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ebay style countdown php function

Posted on 2009-07-09
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Last Modified: 2013-12-13
Hi there, i am looking for a prebuilt php function that displays the time left or time since a mysql time that is passed to it. In terms of days, hours and mins only.

for example a time 2 weeks 4 hours and 32 mins in the future would output:

14d 4h 32m

and so on... This should be pretty simple but im restrained for time so any prebuilt functions provided would be great!




that would be great thanks!
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Question by:cookiejest
3 Comments
 
LVL 11

Accepted Solution

by:
BrianMM earned 250 total points
ID: 24812660
Try this:
$diff = Timesince(strtotime($MySQL));
 

<?php

function Timesince($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;  

 }  

?>

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LVL 9

Assisted Solution

by:xBellox
xBellox earned 250 total points
ID: 24812720
If you want in days, hours, minutes and secons try this (you can erase the seconds part if you want):

<?

     function timeBetween($start_date,$end_date)  

     {  

         $diff = $end_date-$start_date;  

         $seconds = 0;  

         $hours   = 0;  

         $minutes = 0;  

   

         if($diff % 86400 <= 0){$days = $diff / 86400;}  // 86,400 seconds in a day  

         if($diff % 86400 > 0)  

         {  

             $rest = ($diff % 86400);  

             $days = ($diff - $rest) / 86400;  

             if($rest % 3600 > 0)  

             {  

                 $rest1 = ($rest % 3600);  

                 $hours = ($rest - $rest1) / 3600;  

                 if($rest1 % 60 > 0)  

                 {  

                     $rest2 = ($rest1 % 60);  

                 $minutes = ($rest1 - $rest2) / 60;  

                 $seconds = $rest2;  

                 }  

                 else{$minutes = $rest1 / 60;}  

             }  

             else{$hours = $rest / 3600;}  

         }  

   

         if($days > 0){$days = $days.' days, ';}  

         else{$days = false;}  

         if($hours > 0){$hours = $hours.' hours, ';}  

         else{$hours = false;}  

         if($minutes > 0){$minutes = $minutes.' minutes, ';}  

         else{$minutes = false;}  

         $seconds = $seconds.' seconds'; // always be at least one second  

   

         return $days.''.$hours.''.$minutes.''.$seconds;  

     }  
 

	 

	 $date_now = date("U");

	 $date_future = date("U", strtotime("07/10/2009 09:30:00"));

	 

	 echo timeBetween($date_now, $date_future).'';

	 

?>

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Author Closing Comment

by:cookiejest
ID: 31601536
thanks both solutions were good!
0

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