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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