RE: [PHP] take date and convert to day of year

I'm running on WinNT4 w/ PHP 4.0.6 and the code I supplied came back with 333 and for today I get 317 dunno why you're getting 364 anyone, any eye-dears ?? -----Original Message----- From: sundogcurt [mailto:sundogcurt@netscape.net] Sent: Wednesday, November 14, 2001 12:14 PM To: GENERAL PHP LIST Subject: Re: [PHP] take date and convert to day of year I have tried to implement your code Martin, and I do thank you VERY MUCH for the help, but your code seems to have the same trouble as mine, it doesn't matter what date I start with, I end up with 364 as the day of the year, I am on win32 though I don't know if that matters. Here is what I tried: //FORMAT HAS TO BE date("d-M-Y"); //GET DAY OF YEAR FOR DOB utime = UNIX time $utine = strtotime("30-Nov-1971"); $dob = getdate($utime); $dobnum = $dob['yday']; print "dob is " . $dobnum . "<br>"; //GET DAY OF YEAR FOR TODAY //$today = date("d-M-Y"); $utoday = strtotime(date("d-M-Y")); $today = getdate($utoday); $todaynum = $today['yday']; print "today is " . $todaynum . "<br>"; There should be a different of about 17 days here right? Not if this is returning 364 for $dobnum, then it's 48! martin.towell@world.net wrote: >looking at the manual, getdate() is meant to be passed a unix time stamp, >so, you'll need to use strtotime() first thus: > >$utime = strtotime("30-Nov-1971"); >$dob = getdate($utime); >$dobnum = $dob['yday']; >print $dobnum; > >Notice I changed the format of the date, when I tried using the original >format, strtotime() complained, saying it couldn't convert it. > >Martin T > >-----Original Message----- >From: sundogcurt [mailto:sundogcurt@netscape.net] >Sent: Wednesday, November 14, 2001 7:57 AM >To: GENERAL PHP LIST >Subject: [PHP] take date and convert to day of year > > >Hi guys, I know that you can take todays date and display it as the >numeric day of the year, 1 - 365 / 0 - 364 etc. >But can you take a date such as (November-30-1971) and convert that to >the numeric day of the year? > >I have been trying to do this but have had no joy, I don't think my code >is even close. > >$dob = getdate("Nov-30-1971"); >$dobnum = $dob['yday']; >print $dobnum; > >$dob 'should' be an array and 'yday' should be the numeric value for the >day of the year, right?!? > >This is what I am trying, and how I understand it, using 'yday' should >give you basically the same output as date ("z"); > >What I would like is the ability to convert any date to the days numeric >value. Any help would be GREATLY appreciated. > >(C: > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net For additional commands, e-mail: php-general-help@lists.php.net To contact the list administrators, e-mail: php-list-admin@lists.php.net