difference between 2 date

[prokhaled]

Is there any function to give me the difference between 2 date

for exam:

dif beteen 05-24-2002 and 05-27-2002 = 3 days

[The Hobo]

Just use the Date() and strtotime() functions:

echo date("d M Y", strtotime("05-28-2002")); 

the Date() function formats the date, and the strtotime() function converts the text date into a unix timestamp.

[The Hobo]

http://www.php.net/manual/en/function.date.php contains information on the function and many format strings that you might want to know.

[The Hobo]

You changed your question...

[prokhaled]

I'm Sorry.

do you have answer for my new question ??

[scoutt]

if you keep the date like it is and then next year you will have problems. you need to make the date like this 2002-5-24 as you can do comparisopns on dates a lot easier. the reason is that you have a date 5-24-2003 - 5-24-2002 is not 365 as far as php is concerned. but 2003-5-24 - 2002-5-24 will be don't ask me how but I have come across this.

[The Hobo]

This should give you the difference in days:

<?php
 
  $date1 = "05-24-2002";
  $date2 = "05-27-2002";
  
  echo DateDiff($date1, $date2);

  function DateDiff($first, $second) { 
    $t1 = strtotime($first); 
    $t2 = strtotime($first); 
    $dif = $t1 - $t2; 
    $days = floor(($dif / (60*60*24))); 
    
    return $days; 
} 

?>

You could probably expand it farther to get weeks and years (months are too complicated and I don't feel like working it out (since each month has a different number of days)):

<?php

  function DateDiff($first, $second) { 
    $t1 = strtotime($first); 
    $t2 = strtotime($first); 
    $dif = $t1 - $t2; 
    $days = floor(($dif / (60*60*24)));
   
    if ($days > 365) { //not leap year compliant
      $years = floor($days / 365) . " years";
      $days = floor($days % 365);
    if ($days > 7) {
      $weeks = floor($days / 7) . " weeks";
      $days = floor($days % 7) . " days";
    } 
  
    if ($years != 0) {
      $diff = $years;
    }
    if ($weeks != 0) {
      $diff .= ", $weeks";
    }
    if ($days !=0) {
      $diff .= ", $days";
    }
    $diff .= ".";

    return $diff; 
} 

?>

Something like this.

[prokhaled]

It's very good idea.
Thanks

[The Hobo]