我希望数据库连接与SQL中的类，但我面临错误？

Question

<?php
    class Database_Object
    {
        protected static $DB_Name;
        protected static $DB_Open;
        protected static $DB_Conn;

        protected function __construct($hostname, $username, $password, $database)
        {
            self::$DB_Name = $database;
            self::$DB_Conn = mysqli_connect($hostname, $username, $password, $database);
            if (!self::$DB_Conn) { die('Critical Stop Error: Database Error<br />' . mysqli_error()); }
            mysqli_select_db(self::$DB_Name, self::$DB_Conn);
        }

        private function __clone() {}

        public function __destruct(){}

    }

    class DB extends Database_Object
    {
        public static function Open($hostname = 'localhost', $username = 'root', $password =' ', $database = 'anil1')
        {
            if (!self::$DB_Open)
            {
                self::$DB_Open = new self($hostname, $username, $password, $database);
            }
            else
            {
                self::$DB_Open = null;
                self::$DB_Open = new self($hostname, $username, $password, $database);
            }
            return self::$DB_Open;
        }

        public function qry($sql, $return_format = 0)
        {
            $query = mysqli_query($sql, self::$DB_Conn) OR die(mysqli_error());
            switch ($return_format)
            {
                case 1:
                    $query = mysqli_fetch_row($query);
                    return $query;
                    break;
                case 2:
                    $query = mysqli_fetch_array($query);
                    return $query;
                    break;
                case 3:
                    $query = mysqli_fetch_row($query);
                    $query = $query[0];
                    return $query;
                default:
                    return $query;
            }
        }
    }
    //$dbconnect = new DB_Class();
    $DB = DB::Open();
$result = $DB->qry(" {SELECT Firstname FROM persons} ;");
?>

hi Warning: mysqli_connect() [function.mysqli-connect]: (28000/1045): Access denied for user 'root'@'localhost' (using password: YES) in C:\xampp\htdocs\Anil\object.php on line 12

Warning: mysqli_error() expects exactly 1 parameter, 0 given in C:\xampp\htdocs\Anil\object.php on line 13 Critical Stop Error: Database Error

Answer 1

The warning Access denied for user 'root'@'localhost' (using password: YES) means that the connection to the database failed, because you tried to connect with the wrong username or password.

The warning mysqli_error() expects exactly 1 parameter, 0 given means that the function expected 1 parameter (see the docs), but you supplied none. It expects the connection identifier returned by mysqli_connect() as argument.