I'm new at PHP and MySQL and have a problem
| From: | news.php.net | Date: | Tue, 11 May 2004 19:13:30 +0000 |
| Subject: | I'm new at PHP and MySQL and have a problem | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-34368@lists.php.net to get a copy of this message | ||
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;
?>