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Editorial Team
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Asked: 2026-05-31T14:42:16+00:00

<?php //Sessions starten session_start(); //Include stuff include(‘conf.php’); //’Shortcuts’ maken $sname = $site[‘name’]; $shost =

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<?php
//Sessions starten
session_start();
//Include stuff
include('conf.php');
//'Shortcuts' maken
$sname = $site['name'];
$shost = $site['host'];
$shostb = $site['hostb'];
//Verbinding maken met DB
$link = mysql_connect($database['host'], $database['username'], $database['password']) or die (mysql_error());
mysql_select_db($database['db'],$link) or die(mysql_error());
//Gegevens uit db halen
if(isset($_GET['page'])){
    $page['title'] = mysql_query("SELECT title FROM pages WHERE page = $_GET[page]");
}
else{
    $page['title'] = mysql_query("SELECT title FROM pages WHERE page = 'index.php'");
}
?>
<!doctype html>
<html lang="nl">
<head>
    <meta charset="utf-8" />
    <title><?php echo $sname;?>•<?php echo $page['title']?></title>

Does somebody know why this doesn’t work? My DB structure is:

page |title

index.php|index

Sorry but I can’t get this clear. The var dump of $page['title'] = resource(5) of type (mysql result). How can I fix this? Thanks for the help.

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1 Answer

  1. Editorial Team

    mysql_query returns a resource, you need to use mysql_fetch_array (or similar, e.g. mysql_fetch_assoc) on the resource to get the title value.

    For example:

    $res = mysql_query("SELECT title FROM pages WHERE page = 'index.php'");
$page = mysql_fetch_assoc($res);
 
// $page['title'] now contains the value (assuming there's an index.php page in the DB).

    Warning: your query is vulnerable to SQL injection.

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