RE: [PHP] Re: Problem with strtotime() and 2002-03-31

> Does this mean that there is a day with only 23 hours somewhere???? That would be the corresponding day at the end/start of the year (depending on which hemisphere you're in) -----Original Message----- From: John Clarke [mailto:jclarke@premierstateliner.com.au] Sent: Wednesday, March 20, 2002 12:37 PM To: php-general@lists.php.net Subject: [PHP] Re: Problem with strtotime() and 2002-03-31 Thanks guys. You were right. I have been playing around and just found an extra 3600 seconds on 31/3/2002.!!! Thats a trap for us young players. Will develop a creative fix. Does this mean that there is a day with only 23 hours somewhere???? Though this wouldnt cause a problem I guess. Thanks again, John "John Clarke" <jclarke@premierstateliner.com.au> wrote in message news:20020320003933.79770.qmail@pb1.pair.com... > I have used the following script successfully for a year now, but have just > found a problem with the date 2002-03-31. > When I add 0 days to this date it returns 2002-03-31. Correct! > When I add 1 day to this date it still returns 2002-03-31. NOT Correct > n I add 2 days to this date it returns 2002-03-31. NOT > correct > > So far this is the only date that I havefound a problem with. > > Any ideas why this would be happening? > > function addDays($date,$nofdays) { > $dd=strtotime($date); > $dd2=$dd + (3600*24*$nofdays); > return date("Y-m-d", $dd2); > } > $nd=addDays("2002-03-31","2"); > echo $nd > > Regards > > John Clarke > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php