Jag försöker använda mig av något liknande datediff() i ASP men nu i PHP men skriptet jag har returnerar bara 0 hela tiden. Jag undrar om någon kan hjälpa mig eller om någon har ett bättre skript.

$datum1 = "2003-12-21 10:00:00";
$datum2 = "2003-12-12 10:00:00";

echo datediff("d", $datum1, $datum2);


function datediff($interval, $date1, $date2) {
// Function roughly equivalent to the ASP "DateDiff" function
$seconds = $date2 - $date1;

switch($interval) {
case "y":
list($year1, $month1, $day1) = split('-', date('Y-m-d', $date1));
list($year2, $month2, $day2) = split('-', date('Y-m-d', $date2));
$time1 = (date('H',$date1)*3600) + (date('i',$date1)*60) + (date('s',$date1));
$time2 = (date('H',$date2)*3600) + (date('i',$date2)*60) + (date('s',$date2));
$diff = $year2 - $year1;
if($month1 > $month2) {
$diff -= 1;
} elseif($month1 == $month2) {
if($day1 > $day2) {
$diff -= 1;
} elseif($day1 == $day2) {
if($time1 > $time2) {
$diff -= 1;
}
}
}
break;
case "m":
list($year1, $month1, $day1) = split('-', date('Y-m-d', $date1));
list($year2, $month2, $day2) = split('-', date('Y-m-d', $date2));
$time1 = (date('H',$date1)*3600) + (date('i',$date1)*60) + (date('s',$date1));
$time2 = (date('H',$date2)*3600) + (date('i',$date2)*60) + (date('s',$date2));
$diff = ($year2 * 12 + $month2) - ($year1 * 12 + $month1);
if($day1 > $day2) {
$diff -= 1;
} elseif($day1 == $day2) {
if($time1 > $time2) {
$diff -= 1;
}
}
break;
case "w":
// Only simple seconds calculation needed from here on
$diff = floor($seconds / 604800);
break;
case "d":
$diff = floor($seconds / 86400);
break;
case "h":
$diff = floor($seconds / 3600);
break;
case "i":
$diff = floor($seconds / 60);
break;
case "s":
$diff = $seconds;
break;
}
return $diff;
}

Hittade ett script på www.php.net som skall göra detta. Vet dock inte om det fungerar. Det är det första scriptet på sidan där inläggen börjar.
http://se.php.net/manual/en/function.date.php


***cut***

T.E.
07-Jan-2004 05:15
Most functions calculating the difference in days between two days take some assumptions. They often ignore leap-years or decide to set a month to 30 days. This one here should work better and more precise.
I left one thing to fix: I supposed a leap in every four years. This assumption is wrong when passing a xx00-border, if xx00 mod 400 != 0. However, PHPs date-function won't work with 1900 or less and 2100 or more. So for the next few years within the given boarders of PHP, this one should work.

function date_diff($dat1,$dat2)
/* Dat1 and Dat2 passed as "YYYY-MM-DD" */
{
$tmp_dat1 = mktime(0,0,0,
substr($dat1,5,2),substr($dat1,8,2),substr($dat1,0,4));
$tmp_dat2 = mktime(0,0,0,
substr($dat2,5,2),substr($dat2,8,2),substr($dat2,0,4));

$yeardiff = date('Y',$tmp_dat1)-date('Y',$tmp_dat2);
/* a leap year in every 4 years and the days-difference */
$diff = date('z',$tmp_dat1)-date('z',$tmp_dat2) +
floor($yeardiff /4)*1461;

/* remainder */
for ($yeardiff = $yeardiff % 4; $yeardiff>0; $yeardiff--)
{
$diff += 365 + date('L',
mktime(0,0,0,1,1,
intval(
substr(
(($tmp_dat1>$tmp_dat2) ? $dat1 : $dat2),0,4))
-$yeardiff+1));
}

return $diff;
}