PHP Time Bug

[Tomas]

I have a very interesting problem...
Php says that the difference between these times is 1 hour.

<?php
$t1 = "2005-12-31 20:00";
$t2 = "2005-12-31 20:00";
$t1i = strtotime($t1);
$t2i = strtotime($t2);
$diff = abs($t1i - $t2i);
print_r(getdate ($diff));
?>

The Output:
Array
(
[seconds] => 0
[minutes] => 0
[hours] => 1
[mday] => 1
[wday] => 4
[mon] => 1
[year] => 1970
[yday] => 0
[weekday] => Thursday
[month] => January
[0] => 0
)

Can you please suggest something how to calculate the difference
between 2 datetimes?

And another question, a bit off-topic, but how do I get the number of
days of a month with taking leap years in consideration?

Thanks in advence.
Tomas

[Marc van Lieshout]

Re: PHP Time Bug


"Tomas" <tomas.srna@gma il.com> wrote in message
news:1138917713 .942658.239590@ z14g2000cwz.goo glegroups.com.. .[color=blue]
>I have a very interesting problem...
> Php says that the difference between these times is 1 hour.
>
> <?php
> $t1 = "2005-12-31 20:00";
> $t2 = "2005-12-31 20:00";
> $t1i = strtotime($t1);
> $t2i = strtotime($t2);
> $diff = abs($t1i - $t2i);
> print_r(getdate ($diff));
> ?>
>
> The Output:
> Array
> (
> [seconds] => 0
> [minutes] => 0
> [hours] => 1
> [mday] => 1
> [wday] => 4
> [mon] => 1
> [year] => 1970
> [yday] => 0
> [weekday] => Thursday
> [month] => January
> [0] => 0
> )
>
> Can you please suggest something how to calculate the difference
> between 2 datetimes?
>
> And another question, a bit off-topic, but how do I get the number of
> days of a month with taking leap years in consideration?
>
> Thanks in advence.
> Tomas
>[/color]

function is_leap_year($y ear) {
return (($year % 4 == 0) && ($year % 100 != 0)) || ($year % 400 == 0);
}

function days_in_month($ month, $year) {

switch ($month) {
case 4:
case 6:
case 9:
case 11:
return 30;

case 2:
return is_leap_year($y ear) ? 29 : 28;

default:
return 31;
}
}


Marc.

[Andy Hassall]

Re: PHP Time Bug

On 2 Feb 2006 14:01:54 -0800, "Tomas" <tomas.srna@gma il.com> wrote:
[color=blue]
>I have a very interesting problem...
>Php says that the difference between these times is 1 hour.
>
><?php
>$t1 = "2005-12-31 20:00";
>$t2 = "2005-12-31 20:00";
>$t1i = strtotime($t1);
>$t2i = strtotime($t2);
>$diff = abs($t1i - $t2i);
>print_r(getdat e($diff));
>?>
>
>The Output:
>Array
>(
> [seconds] => 0
> [minutes] => 0
> [hours] => 1
> [mday] => 1
> [wday] => 4
> [mon] => 1
> [year] => 1970
> [yday] => 0
> [weekday] => Thursday
> [month] => January
> [0] => 0
>)[/color]

What you've ended up asking with the code above is "what is the year, month,
day and time of the difference of two identical times?" - which doesn't make
much sense.

You end up with "what is the date represented by the value zero", which is
defined as the "UNIX epoch". Apparently you're in a central European country
currently on GMT+1 (Slovakia from the looks of your message headers?); the UNIX
epoch is midnight 1st Jan 1970 GMT - the result you've got is 1am on the same
day, as adjusted for your timezone.
[color=blue]
>Can you please suggest something how to calculate the difference
>between 2 datetimes?[/color]

Depends what you really want. You already had the difference in seconds -
zero. You can divide this down to get hours or days, but it depends how you
want to deal with summer time transitions.
[color=blue]
>And another question, a bit off-topic, but how do I get the number of
>days of a month with taking leap years in consideration?[/color]

The "t" option in http://uk2.php.net/date
Or http://uk2.php.net/calendar

--
Andy Hassall :: andy@andyh.co.u k :: http://www.andyh.co.uk
http://www.andyhsoftware.co.uk/space :: disk and FTP usage analysis tool

[Pedro Graca]

Re: PHP Time Bug

Tomas wrote:[color=blue]
> I have a very interesting problem...
> Php says that the difference between these times is 1 hour.[/color]

No it doesn't
[color=blue]
> <?php
> $t1 = "2005-12-31 20:00";
> $t2 = "2005-12-31 20:00";
> $t1i = strtotime($t1);
> $t2i = strtotime($t2);
> $diff = abs($t1i - $t2i);
> print_r(getdate ($diff));
> ?>[/color]

What PHP says is that getdate(0) returns the following
[color=blue]
> The Output:
> Array
> (
> [seconds] => 0
> [minutes] => 0
> [hours] => 1
> [mday] => 1
> [wday] => 4
> [mon] => 1
> [year] => 1970
> [yday] => 0
> [weekday] => Thursday
> [month] => January
> [0] => 0
> )
>
> Can you please suggest something how to calculate the difference
> between 2 datetimes?[/color]

You already did that.
The difference between $t1 and $t2 is 0 (seconds); if you want to
convert seconds to hours divide by 3600

echo 'Difference is ', $diff / 3600, 'hours';
[color=blue]
> And another question, a bit off-topic, but how do I get the number of
> days of a month with taking leap years in consideration?[/color]

$month = 2;
$year = 2000;
$number_of_days 2000 = date('t', gmmktime(12, 0, 0, $month, 15, $year));
$year = 2004;
$number_of_days 2004 = date('t', gmmktime(12, 0, 0, $month, 15, $year));

--
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