bekomme folgende Fehlermeldung:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /www/zxq.net/t/e/s/test5678/htdocs/inc/grund.php on line 12
PHP Code:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en">
<head>
<title>ToolStore Stealer Panel 4.1</title>
<link rel="shortcut icon" href="img/favicon/money.ico" type="image/x-icon">
<link rel="stylesheet" type="text/css" href="css/facebox.css" />
<?php
require_once('inc/config.php');
$query1 = mysql_query("SELECT * FROM styles WHERE id LIKE '1'");
while($row = mysql_fetch_array($query1))
{
if($row['style'] == '1'){
echo '<link rel="stylesheet" type="text/css" href="css/style1.css"/>';
}elseif($row['style'] == '2'){
echo '<link rel="stylesheet" type="text/css" href="css/style2.css"/>';
}else{
echo '<link rel="stylesheet" type="text/css" href="css/style3.css"/>';
}
}
PHP Code:
$result = mysql_query($query)
Code:
parse error syntax error unexpected '{' in Line 14.