is calculated: > > > > $date1 = strtotime($date1); (March 21st 2008) > > $date2 = strtotime($date2); (December 18th 2007) > > > > echo $date1 => 1206072000 > > echo $date2 => 1197954000 > > > > #86400 is 60 seconds x 60 minutes x 24 hours (in other words 1 days > > worth of seconds) > > > > $factor = 86400; > > > > $difference = (($date1 - $date2) / $factor); > > > > > > > > As Casey suggested, it is a timestamp issue. > > Checkout my test script. > > http://www.cmsws.com/examples/php/testscripts/[EMAIL PROTECTED]/0001.php > > <plaintext><?php > > $date1 = strtotime('March 21st 2008'); //(March 21st 2008) > echo "date1 = {$date1}\n"; > > $date2 = strtotime('December 18th 2007'); //(December 18th 2007) > echo "date2 = {$date2}\n"; > > $date1 = 1206072000; > echo date('c', $date1)."\n"; > > $date2 = 1197954000; > > echo date('c', $date2)."\n"; > > > #86400 is 60 seconds x 60 minutes x 24 hours (in other words 1 days worth of > seconds) > > $factor = 86400; > > $difference = (($date1 - $date2) / $factor); > > echo $difference."\n"; > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
