maybe its best to echo your entire query string and post back as $searchType 
could be anything , also a good way to check your query , is echo your query 
string then do an sql query in phpmyadmin.

>===== Original Message From "Lane Johnston" <[EMAIL PROTECTED]> =====
>Ok php gurus,
>Little info on system: XP, localhost, port 3306
>i am tring to connect to a db in mysql and i looks like that I am doing
>something wrong?
>Here is the code...
>
>@ $db = mysql_pconnect("localhost:3306", "omegaweb", "jesus316");
>
>if( $db ) { echo "Connected"; }
>if( !$db )
>{
>   echo "Error: Could not connect to database.  Please try again later.";
>   exit;
>}
>
>mysql_select_db("books");
>$query = "SELECT * FROM books WHERE ".$searchType." LIKE
>'%".$searchTerm."%'";
>$result = mysql_query($query);
>
>$num_results = mysql_num_rows($result); //This is where I am getting the
>error actually: "Warning: mysql_num_rows(): supplied argument is not a valid
>MySQL result resource in c:\inetpub\wwwroot\php testing\results.php on line
>33"
>
>I figured that if $result is NULL then of course there would be an error.
>
>Could someone shed some light on this newbie?
>Thanx
>
>
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>
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