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Home/ Questions/Q 6007713
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T01:43:13+00:00 2026-05-23T01:43:13+00:00

I’ve made this simple logout script: <?php session_start(); $db_connect = mysql_connect(‘localhost’, ‘root’, ‘*****’); if(!$db_connect)

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I’ve made this simple logout script:

<?php
session_start();
$db_connect = mysql_connect('localhost', 'root', '*****');
    if(!$db_connect)
        {
         die('Не може да се осъществи връзка с базата данни' . mysql_error());
        }
mysql_select_db("chat", $db_connect);
mysql_query("DELETE FROM activeusers WHERE au_id = '$_SESSION['UserId']'");
mysql_close($db_connect);
session_unset();
session_destroy();
?>

But when I put session_unset() and session_destroy() at the end my editor shows an error with the mysql_query I haven’t tried this yet but I think that probably written this way I empty the $_SESSION array() and thus $_SESSION['UserId'] is destroyed before the query.Am I right here and how should I do it right?

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  1. Editorial Team
    Editorial Team
    2026-05-23T01:43:14+00:00Added an answer on May 23, 2026 at 1:43 am

    Format your mysql_query-command like this:

    mysql_query("DELETE FROM activeusers WHERE au_id = '".$_SESSION['UserId']."'");
    

    This makes sure it is properly embedded into the SQL-part of the command.

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