OK GOT IT!!
$utine = strtotime("26-Nov-2001");
$bday=(int)$utine;
$utoday = strtotime("");
$btoday = (int)$utoday;
if((($bday-$btoday) <= 1209600 && ($bday-$btoday)>0)){
print "you have " . round(($bday-$btoday)/86400)." Days till
'blanks' bday<br>";
print "you are within the two week range.";
}else{
print "you are NOT within the date range.";
}
It's messy but it works, thanks everybody that helped me out on this.
(C:
[EMAIL PROTECTED] wrote:
>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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>>
>
