mysql_db_query(): PROBLEM

hello.

i got this error
Warning: mysql_db_query(): supplied argument is not a valid MySQL-Link resource in /home/motorbik/public_html/cat page.php on line 49

here is my code

<?php
    $categories = "Gear Reviews";

    // setup SQL statement
    $SQL = " SELECT * FROM Gear Reviews ";
    $SQL = $SQL . " WHERE categories = '$categories' ";

    // execute SQL statement
    $retid = mysql_db_query($db, $SQL, $cid);

    // check for errors
    if (!$retid) { echo( mysql_error()); }
    else {

        // display results
        echo ("<p><dt><b>$categories</b><br>\n");
        while ($row = mysql_fetch_array($retid)) {
            $url = $row["url"];
            $name = $row["name"];
            $description = $row["description"];
            
            echo ("<p><a href='$url'>$name</a></p>\n");
        }
        echo ("</dt></p>");
    }
?>

PHP:

 

I think the error is in one of the lines

$SQL = " SELECT * FROM Gear Reviews ";
$SQL = $SQL . " WHERE categories = '$categories' ";

PHP:

Why don't you simplify the query

$SQL = "SELECT * FROM Gear Reviews WHERE categories = '$categories'";

PHP:

I think also that 'categories' is a RESERVED word in MySql, try to rename your table field...

 

okay i changed the table to categ.

and got this error now
Warning: mysql_db_query(): supplied argument is not a valid MySQL-Link resource in /home/motorbik/public_html/cat page.php on line 48


this is the code changed

<?php
    $categ = "Gear Reviews";

    // setup SQL statement
    $SQL = "SELECT * FROM Gear Reviews WHERE categ = '$categ'";

    // execute SQL statement
    $retid = mysql_db_query($db, $SQL, $cid);

    // check for errors
    if (!$retid) { echo( mysql_error()); }
    else {

        // display results
        echo ("<p><dt><b>$categ</b><br>\n");
        while ($row = mysql_fetch_array($retid)) {
            $url = $row["url"];
            $name = $row["name"];
            $description = $row["description"];
            
            echo ("<p><a href='$url'>$name</a></p>\n");
        }
        echo ("</dt></p>");
    }
?>                            

PHP: