Warning: mysql_fetch_array(): supplied argument is not a valid [duplicate] - php

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Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
I have the following mysql query which is generating the above error and i can not work out why, any help would be appreciated.
$tableName="livetrack";
$targetpage = "visitors.php";
$limit = "$pagination";
$query = "SELECT COUNT(*) as num FROM $tableName where member_id = '$site_id' and display = 'yes'";
$total_pages = mysql_fetch_array(mysql_query($query));
$total_pages = $total_pages[num];
$stages = 3;
$page = mysql_escape_string($_GET['page']);
if($page){
$start = ($page - 1) * $limit;
}else{
$start = 0;
}
// Get page data
$query1 = "SELECT * FROM $tableName WHERE date >= ( NOW() - '$fdate' ) and ip ='$ip' and member_id = '$site_id' and display = 'yes' and category= '$categories' and type ='$vtype' order by id DESC LIMIT $start, $limit";
$result = mysql_query($query1);
// Initial page num setup
if ($page == 0){$page = 1;}
$prev = $page - 1;
$next = $page + 1;
$lastpage = ceil($total_pages/$limit);
$LastPagem1 = $lastpage - 1;
$paginate = '';
if($lastpage > 1)
{
The code below is where the error message is generating from.
while($row = mysql_fetch_array($result))
{

Answer 1

You should print out the raw query and then run it manually in mysql (either through mysql tools or something like phpMyAdmin).
Do you get an error? (probably)
If yes, fix the query.
If not, you should always do a check for mysql errors whenever you run a query and before you try to access the results.
For instance:
$query = 'SELECT * FROM users;';
$result = mysql_query($query);
if(mysql_error() != '') { echo 'error on mysql query: #'.mysql_errno().' - '.mysql_error();
else { $data = mysql_fetch_array($result); }
A second likely possibility is that you don't have a valid connection to a database, or you are connected to the wrong database.