Код:
..............
// Patched function to detect REAL IP address if it's valid
function getip() {
if (isset($_SERVER)) {
if (isset($_SERVER['HTTP_X_FORWARDED_FOR']) && validip($_SERVER['HTTP_X_FORWARDED_FOR'])) {
$ip = $_SERVER['HTTP_X_FORWARDED_FOR'];
} elseif (isset($_SERVER['HTTP_CLIENT_IP']) && validip($_SERVER['HTTP_CLIENT_IP'])) {
$ip = $_SERVER['HTTP_CLIENT_IP'];
} else {
$ip = $_SERVER['REMOTE_ADDR'];
}
} else {
if (getenv('HTTP_X_FORWARDED_FOR') && validip(getenv('HTTP_X_FORWARDED_FOR'))) {
$ip = getenv('HTTP_X_FORWARDED_FOR');
} elseif (getenv('HTTP_CLIENT_IP') && validip(getenv('HTTP_CLIENT_IP'))) {
$ip = getenv('HTTP_CLIENT_IP');
} else {
$ip = getenv('REMOTE_ADDR');
}
}
return $ip;
}
function randompw($length = dirol.gif
{
$chars = 'abdefhiknrstyzaaoABDEFGHKNQRSTYZAAO23456789';
$numChars = strlen($chars);
................
Пише ми това
Код:
Parse error: parse error, unexpected '.', expecting ')' in C:\Program Files\xampp\htdocs\include\файла.php on line 234
А на 234 ред е } под return $ip; Кажете как да го оправя че се измъчих