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我无法登录到我的受限区域。我正在关注symfony.com上的教程。Symfony3使用UserLoaderInterface扩展UserRepository并使用加密，登录不起作用
我会提供我的security.yml文件。
# To get started with security, check out the documentation:
# http://symfony.com/doc/current/book/security.html
security:
role_hierarchy:
ROLE_ADMIN: ROLE_USER
ROLE_SUPER_ADMIN: [ROLE_ADMIN, ROLE_ALLOWED_TO_SWITCH]
encoders:
Symfony\Component\Security\Core\User\User:
algorithm: bcrypt
cost: 12
AppBundle\Entity\User:
algorithm: bcrypt
cost: 12
providers:
# http://symfony.com/doc/current/book/security.html#where-do-users-come-from-user-providers
our_db_provider:
entity:
class: AppBundle:User
# if you're using multiple entity managers
# manager_name: customer
firewalls:
# disables authentication for assets and the profiler, adapt it according to your needs
dev:
pattern: ^/(_(profiler|wdt)|css|images|js)/
security: false
main:
pattern: ^/
http_basic: ~
provider: our_db_provider
anonymous: ~
# activate different ways to authenticate
#http_basic: ~
# http://symfony.com/doc/current/book/security.html#a-configuring-how-your-users-will-authenticate
# form_login: ~
# http://symfony.com/doc/current/cookbook/security/form_login_setup.html
logout:
path: /logout
target:/
access_control:
# require ROLE_ADMIN for /admin*
#- { path: ^/admin, roles: ROLE_ADMIN }
下一页这里是我的User类
<?php
// src/AppBundle/Entity/User.php
namespace AppBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
//use Symfony\Component\Security\Core\User\UserInterface;
use Symfony\Component\Security\Core\User\AdvancedUserInterface;
/**
* @ORM\Table(name="app_users")
* @ORM\Entity(repositoryClass="AppBundle\Repository\UserRepository")
*/
class User implements AdvancedUserInterface, \Serializable
{
/**
* @ORM\Column(type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* @ORM\Column(type="string", length=25, unique=true)
*/
private $username;
/**
* @ORM\Column(type="string", length=64)
*/
private $password;
/**
* @ORM\Column(type="string", length=60, unique=true)
*/
private $email;
/**
* @ORM\Column(name="is_active", type="boolean")
*/
private $isActive;
public function __construct()
{
$this->isActive = true;
// may not be needed, see section on salt below
// $this->salt = md5(uniqid(null, true));
}
public function isAccountNonExpired()
{
return true;
}
public function isAccountNonLocked()
{
return true;
}
public function isCredentialsNonExpired()
{
return true;
}
public function isEnabled()
{
return $this->isActive;
}
public function getUsername()
{
return $this->username;
}
public function getSalt()
{
// you *may* need a real salt depending on your encoder
// see section on salt below
return null;
}
public function getPassword()
{
return $this->password;
}
public function getRoles()
{
return array('ROLE_USER','ROLE_ADMIN');
}
public function eraseCredentials()
{
}
/** @see \Serializable::serialize() */
public function serialize()
{
return serialize(array(
$this->id,
$this->username,
$this->password,
$this->active
// see section on salt below
// $this->salt,
));
}
/** @see \Serializable::unserialize() */
public function unserialize($serialized)
{
list (
$this->id,
$this->username,
$this->password,
$this->active
// see section on salt below
// $this->salt
) = unserialize($serialized);
}
/**
* Get id
*
* @return integer
*/
public function getId()
