php建立连接失败

我正在尝试一些非常基本的东西，但由于某种原因，它无法工作！基本上，我正在尝试为我的网站创建一个新的用户名和密码。这只是为了测试。当我提交请求时，我得到的是：

警告：mysql_real_eescape_string（）[function.mysql real-eescape-string]：拒绝用户"santosh'@'localhost'"（使用密码：NO）在/home/kali197/public_html/test/enroll.php的第27行中访问

我不知道谁是桑托斯显示在这里@localhost。在家里，它展示了不同的人。可能是什么？

我有以下代码来尝试实现我的目标。。。真的很简单：

<?php
class ServerConnection{
    function connect(){
        $mysqli = new mysqli("localhost", "dsdssd_admin", "sddsdsd", 
 "khadsdsli197_dsd");
        if ($mysqli->connect_errno) {
                echo "Failed to connect to MySQL: (" . 
$mysqli->connect_errno . ") " .
                    $mysqli->connect_error;
        }
        //echo $mysqli->host_info . "'n";
    }
}
?>

register.php格式：

<form name="register" action="enroll.php" method="post">
Username: <input type="text" name="username" maxlength="30" />
Password: <input type="password" name="pass1" />
Password Again: <input type="password" name="pass2" />
<input type="submit" value="Register" />
</form>

以及处理该问题的表格，即enroll.php：

<?php
include("ServerConnection.php");
//retrieve our data from POST
$username = $_POST['username'];
$pass1 = $_POST['pass1'];
$pass2 = $_POST['pass2'];
if($pass1 != $pass2)
    header('Location: register_form.php');
if(strlen($username) > 30)
    header('Location: register_form.php');
$hash = hash('sha256', $pass1);
//creates a 3 character sequence
function createSalt()
{
    $string = md5(uniqid(rand(), true));
    return substr($string, 0, 3);
}

$salt = createSalt();
$hash = hash('sha256', $salt . $hash);
$connection = new ServerConnection();
$connection->connect();
$username = mysql_real_escape_string($username);
$query = "INSERT INTO users ( username, password, salt )
        VALUES ( '$username' , '$hash' , '$salt' );";
mysql_query($query);
mysql_close();
header('Location: index.php');
?>

如有任何帮助，我们将不胜感激！提前感谢