[Resolvido] Supplied argument is not a valid MySQL result resourc

Bom dia a todos.

Galera, não "manjo" nada em PHP e estou com os seguintes erros em minha página:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/storage/b/58/d2/arterz1/public_html/index.php on line 9 

Warning: extract() [function.extract]: First argument should be an array in /home/storage/b/58/d2/arterz1/public_html/index.php on line 10

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/storage/b/58/d2/arterz1/public_html/index.php on line 14

Warning: extract() [function.extract]: First argument should be an array in /home/storage/b/58/d2/arterz1/public_html/index.php on line 15

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/storage/b/58/d2/arterz1/public_html/index.php on line 25

Warning: extract() [function.extract]: First argument should be an array in /home/storage/b/58/d2/arterz1/public_html/index.php on line 26

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/storage/b/58/d2/arterz1/public_html/index.php on line 43

Warning: extract() [function.extract]: First argument should be an array in /home/storage/b/58/d2/arterz1/public_html/index.php on line 44

Esse é meu código:

::<?php
require_once("./bluadmin/flash/databaseinfo.php");
include('./bluadmin/browserdetect.php');

$dir = './v1site_images/';

$query = "SELECT MetaTitle, MetaKey, MetaDescription FROM v1Settings";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
extract($row);

$query = "SELECT MetaStats FROM Settings";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
extract($row);

$bgColor = "000000";
$txtColor = "FFFFFF";
$accentColor = "FFFFFF";
$frameColor = "666666";

$query2 = "SELECT * FROM Splash";
$result2 = mysql_query($query2);

$row2 = mysql_fetch_array($result2, MYSQL_BOTH);
extract($row2);
$bgColor = $row2["BackColor"];
$txtColor = $row2["TextColor"];
$accentColor = $row2["AccentColor"];
$frameColor = $row2["FrameColor"];

for ($i=1; $i<=4; $i++){
if ($row2["URL".$i] == "index2.php"){
	$new_url = "index2.php?v=v1";
	$query3 = "update Splash set URL".$i." = '$new_url'";
	$result3 = mysql_query($query3);
}
}

$query2 = "SELECT * FROM Splash";
$result2 = mysql_query($query2);

$row2 = mysql_fetch_array($result2, MYSQL_BOTH);
extract($row2);

$TextFont = "Arial, Helvetica, sans-serif";
?>

Alguém pode me ajudar ??