mengatasi extract() expects parameter 1 to be array, null given in

Assalamualaikum Warahmatullahi Wabarakatuh...
mohon izin bertanya bang.
saya sedang belajar php mysqli membuat login dn mendapat kendala. jika input nik dan password bener bisa masuk dasboard. namun jika nik bener dan password salah dia tampil Warning: extract() expects parameter 1 to be array, null given in. itu kesalahan script dimana ya?
 
<?php  if (isset($_POST['login'])) {
$error = array();
if (empty($_POST['ppdb_nik'])) { 
        $error[] = 'NIK tidak boleh kosong'; 
	} else if (!preg_match("/^[0-9]*$/",$_POST['ppdb_nik'])){
		$error[] = "NIK Hanya boleh diisi Angka"; 
    } else if(strlen($_POST['ppdb_nik']) < 16 || strlen($_POST['ppdb_nik']) > 16){
	 $error['password'] = "NIK berjumlah 16 Didit";
} else {
    $ppdbnik=mysqli_real_escape_string($mysqli, $_POST['ppdb_nik']);
 }
 
if (empty($_POST['password'])) { 
     $error[] = 'Password tidak boleh kosong'; 
  } else {
  $password=mysqli_real_escape_string($mysqli, md5($_POST['password']));
 }
 
if (empty($error)){
$login="SELECT * FROM pendaftar WHERE nik='$ppdbnik' AND password='$password'";
$result = $mysqli->query($login) or die($mysqli->error.__LINE__);
$rows= $result->fetch_assoc();
extract($rows);
// Apabila username dan password ditemukan
if($result->num_rows > 0){
$_SESSION['nik'] = $ppdbnik;
header('location:./formulir.php');
echo "<meta http-equiv='refresh' content='0; url=./formulir.php";
$update="UPDATE pendaftar SET status_siswa='Online', login_terakhir=now() WHERE nik='$ppdbnik'" or die($conn->error.__LINE__);
$mysqli->query($update);
} 
else{
echo'<div class="alert alert-warning alert-dismissable">
<button type="button" class="close" data-dismiss="alert" aria-hidden="true">×</button>NIK atau Password salah..!</div>';
}
}
else {echo'<div class="alert alert-danger alert-dismissable"><ul>';
foreach ($error as $key => $values) {            
echo'<li>'.$values.'</li>';}
echo'</ul></div>';
}
unset($_POST['login']);}