Re: Resource id #2

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Umm, he is putting them into an array, I quote:

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'];
> > 
> > }

See the while condition?

On Wed, 2002-11-27 at 06:03, Chris Barnes wrote:
> You need to put your $result into an array. you can use:
> 
> $result_array = mysql_fetch_array($result);
> 
> then, if you know the field names in the array, print them like so:
> 
> echo $result_array["field1"];
> echo $result_array["field2"];
> 
> or if you dont know their names you can refer to their position numbers
> starting from 0 e.g.
> 
> echo $result_array[0];
> echo $result_array[1];
> 
> using the position numbers you could put together a quick script to
> crawl through the array and print all the fields with a few lines of
> code.
> 
> On Wed, 2002-11-27 at 05:09, 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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> > 
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> > 
> 
-- 
Adam Voigt (adam@cryptocomm.com)
The Cryptocomm Group
My GPG Key: http://64.238.252.49:8080/adam_at_cryptocomm.asc

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