PHP timeago function like stackoverflow - php

Question

I have this function for show timeago from unix timestamp:
function Timesince($original) {
$original = strtotime($original);
// array of time period chunks
$chunks = array(
array(60 * 60 * 24 * 365 , 'year'),
array(60 * 60 * 24 * 30 , 'month'),
array(60 * 60 * 24 * 7, 'week'),
array(60 * 60 * 24 , 'day'),
array(60 * 60 , 'hour'),
array(60 , 'min'),
array(1 , 'sec'),
);
$today = time(); /* Current unix time */
$since = $today - $original;
// $j saves performing the count function each time around the loop
for ($i = 0, $j = count($chunks); $i < $j; $i++) {
$seconds = $chunks[$i][0];
$name = $chunks[$i][1];
// finding the biggest chunk (if the chunk fits, break)
if (($count = floor($since / $seconds)) != 0) {
break;
}
}
$print = ($count == 1) ? '1 '.$name : "$count {$name}s";
if ($i + 1 < $j) {
// now getting the second item
$seconds2 = $chunks[$i + 1][0];
$name2 = $chunks[$i + 1][1];
// add second item if its greater than 0
if (($count2 = floor(($since - ($seconds * $count)) / $seconds2)) != 0) {
$print .= ($count2 == 1) ? ', 1 '.$name2 : " $count2 {$name2}s";
}
}
return $print;
}
This Worked for me. But i need to print timeago only for 3 days, after 3 days need to show normal date like :01/01/2015. in action stackoverflow work with this method.
how do create this ?

Answer 1

step1 : calulate difference
$date1 = new DateTime("2007-03-24");
$date2 = new DateTime("2009-06-26");
$interval = $date1->diff($date2);
//echo "difference " . $interval->y . " years, " . $interval->m." months, //".$interval->d." days ";
Step 2:
use if condition to show off your date
if($interval->d > 3 || $interval->m >0 || $interval->y >0 ){
// echo out your date in your required format
}
thanks to SO
Use this as an idea and approach....