Re: [PHP] Variable won't work in function, even when I global it?

[Leif K-Brooks]

Ok, I guess I'll send it.  Thanks for trying to help.  Here it is:

    $ip = $REMOTE_ADDR;
    $query = mysql_query("select COUNT(*) as rowexists from ips where ip
    = '$ip'");
    $result = mysql_fetch_array($query);
    if(($result['rowexists'] == 0) && ($loggedin)){
    mysql_query("INSERT INTO `ips` (ip,username) VALUES
    ('$ip','$userinfo[username]')") or print "Error: ".mysql_error();
    }
    $time_end = getmicrotime();
    $time_taken = $time_end - $time_start;
    $query = mysql_query("select COUNT(*) as cnt from ips");
    $result = mysql_fetch_array($query);
    $visits = $result['cnt'] + 2800;
    eval($getlayout['footer']);


Miguel Cruz wrote:

>Rather than continuing to shroud this in mystery, could you at least give 
>us a taste of what is in $footertext? 
>