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Editorial Team
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Asked: 2026-06-10T00:34:17+00:00

So I have a log in system that generates a random token for each

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So I have a log in system that generates a random token for each log in attempt and saves it in $_session['loginToken'] and after post form checks if session value is equal to posted input or not.
I also found manually set timeout after certain time in here : How do I expire a PHP session after 30 minutes?

Problem is on first log in attempt or after session destroy (timeout) $_SESSION is an empty array and nothing is set but after second try it works fine.

<?php
if(!isset($_SESSION))
    session_start();

print_r($_SESSION);
/*
first try output : Array ( )
second try output : Array ( [LAST_ACTIVITY] => 1345402023 [loginToken] => e3d997090751883feadfed3ae4d8b63e )
*/
if (isset($_SESSION['LAST_ACTIVITY']) && (time() - $_SESSION['LAST_ACTIVITY'] > 10)) {
    session_destroy();
    $_SESSION = array();
}

$_SESSION['LAST_ACTIVITY'] = time();
$token = $_SESSION['loginToken'] = md5(uniqid(mt_rand(), true));
?>
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<form method="post" action="<?=$_SERVER['PHP_SELF'];?>">
    <input type="hidden" name="token" value="<?=$token;?>" />
    <button type="submit" value="login" id="login" name="login">Click</button>
</form>
<body>
</body>
</html>
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1 Answer

  1. Editorial Team

    I’m not sure (and can’t test it now), but

    if(!isset($_SESSION))
    session_start();

    seems to never happen because $_SESSION is always set. Try it without if:

    session_start();

    and don’t do

    $_SESSION = array();

    because it’s bad practice.

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