date("z"); // the today's ordinal number
z - day of the year; i.e. "0" to "365"
if you want the ordinal number of other days different than the current one
you can play with mktime or strftime or strtotime to make an appropiate
timestamp as you did in your earlier codes.
----- Original Message -----
From: "Martin Towell" <[EMAIL PROTECTED]>
To: "GENERAL PHP LIST" <[EMAIL PROTECTED]>
Sent: Wednesday, November 14, 2001 3:19 AM
Subject: 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:[EMAIL PROTECTED]]
> 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 $dobnum, then it's 48!
>
>
> [EMAIL PROTECTED] wrote:
>
> >looking at the manual, getdate() is meant to be passed a unix time stamp,
> >so, you'll need to use strtotime() first thus:
> >
> >$utime = strtotime("30-Nov-1971");
> >$dob = getdate($utime);
> >$dobnum = $dob['yday'];
> >print $dobnum;
> >
> >Notice I changed the format of the date, when I tried using the original
> >format, strtotime() complained, saying it couldn't convert it.
> >
> >Martin T
> >
> >-----Original Message-----
> >From: sundogcurt [mailto:[EMAIL PROTECTED]]
> >Sent: Wednesday, November 14, 2001 7:57 AM
> >To: GENERAL PHP LIST
> >Subject: [PHP] take date and convert to day of year
> >
> >
> >Hi guys, I know that you can take todays date and display it as the
> >numeric day of the year, 1 - 365 / 0 - 364 etc.
> >But can you take a date such as (November-30-1971) and convert that to
> >the numeric day of the year?
> >
> >I have been trying to do this but have had no joy, I don't think my code
> >is even close.
> >
> >$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:
> >
> >
>
>
>
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