4

当我尝试登录 symfony2 时，我总是收到消息：“凭据错误”。我这样做是基于http://symfony.com/doc/current/cookbook/security/custom_provider.html。请帮我弄清楚，问题出在哪里？提前致谢。

security.yml 看起来像这样

security:
    encoders:
        Zags\UserBundle\Security\User\User: plaintext

    role_hierarchy:
        ROLE_ADMIN:       ROLE_USER
        ROLE_SUPER_ADMIN: [ROLE_USER, ROLE_ADMIN, ROLE_ALLOWED_TO_SWITCH]

    providers:
        webservice:
            id: zags_user_provider

    firewalls:
        dev:
            pattern:  ^/(_(profiler|wdt)|css|images|js)/
            security: false

        login_firewall:
            pattern:    ^/login$
            anonymous:  ~

        secured_area:
            pattern:    ^/
            anonymous: ~
            form_login:
                login_path:  /login
                check_path:  /login_check
            #anonymous: ~
            #http_basic:
            #    realm: "Secured Demo Area"

    access_control:
        - { path: ^/gender_type, roles: ROLE_USER }
        #- { path: ^/_internal/secure, roles: IS_AUTHENTICATED_ANONYMOUSLY, ip: 127.0.0.1 }

我已将这些行添加到 routing.yml

login:
    pattern:   /login
    defaults:  { _controller: ZagsUserBundle:Security:login }
login_check:
    pattern:   /login_check

User.php 类看起来像这样

<?php    
namespace Zags\UserBundle\Security\User;

use Symfony\Component\Security\Core\User\UserInterface;

class User implements UserInterface
{
    private $username;
    private $password;
    private $salt;
    private $roles;

    public function __construct($username, $password, $salt, array $roles)
    {
        $this->username = $username;
        $this->password = $password;
        $this->salt = $salt;
        $this->roles = $roles;
    }

    public function getRoles()
    {
        return $this->roles;
    }

    public function getPassword()
    {
        return $this->password;
    }

    public function getSalt()
    {
        return $this->salt;
    }

    public function getUsername()
    {
        return $this->username;
    }

    public function eraseCredentials()
    {
    }

    public function equals(UserInterface $user)
    {
        return $user->getUsername() === $this->username;
    }
}
?>

所以这是我的 UserProvider.php 类

<?php

namespace Zags\UserBundle\Security\User;

use Symfony\Component\Security\Core\User\UserProviderInterface;
use Symfony\Component\Security\Core\User\UserInterface;
use Symfony\Component\Security\Core\Exception\UsernameNotFoundException;
use Symfony\Component\Security\Core\Exception\UnsupportedUserException;

class UserProvider implements UserProviderInterface
{
    public function loadUserByUsername($username)
    {
        // make a call to your webservice here
        $userData = array("username" => "latysh", "password" => "123", "salt" => "123", "roles" => array('ROLE_USER'));
        // pretend it returns an array on success, false if there is no user

        if ($userData) {
            $username = $userData['username'];
            $password = $userData['password'];
            $salt = $userData['salt'];
            $roles = $userData['roles'];
            // ...

            return new User($username, $password, $salt, $roles);
        }

        throw new UsernameNotFoundException(sprintf('Username "%s" does not exist.', $username));
    }

    public function refreshUser(UserInterface $user)
    {
        if (!$user instanceof User) {
            throw new UnsupportedUserException(sprintf('Instances of "%s" are not supported.', get_class($user)));
        }

        return $this->loadUserByUsername($user->getUsername());
    }

    public function supportsClass($class)
    {
        return $class === 'Zags\UserBundle\Security\User\User';
    }
}
?>

services.yml 看起来像这样

parameters:
    zags_user_provider.class: Zags\UserBundle\Security\User\UserProvider

services:
    zags_user_provider:
        class: "%zags_user_provider.class%"

安全控制器.php

<?php
namespace Zags\UserBundle\Controller;

use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Symfony\Component\Security\Core\SecurityContext;

class SecurityController extends Controller
{
    public function loginAction()
    {
        $request = $this->getRequest();
        $session = $request->getSession();

        // get the login error if there is one
        if ($request->attributes->has(SecurityContext::AUTHENTICATION_ERROR)) {
            $error = $request->attributes->get(SecurityContext::AUTHENTICATION_ERROR);
        } else {
            $error = $session->get(SecurityContext::AUTHENTICATION_ERROR);
            $session->remove(SecurityContext::AUTHENTICATION_ERROR);
        }

        return $this->render('ZagsUserBundle:Security:login.html.twig', array(
            // last username entered by the user
            'last_username' => $session->get(SecurityContext::LAST_USERNAME),
            'error'         => $error,
        ));
    }
}
?>

和 login.html.twig

{% if error %}
    <div>{{ error.message }}</div>
{% endif %}

<form action="{{ path('login_check') }}" method="post">
    <label for="username">Username:</label>
    <input type="text" id="username" name="_username" value="{{ last_username }}" />

    <label for="password">Password:</label>
    <input type="password" id="password" name="_password" />

    {#
        If you want to control the URL the user is redirected to on success (more details below)
        <input type="hidden" name="_target_path" value="/account" />
    #}

    <button type="submit">login</button>
</form>
4

2 回答 2

5

找到了我的问题的答案。感谢machour的回应。问题出在 SALT 上。所以我将 User.php 类更新为

public function getSalt()
    {
        return '';
    }

然后它正确登录，或者我应该用盐编码密码才能成功登录。如果有人知道该怎么做，请将其写为答案，YAHOO ))

于 2012-12-01T05:08:48.920 回答
2 
$factory = $this->get('security.encoder_factory');
$encoder = $factory->getEncoder($user);
$pass = $encoder->encodePassword($user->getPassword(), $user->getSalt());
于 2013-10-19T21:25:28.123 回答