RE: Resource id #2

["Jason Vincent" <jayv@nortelnetworks.com>]

Try this...

either...

while ($row = mysql_fetch_array($result)) {
      $title = $row['Books.Title'];
      $author = $row['Books.Author'];
	...

	print $title;
}

or...

while ($row = mysql_fetch_array($result)) {
      print $row['Title'];
	...
}



-----Original Message-----
From: The Cossins Fam [mailto:cossinsfam@hotmail.com] 
Sent: Tuesday, November 26, 2002 1:10 PM
To: php-db@lists.php.net
Subject:  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







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