RE: [PHP] Within the date format
| From: | Demitrious S. Kelly | Date: | Fri, 19 Apr 2002 15:54:38 +0000 |
| Subject: | RE: [PHP] Within the date format | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-93724@lists.php.net to get a copy of this message | ||
function is_leap($yyyy) {
$leap=0;
$refy=2000;
while ( $yyyy >= $refy ) {
if ( $refy == $yyyy ) {
$leap=1;
} else if ( $refy > $yyyy ) {
break;
}
$refy++;
$refy++;
$refy++;
$refy++;
}
return($leap);
}
-----Original Message-----
From: Miguel Cruz [mailto:mnc@stoic.net]
Sent: Friday, April 19, 2002 8:52 AM
To: Demitrious S. Kelly
Cc: php-general@lists.php.net
Subject: RE: [PHP] Within the date format
Like I said, if it crosses through a leap year, your calculation may be
off. How many days are there in February when you don't know what year
it
is?
miguel
On Fri, 19 Apr 2002, Demitrious S. Kelly wrote:
> Wouldn't it be easier to convert each date into a unix timestamp, then
> subtract... the resulting number is the difference in seconds. Then
> devide by 60 for minutes, again for hours 24 for days, etc, etc
>
>
> -----Original Message-----
> From: Miguel Cruz [mailto:mnc@stoic.net]
> Sent: Friday, April 19, 2002 8:44 AM
> To: Ron Allen
> Cc: php-general@lists.php.net
> Subject: Re: [PHP] Within the date format
>
> On Fri, 19 Apr 2002, Ron Allen wrote:
> > This is what I have right now
> >
> > $totaltime= date(":H:i:s", mktime(0,0,$totaltime));
> >
> > This is the result
> >
> > 04:20:46
> >
> > from the following dates
> > 2002-04-25 16:30:16
> > 2002-04-19 12:09:30
> > 534046 seconds
> >
> > I would like to be able to get the days and, if needed, the number
of
> months
> > and years
>
> Well, I think the easiest way is going to be to split your date apart
> into
> $day, $month, $year and then subtract.
>
> You can't just do a cascading modulus calculation on the delta between
> the
> timestamps, because that won't take leap years into account.
>
> miguel
>
>
>
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