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
> >
> >
> >
> >
> >
> >
> >
> > _________________________________________________________________
> > The new MSN 8: advanced junk mail protection and 2 months FREE*
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> >
> >
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> >
>
--
Adam Voigt (adam@cryptocomm.com)
The Cryptocomm Group
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