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:[EMAIL PROTECTED]]
Sent: Tuesday, November 26, 2002 1:10 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] 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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