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Home/ Questions/Q 7737111
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T07:54:50+00:00 2026-06-01T07:54:50+00:00

My included file (include.php) is this: <?php $myarray=(a,b,c); shuffle($myarray); ?> My main php file

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My included file (include.php) is this:

<?php
$myarray=(a,b,c);
shuffle($myarray);
?>

My main php file is this:

include('include.php');

if isset($_POST['submit_button']){
      echo "Button was clicked";
      echo $myarray;
      }
else {
     echo "Not clicked."; 
     echo $myarray;
     }
?>

<form method='POST'><input type='submit' name='submit_button'></form>

Why are the elements of $myarray displayed in a different order after I clicked the button? Isn’t it shuffled only once?

How can I prevent the shuffle from being executed more than one time? (so that I can display the elements of myarray in the same order, before and after the button was clicked)

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  1. Editorial Team
    Editorial Team
    2026-06-01T07:54:51+00:00Added an answer on June 1, 2026 at 7:54 am

    Your PHP files are interpreted upon every request. As you have it now, there is no memory in your system, so there’s no way for your files to “remember” that the array has already been shuffled. Furthermore, if you shuffle the array once, and then load the page a second time, and managed not to shuffle it, the array would be (a,b,c), as the variable is initialized to (a,b,c) and never shuffled.

    To do what you want, if I understand it correctly, you could use sessions.

    $myarray=(a,b,c);
    
    if (!isset($_SESSION['shuffled'])) {
        shuffle($myarray);
        $_SESSION['shuffled'] = $myarray;
    } else {
        $myarray = $_SESSION['shuffled'];
    }
    
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