Caranya gimana ya?

di file koneksi.php seperti ini :

<?php

$koneksi = mysqli_connect("localhost", "root", "password", "payrollhs");

if (!$koneksi) {

echo "Koneksi ke MySQL Gagal...";

}

dan di file login.php seperti ini :

<?php

if ($_SERVER['REQUEST_METHOD'] == 'POST') {

$user  = $_POST['username'];

$pass = $_POST['password'];

$p       = md5($pass);

if ($user == '' || $pass == '') {

?>

<div class="alert alert-warning"><b>Peringatan!</b> Form anda belum lengkap.</div>

<?php

} else {

include 'koneksi.php';

$sqlLogin = mysqli_query($koneksi, "SELECT * FROM admin WHERE username = '$user' AND password = '$p'");

$jml      = mysqli_num_rows($sqlLogin);

$d        = mysqli_fetch_array($sqlLogin);

if ($jml > 0) {

session_start();

$_SESSION['login']               = TRUE;

$_SESSION['id']                     = $d['id_admin'];

$_SESSION['username']       = $d['username'];

$_SESSION['namalengkap'] = $d['namalengkap'];

header('Location:./index.php');

} else {

?>

<div class="alert alert-danger"><b>Gagal!</b> Username dan Password anda salah.</div>

<?php

}

}

}

?>

gagal nya seperti ini :

Warning: mysqli_connect(): (HY000/1045): Access denied for user 'root'@'localhost' (using password: YES) in C:\xampp\htdocs\hspayroll\koneksi.php on line 2

Koneksi ke MySQL Gagal...

Warning: mysqli_query() expects parameter 1 to be mysqli, boolean given in C:\xampp\htdocs\hspayroll\login.php on line 59

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\hspayroll\login.php on line 60

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\hspayroll\login.php on line 61