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Posted by Andy Hassall on 02/03/06 00:42
On 2 Feb 2006 14:01:54 -0800, "Tomas" <tomas.srna@gmail.com> wrote:
>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
>)
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.
>Can you please suggest something how to calculate the difference
>between 2 datetimes?
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.
>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?
The "t" option in http://uk2.php.net/date
Or http://uk2.php.net/calendar
--
Andy Hassall :: andy@andyh.co.uk :: http://www.andyh.co.uk
http://www.andyhsoftware.co.uk/space :: disk and FTP usage analysis tool
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