MySQLI 28000/1045用户'root'@'localhost'的访问被拒绝 [英] MySQLI 28000/1045 Access denied for user 'root'@'localhost'

问题描述

我遇到了麻烦.因此，我尝试使用MySQLi连接到我的数据库，但出现此错误:

I am having some trouble. So I am trying to connect to my database using MySQLi, but I am getting this error:

Warning: mysqli::mysqli(): (28000/1045): Access denied for user 'root'@'localhost' (using password: NO) in /home/venge/public_html/library/classes/database.class.php on line 16

Warning: Missing argument 1 for Users::__construct(), called in /home/venge/public_html/routing.php on line 4 and defined in /home/venge/public_html/library/classes/users.class.php on line 3

Warning: mysqli::mysqli(): (28000/1045): Access denied for user 'root'@'localhost' (using password: NO) in /home/venge/public_html/library/classes/database.class.php on line 16

Warning: Cannot modify header information - headers already sent by (output started at /home/venge/public_html/library/classes/database.class.php:16) in /home/venge/public_html/routing.php on line 11

.我不知道为什么它作为用户说"root"，我的代码在下面.我可以用该信息登录到PHPMyAdmin.

. I have no idea why it says "root" as the user, my code is below. I can login to PHPMyAdmin with that info fine.

<?php

$db['host'] = 'localhost';
$db['user'] = 'venge_main';
$db['pass'] = 'fakepassword';
$db['name'] = 'venge_panel';


class DB {
    function __construct($db) {
        $this->mysqli = new mysqli($db['host'], $db['user'], $db['pass'], $db['name']);
    }
    function query($i) {
        return $this->mysqli->query($i);
    }
    function fetch_array($i) {
        return $i->fetch_array(MYSQLI_ASSOC);
    }
    function num($i) {
        return $i->num_rows;
    }
}
?>

这是我的global.php文件:

Here is my global.php file:

<?php
session_start();

$venge['library'] = 'library/classes/';

include_once($venge['library'] . 'users.class.php');
include_once($venge['library'] . 'database.class.php');
$users = new Users($database);
?>

这是我的用户课程:

<?php
class Users {
    public function __construct($db) {
        $this->db = new DB($db);
    }
    public function uidset() {
        if (isset($_SESSION['uid'])) {
            return true;
        } else {
            return false;
        }
    }
    public function securitycheck() {
        if (isset($_SESSION['uid'])) {
            //passed
            return true;
        } else {
            //failed
            die('No permissions');
        }
    }
}
?>

这是routing.php:

Here is routing.php:

<?php
class Routing {
    public function __construct($route) {
        $this->users = new Users();
        $this->route = $route;
    }
    public function File() {
        if (!$this->users->uidset()) {
            switch ($this->route) {
                default:
                    header("Location: /user/login");
                    break;
                case '/user/login':
                    include_once('library/pages/login.page.php');
                    break;
            }
        } else {
            switch ($this->route) {
                default:
                    header("Location: /venge");
                    break;
                case '/venge':
                    echo 'Welcome to <strong>Venge</strong> .';
                    break;
            }
        }
    }
}

$route = new Routing($_SERVER['ORIG_PATH_INFO']);

$route->File();
?>

推荐答案

错误原因在这里:

class Routing {
    public function __construct($route) {
        $this->users = new Users();//<-?
        $this->route = $route;
    }

您没有将参数传递给Users .__ construct($ db)

You are not passing a parameter to Users.__construct($db)

使用凭据定义数组并传递如下:

Define an array with credentials and pass is like this:

class Routing {
    public function __construct($route) {
        $db = array();
        $db['host'] = 'localhost';
        $db['user'] = 'venge_main';
        $db['pass'] = 'fakepassword';
        $db['name'] = 'venge_panel';

        $this->users = new Users($db);
        $this->route = $route;
    }

或者使用全局$ db变量而不是像我一样在本地定义它.但是创建用户对象时必须将其传递给构造函数.

Or use a global $db variable instead of defining it locally, like I did. But you must pass it to constructor when creating Users object.

$users = new Users($database);

我认为必须是

$users = new Users($db);

如果$ db在该文件中定义.

If $db is defined in that file.

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