[RESOLVED] PHP error

[aksdad]

When I run the script like this

<!-- BEGIN - EMBED Channel FROM Come2Play.com -->
<iframe src="http://www.come2play.com/channel_auth.asp?channel_id=50562&uid=<?php echo $_COOKIE['ava_userid'];?>&nick_name=<?php echo $_COOKIE["ava_username"];?>&auth_sig=ea0c9cab64bc52b53789d78e47750db2 width="990" height="537" frameborder="0" scrolling="no" allowtransparency="true" ></iframe>
<div id="c2p" style="text-align:center;width:990;font-family:Tahoma,Arial,sans-serif;font-size:11px;"><a href="http://www.come2play.com/shared/affiliation/setAffiliate.asp?channel_id=50562&next_url=http%3A%2F%2Fwww%2Ecome2play%2Ecom%2Fpublisher%2Easp" target="_blank" >Come<span>2</span>Play</a> - create your own multiplayer channel</div>
<!-- END - EMBED Channel FROM Come2Play.com -->

It works fine
But if I add this above that code

<?php
$strauthsig = $_COOKIE['ava_userid'].196975eb3d26cb24fc37df4b9ca36730;
echo '$strauthsig';
?>

It stops working.
What is it that i'm doing wrong ?

[Pikachu2000]

It looks as though you're trying to append a string to the variable, is that correct?
if so,
$strauthsig = $_COOKIE['ava_userid'] . "196975eb3d26cb24fc37df4b9ca36730";

[aksdad]

Thanks for that but now theres another problem:
I modified the code to this :

<?php

$server = 'localhost';                   

$dbusername = 'user';                      

$password = 'pass' ;                      

$database = 'intergal_arcade';                  

// The following should not be edited
  $con = mysql_connect("$server","$dbusername","$password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("$database", $con);

$q = mysql_query("SELECT username, avatar FROM ava_users WHERE username='$_COOKIE['ava_username']' LIMIT 1");
$row = mysql_fetch_array($q);
$avatar = 'http://www.intergalacticwars.info/games/uploads/avatars/' . $row['avatar'];
$strauthsig = md5($_COOKIE['ava_userid'] . "196975eb3d26cb24fc37df4b9ca36730");
?>


<!-- BEGIN - EMBED Channel FROM Come2Play.com -->
<iframe src="http://www.come2play.com/channel_auth.asp?channel_id=50562&uid=<?php echo $_COOKIE['ava_userid'];?>&nick_name=<?php echo $_COOKIE['ava_username'];?>&auth_sig=<?php echo $strauthsig;?>" width="990" height="537" frameborder="0" scrolling="no" allowtransparency="true" ></iframe>
<!-- END - EMBED Channel FROM Come2Play.com -->

And it stopped working again -_-

[Pikachu2000]

Are any errors displayed or logged?

[aksdad]

Nope. If you have time can you check the syntax?
I usually make mistakes there

[bradgrafelman]

You do have a parse error in your code... which begs the question, why don't you have display_errors set to On, or utilize the log_errors directive?

[aksdad]

in this line :

$q = mysql_query("SELECT username, avatar FROM ava_users WHERE username='$_COOKIE['ava_username']' LIMIT 1");

If I replace $_COOKIE['ava_username'] with a name in my db like exo it works but if I put it in like shown in the one above it doesn't work.
What could be the problem?

[bradgrafelman]

You can't use quoted array indexes inside of a double-quoted string. Either use complex syntax (e.g. surround the variable with curly braces) or use concatenation. Examples can be found in the manual here (scroll down a bit to the second code example with arrays).

Also note that user-supplied data should never be placed directly into a SQL query. Instead, it must first be sanitized with a function such as [man]mysql_real_escape_string/man.

[aksdad]

ok So I edited the code to this:

$server = 'lxxxt';                   

$dbusername = 'inxxx';                      

$password = 'axxx';                      

$database = 'xxxxx';
$userid = mysql_real_escape_string($_COOKIE['ava_username'])

  $con = mysql_connect("$server","$dbusername","$password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("$database", $con);

$q = mysql_query("SELECT username, avatar FROM ava_users WHERE username='{$userid}' LIMIT 1");

Still it doesnt showup

[aksdad]

Found the problem I didn't put ";" after $userid = mysql_real_escape_string($_COOKIE['ava_username'])

Sorry for wasting your time lol 🙂