session error

[JoeyWong]

Warning: Unknown: Your script possibly relies on a session side-effect which existed until PHP 4.2.3. Please be advised that the session extension does not consider global variables as a source of data, unless register_globals is enabled. You can disable this functionality and this warning by setting session.bug_compat_42 or session.bug_compat_warn to off, respectively in Unknown on line 0

Hi all, i need your help!! how to solve this error?? thanks~

This is my php.ini. What should I change it so that it won't come out the warning?? thanks~

; session.bug_compat_42
; Default Value: On
; Development Value: On
; Production Value: Off

; session.bug_compat_warn
; Default Value: on
; Development Value: on
; Production Value: Off

[P0lT10n]

change register_globals... but show your code !

[JoeyWong]

change register_globals... but show your code !

This is my code... thanks~

<h2><?php 
$Iname=$_GET["Iname"];
$sql = "SELECT i_name, description FROM institution WHERE i_id='$Iname'";
$result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_assoc($result);
$IName=$row["i_name"];
$Description=$row["description"];
echo "$IName";?></h2><br /><br />
<?php 
echo "<p>$Description</p>";
session_start();
$_SESSION[IName] = $_GET["IName"];
?>
<table width="100%" border="0">
   <?php
	if(isset($_GET["Iname"]))
    $Iname = $_GET["Iname"];
	$sql = "SELECT * FROM school WHERE i_id='$Iname' ORDER BY s_name";
	$result = mysql_query($sql) or die(mysql_error());
	$row = mysql_fetch_assoc($result);
	?>