Resource id #2

Hello. I am using MySQL as a database for a departmental library. I have written a quick search script, but keep getting "resource id #2" as a result to my search. I have read the online documentation for the mysql_fetch_array() function and must say, I don't see that I'm missing anything. However, I've only started programming, much less working with PHP, so perhaps someone can help me out. Here's my code: <? $quickSearch = "mcse"; $table1 = "Books"; $table2 = "BookList"; $table3 = "BoxSet"; $table4 = "Category"; $table5 = "Publisher"; $table6 = "AuthUsers"; $table7 = "CD"; $connection = mysql_connect("localhost", "root") or die("Couldn't connect to the library database."); $db_select = mysql_select_db("library", $connection) or die("Couldn't select the library database."); $search = "SELECT * FROM $table1 LEFT JOIN $table2 ON Books.BookListID = BookList.BookListID

       LEFT JOIN $table3 ON Books.BoxSetID = BoxSet.BoxSetID
       LEFT JOIN $table4 ON Books.CategoryID = Category.CategoryID
       LEFT JOIN $table5 ON Books.PublisherID = Publisher.PublisherID
       LEFT JOIN $table6 ON Books.auID = AuthUsers.auID
       LEFT JOIN $table7 ON Books.CD = CD.CD_ID
       WHERE Books.Title LIKE \"%'$quickSearch'%\"
       OR Books.Author LIKE \"%'$quickSearch'%\"
       OR Books.ISBN LIKE \"%'$quickSearch'%\"
       OR BookList.dbase LIKE \"%'$quickSearch'%\"
       OR BookList.dbase_user LIKE \"%'$quickSearch'%\"
       OR BoxSet.BoxSet LIKE \"%'$quickSearch'%\"
       OR Category.Category LIKE \"%'$quickSearch'%\"
       OR Category.Sub_category LIKE \"%'$quickSearch'%\"
       OR Publisher.Publisher LIKE \"%'$quickSearch'%\"";

$result = mysql_query($search, $connection) or die("Couldn't search the library."); while ($row = mysql_fetch_array($result)) {

       $row['Books.Title'];
       $row['Books.Author'];
       $row['Books.ISBN'];
       $row['BookList.dbase'];
       $row['BookList.dbase_user'];
       $row['BoxSet.BoxSet'];
       $row['Category.Category'];
       $row['Category.Sub_category'];
       $row['Publisher.Publisher'];
       $row['AuthUsers.email'];

} ?> I then have some HTML to display the result of the search. I don't receive any error messages - I just see an empty table from the HTML code I wrote. I added an echo of the $result to find the "resouce id #2". Thanks for any help you can provide. --joel _________________________________________________________________ The new MSN 8: advanced junk mail protection and 2 months FREE* http://join.msn.com/?page=features/junkmail