[SOLVED] Use of undefined constant

[cheechm]

Hello,

Whenever I run this I get:

Notice: Use of undefined constant online - assumed 'online'

function update_sessions()
{
        $sid = session_id();
        if($_SESSION[online] == "1")
        {
                mysql_query("UPDATE `sessions` SET `time` = '". time() ."' WHERE `sid` = '$sid'") or die(mysql_error());
        }
        else
        {
                $_SESSION[online] = 1;
                mysql_query("INSERT INTO `sessions` SET `time` = '". time() ."', `sid` = '$sid'") or die(mysql_error());
        }
}

function get_onlineusers()
{
        $min = time() - 301;
        mysql_query("DELETE FROM `sessions` WHERE `time` <= '$min'") or die(mysql_error());
        $query = mysql_query("SELECT COUNT(sid) FROM `sessions`");
        $num = mysql_fetch_row($query);
        return($num[0]);
}

 

What is the problem?

 

Thanks

[GingerRobot]

Array keys which are not integers should be enclosed in single quotes, e.g.:

 

$_SESSION['online']

 

You'll see them not done like this a lot of the time, since you only recieve a notice for this kind of minor error, and a lot of people have PHP set up not to report notices.

[cheechm]

Cool thanks. How do I turn off error reporting?

 

Thanks

[GingerRobot]

You can either set in your php.ini file or you can use the error_reporting() function if you only want to make the changes for one particular script.

[cheechm]

Also what is cross site scripting?

 

Thanks

 

[GingerRobot]

Slight change of subject there. I suggest you google - im sure you'll find what you need.