mysql_num_rows() not a valid resource - mysql_error() shows nothing - php

Question

I have the following code.
include("DBHeader.inc.php");
include("libs/ps_pagination.php");
$sql = "SELECT * FROM Products P, Manufacturers M WHERE M.sManufacturerCode='$LC' AND M.iManufacturerID=P.iManufacturerID";
$rs = mysql_query($sql);
echo $sql;
$pager = new PS_Pagination( $conn, $sql, 3, 4, null );
$rs = $pager->paginate();
$num = mysql_num_rows( $rs ) or die('Database Error: ' . mysql_error());
if ($num >= 1 ) {
echo "<table border='0' id='tbProd' class='tablesorter' style='width:520px;'>
<thead>
<tr>
<th>Product Code</th>
<th>Product Name</th>
<th> </th>
</tr>
</thead>
<tbody>";
//Looping through the retrieved records
while($row = mysql_fetch_array($rs))
{
echo "<tr class='prodRow'>";
echo "<td>" . $row['sProductCode'] . "</td>";
echo "<td>" . $row['sProductName'] . "</td>";
echo "<td><a href='ProdEdit.php?=" . $row['sProductCode'] . "'><img src='images/manage.gif' alt='Edit " . $row['sProductName'] . "' /></a></td>";
echo "</tr>";
}
echo "</tbody></table>";
}
else {
//if no records found
echo "No records found!";
}
And instead of it giving me the data from the table, it spits out on the screen:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/nyksys/www/regserver2/search_results.php on line 37
mysql_error() is actually returning nothing at all, so I'm very confused as to what the error is. The SQL when echo'd:
SELECT * FROM Products P, Manufacturers M WHERE M.sManufacturerCode='216E3ACAC673DE0260083B5FF809B102B3EC' AND M.iManufacturerID=P.iManufacturerID
I'm baffled here! Am I overlooking something simple here?
I've double checked my database information, I'm certain that isn't the problem.
EDIT- I'm following the tutorial Paginating Your Data with AJAX and Awesome PHP Pagination Class.

Answer 1

$sql = "SELECT * FROM Products P, Manufacturers M WHERE M.sManufacturerCode='$LC' AND M.iManufacturerID=P.iManufacturerID";
$rs = mysql_query($sql);
echo $sql;
$rs is a MySQL result resource that you could use with mysql_num_rows.
$pager = new PS_Pagination( $conn, $sql, 3, 4, null );
$rs = $pager->paginate();
Now it's not1!
$num = mysql_num_rows( $rs ) or die('Database Error: ' . mysql_error());
Oops!
1 Or, if it is, [a] you didn't show us that in your question, and [b] the original query was entirely pointless.

Answer 2

You are overwriting the $rs variable

Answer 3

My guess is whatever the PS_Pagination class is doing, it is not returning a MySQL resource. You are overwriting your $rs resource variable with that object, and it ceases to be a valid resource, even if your query succeeds.
$rs = mysql_query($sql);
echo $sql;
$pager = new PS_Pagination( $conn, $sql, 3, 4, null );
// Use a different variable than $rs here.
$rs = $pager->paginate();