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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