Add: print_r($row)

In your while loop, that will show you everything that is being returned
with both it's numeric and text based position.

On Tue, 2002-11-26 at 13:09, The Cossins Fam wrote:
> 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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-- 
Adam Voigt ([EMAIL PROTECTED])
The Cryptocomm Group
My GPG Key: http://64.238.252.49:8080/adam_at_cryptocomm.asc

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