{
return $this->id;
}
/**
* Set username
*
* @param string $username
*
* @return User
*/
public function setUsername($username)
{
$this->username = $username;
return $this;
}
/**
* Set password
*
* @param string $password
*
* @return User
*/
public function setPassword($password)
{
$this->password = $password;
return $this;
}
/**
* Set email
*
* @param string $email
*
* @return User
*/
public function setEmail($email)
{
$this->email = $email;
return $this;
}
/**
* Get email
*
* @return string
*/
public function getEmail()
{
return $this->email;
}
/**
* Set isActive
*
* @param boolean $isActive
*
* @return User
*/
public function setIsActive($isActive)
{
$this->isActive = $isActive;
return $this;
}
/**
* Get isActive
*
* @return boolean
*/
public function getIsActive()
{
return $this->isActive;
}
}
，最终用户系统信息库类
<?php
// src/AppBundle/Repository/UserRepository.php
namespace AppBundle\Repository\Entity;
use Symfony\Bridge\Doctrine\Security\User\UserLoaderInterface;
use Symfony\Component\Security\Core\User\UserInterface;
use Symfony\Component\Security\Core\Exception\UsernameNotFoundException;
use Doctrine\ORM\EntityRepository;
class UserRepository extends EntityRepository implements UserLoaderInterface
{
public function loadUserByUsername($username)
{
$user = $this->createQueryBuilder('u')
->where('u.username = :username OR u.email = :email')
->setParameter('username', $username)
->setParameter('email', $username)
->getQuery()
->getOneOrNullResult();
if (null === $user) {
$message = sprintf(
'Unable to find an active admin AppBundle:User object identified by "%s".',
$username
);
throw new UsernameNotFoundException($message);
}
return $user;
}
}
?>
我对管理的路线看起来象下面这样：
<?php
namespace AppBundle\Controller;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\Response;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Security;
class DefaultController extends Controller
{
/**
* @Route("/", name="homepage")
*/
public function indexAction(Request $request)
{
// replace this example code with whatever you need
return $this->render('default/index.html.twig', [
'base_dir' => realpath($this->container->getParameter('kernel.root_dir').'/..'),
]);
}
/**
* @Route("/admin")
* @Security("has_role('ROLE_USER') or has_role('ROLE_ADMIN')")
*/
public function adminAction()
{
return new Response('<html><body>Admin page!</body></html>');
}
}
我试图按照说明在这里登录http://symfony.com/doc/current/cookbook/security/entity_provider.html#using-a-custom-query-to-load-the-user，以便我可以用用户名或电子邮件登录。当我去/管理但是不管我怎么输入电子邮件或用户名到http_basic提示我似乎无法进入。
我猜它可能有一些与编码可能吗？所以在security.yml文件中我使用了两种编码器类型，但它只用1或另一种就不起作用。
记住/管理员具有安全角色的用户或管理员，这样以来我默认的类ROLE_USER的恢复作用应该还是能够得到。
如果有错误日志的地方我米不知道如何找到它。因为我对Symphony还是个新手，所以在这里可以获得帮助。
编辑：
忘记在数据库中提到，目前，我有以下::
1 admin $2y$12$qvLb/T2Xs4aWsFU6D4U2f.gmZi/blKYtspXeqiPLrHPFOPxwMaHY. [email protected] 1
2 joe $2y$12$M/7nTIEfQ1Ajpr/IhmEVoejskvt0dIb/FfIvT8i9LXdSR95zjT5OS [email protected] 1
的列是ID，用户名，密码，电子邮件，IS_ACTIVE
加密我手动使用：bin/console security:encode-password，然后放入数据库字段。这工作以前的一些其他记录在我做的测试，但以防万一这是在这里把这个问题。我试着在数据库中放入纯文本，然后输入以登录并且不起作用。
谢谢！
在你的实体配置你的用户仓库是'AppBundle \ Repository \ UserRepository'，但是'UserRepository'类中的名字空间是'AppBundle \ Entity'。 – xabbuh
我改变了命名空间，但仍然无法登录。当我不断尝试通过http_basic提示符输入用户名和密码时，提示符就会继续进入无限循环。好，但是，有什么想法？ –
要清楚我将命名空间更改为... namespace AppBundle \ Repository \ Entity; –