RE: [PHP] Within the date format
On Fri, 19 Apr 2002, Demitrious S. Kelly wrote: > Whatever works > > And the function works fine for any year after 2000 Not for 2100. miguel -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Within the date format
Whatever works
And the function works fine for any year after 2000
Besides... it was just a quick and dirty example
-Original Message-
From: Miguel Cruz [mailto:[EMAIL PROTECTED]]
Sent: Friday, April 19, 2002 9:14 AM
To: Demitrious S. Kelly
Cc: [EMAIL PROTECTED]
Subject: RE: [PHP] Within the date format
1) That function isn't correct.
2) It's slow.
3) You still would have to call it once for each year in the range in
order to figure out how many intervening leap years there are, and you'd
have to check whether the dates in question fell before or after Feb 29.
Much easier to just calculate the difference of days, months, and years.
BTW, off the top of my head (untested), here's a more accurate (and a
million times faster) way to check for a leap year:
function is_leap ($year)
{
return (!($year % 4) && (($year % 100) || !($year % 400)));
}
miguel
On Fri, 19 Apr 2002, Demitrious S. Kelly wrote:
> function is_leap($) {
> $leap=0;
> $refy=2000;
> while ( $ >= $refy ) {
> if ( $refy == $ ) {
> $leap=1;
> } else if ( $refy > $ ) {
> break;
> }
> $refy++;
> $refy++;
> $refy++;
> $refy++;
> }
> return($leap);
> }
>
> -Original Message-
> From: Miguel Cruz [mailto:[EMAIL PROTECTED]]
> Sent: Friday, April 19, 2002 8:52 AM
> To: Demitrious S. Kelly
> Cc: [EMAIL PROTECTED]
> 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:[EMAIL PROTECTED]]
> > Sent: Friday, April 19, 2002 8:44 AM
> > To: Ron Allen
> > Cc: [EMAIL PROTECTED]
> > 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, $year and then subtract.
> >
> > You can't just do a cascading modulus calculation on the delta
between
> > the
> > timestamps, because that won't take leap years into account.
> >
> > miguel
> >
> >
> >
>
>
>
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RE: [PHP] Within the date format
1) That function isn't correct.
2) It's slow.
3) You still would have to call it once for each year in the range in
order to figure out how many intervening leap years there are, and you'd
have to check whether the dates in question fell before or after Feb 29.
Much easier to just calculate the difference of days, months, and years.
BTW, off the top of my head (untested), here's a more accurate (and a
million times faster) way to check for a leap year:
function is_leap ($year)
{
return (!($year % 4) && (($year % 100) || !($year % 400)));
}
miguel
On Fri, 19 Apr 2002, Demitrious S. Kelly wrote:
> function is_leap($) {
> $leap=0;
> $refy=2000;
> while ( $ >= $refy ) {
> if ( $refy == $ ) {
> $leap=1;
> } else if ( $refy > $ ) {
> break;
> }
> $refy++;
> $refy++;
> $refy++;
> $refy++;
> }
> return($leap);
> }
>
> -Original Message-
> From: Miguel Cruz [mailto:[EMAIL PROTECTED]]
> Sent: Friday, April 19, 2002 8:52 AM
> To: Demitrious S. Kelly
> Cc: [EMAIL PROTECTED]
> 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:[EMAIL PROTECTED]]
> > Sent: Friday, April 19, 2002 8:44 AM
> > To: Ron Allen
> > Cc: [EMAIL PROTECTED]
> > 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, $year and then subtract.
> >
> > You can't just do a cascading modulus calculation on the delta between
> > the
> > timestamps, because that won't take leap years into account.
> >
> > miguel
> >
> >
> >
>
>
>
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RE: [PHP] Within the date format
function is_leap($) {
$leap=0;
$refy=2000;
while ( $ >= $refy ) {
if ( $refy == $ ) {
$leap=1;
} else if ( $refy > $ ) {
break;
}
$refy++;
$refy++;
$refy++;
$refy++;
}
return($leap);
}
-Original Message-
From: Miguel Cruz [mailto:[EMAIL PROTECTED]]
Sent: Friday, April 19, 2002 8:52 AM
To: Demitrious S. Kelly
Cc: [EMAIL PROTECTED]
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:[EMAIL PROTECTED]]
> Sent: Friday, April 19, 2002 8:44 AM
> To: Ron Allen
> Cc: [EMAIL PROTECTED]
> 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
>
>
>
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
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:[EMAIL PROTECTED]]
> Sent: Friday, April 19, 2002 8:44 AM
> To: Ron Allen
> Cc: [EMAIL PROTECTED]
> 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
>
>
>
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Within the date format
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:[EMAIL PROTECTED]]
Sent: Friday, April 19, 2002 8:44 AM
To: Ron Allen
Cc: [EMAIL PROTECTED]
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
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
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
--
PHP General Mailing List (http://www.php.net/)
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[PHP] Within the date format
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
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
