Re: [PHP] ERROR IN THIS CODE

[Jorge]

I resolve the problem.
Thank you for all.
----- Original Message ----- 
From: "Jorge" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Sunday, September 05, 2004 6:23 PM
Subject: [PHP] 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" <[EMAIL PROTECTED]>
To: "Jorge" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
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
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php