Linux, Apache, MySQL, PHP

<?php
$date1 = "2012-10-04 22:45:00";
$date2 = "2013-12-05 13:44:01";
$diff = abs(strtotime($date2) - strtotime($date1));
$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24) / (30*60*60*24));
$days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));
$hours = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24)/ (60*60));
$minuts = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24 - $hours*60*60)/ 60);
$seconds = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24 - $hours*60*60 - $minuts*60));
printf("%d years, %d months, %d days, %d hours, %d minuts\n, %d seconds\n", $years, $months, $days, $hours, $minuts, $seconds);
die;
?>

<?php
$td = time_diff($timestamp1-$timestamp2); // has to be a positive result
$td .= ($td=="now")? "":" ago"; // in this example "ago"
echo $td;

function time_diff($s){
   
    $time = get_post_time('G', true, $post);
    $s = time() - $time;

    if($s>=1) {
    $td = "$s sec";
    }  

    if($s>59) {
        $m = (int)($s/60);
        $s = $s-($m*60); // sec left over
        $td = "$m min"; if($s>1) $td .="s";
    }
    if($m>59){
        $hr = (int)($m/60);
        $m = $m-($hr*60); // min left over
        $td = "$hr hr"; if($hr>1) $td .= "s";
        if($m>0) $td .= ", $m min";
    }
    if($hr>23){
        $d = (int)($hr/24);
        $hr = $hr-($d*24); // hr left over
        $td = "$d day"; if($d>1) $td .= "s";
        if($d<3){
            if($hr>0) $td .= ", $hr hr"; if($hr>1) $td .= "s";
        }
    }
    return $td;
}
?>

Stephan A. wrote:

Very smart reply right?

Stephan A. wrote:

Can anyone please tell me what is the best and fastest way to learn PHP?

Stephan A. wrote:

a link to some nice tutorial or a nice book which is available online.

Stephan A. wrote:

ps. How long do you think it could take to learn it?

Andrei Straut wrote:

I think you should actually cut him some slack