ERROR IN THIS CODE

the problem is that when i do a echo shows me: "SELECT * FROM oposicions WHERE id = ''" ----- Original Message ----- From: "raditha dissanayake" <jabber@raditha.com> To: "Jorge" <jorvgl@telefonica.net> Cc: <php-general@lists.php.net> Sent: Sunday, September 05, 2004 6:01 PM Subject: Re: Fw: [PHP] ERROR IN THIS CODE > Is this part of a guessing game? > do you want us to guess the error message you are seeing? > > Jorge wrote: > >> >> include('include/conexion.php'); >> include('include/conf.php'); >> >> $sql = "SELECT * FROM >> oposicions WHERE id = '$id'";----------------> I think that the >> problem is here. >> >> >> $result = mysql_query($sql, >> $link); >> >> while( >> ($row=mysql_fetch_object($result)) ) { >> $nome = $row->nome; >> >> >> In my database exist the table oposicions with a field called id.My >> php version es 4.3.38 and of mysql is 3.23.58 >> >> thank you in advance >> > > > -- > Raditha Dissanayake. > ------------------------------------------------------------------------ > http://www.radinks.com/sftp/ | > http://www.raditha.com/megaupload > Lean and mean Secure FTP applet with | Mega Upload - PHP file uploader > Graphical User Inteface. Just 128 KB | with progress bar. > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php >