"News.Php.Net" <Adam.Freden@xxxxxxxxxxxxx> wrote in message news:20040511191338.97262.qmail@xxxxxxxxxxxxxxx > To help me learn using these awesome tools together, I'm creating a sample > application for myself - a movie database. In this database (which is named > "movies"), I have five tables. One named "moviemain", one named "actors", > one named "genre", and then two tables that are designed to link these > tables together "moviegenre" and "movieactor". I've created a MySQL query > to get all records, and it works great if I use it directly with MySQL, > however every time I try it with PHP it gives me a "Resource id #2" message. > It seems to have something to do with creating a user connection twice, or > something liek that, but I'm not sure. Here's the simple code snippet: > > <?php > $dbcxn = @mysql_connect('localhost','root',''); > if (!$dbcxn) { > echo('<p>Unable to connect to the database at this time</p>'); > exit(); > } > mysql_select_db('movies',$dbcxn); > if (! @mysql_select_db('movies') ) { > die('<p>Unable to locate the Movie database within the database > server</p>'); > } > $result = @mysql_query("SELECT * from > MovieMain,movieactor,moviegenre,genre,actors where > MovieMain.movieid=movieactor.movieid and > MovieMain.movieid=moviegenre.movieid and genre.genreid=moviegenre.genreid > and actors.actorid=movieactor.actorid"); > if (!$result) { > die('<p>Error performing query: ' . mysql_error() . '</p>'); > } > echo $result; > ?> $result is a MySQL result resource handle, you cannot just echo it out. You have to use a loop to iterate through every row of the result. Try this: while ($row = mysql_fetch_assoc($result)) { // now you have an array with all field values of the current record, do a print_r($row) to see how the array is structured } See here: http://de.php.net/mysql_fetch_assoc Regards, Torsten -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
