如何解决期望“Symfony\Component\Security\Core\User\UserInterface”的实例作为第一个参数
我正在尝试为我的用户添加散列到密码中,我遵循了 this symfony 5.3 指南以及何时使用
->setPassword($passwordHasher->hashPassword(
$user,'contraseña'
))
在测试它是否有效时,我收到错误:
期望“Symfony\Component\Security\Core\User\UserInterface”的实例作为第一个参数,但得到“App\Entity\Usuario”。
我不明白,因为它是按照指南显示的字面写的。
这些是我的文件:
用户实体
namespace App\Entity;
use App\Repository\UsuarioRepository;
use Doctrine\ORM\Mapping as ORM;
use Symfony\Component\Security\Core\User\UserInterface;
/**
* @ORM\Entity(repositoryClass=UsuarioRepository::class)
*/
class Usuario
{
/**
* @ORM\Id
* @ORM\GeneratedValue
* @ORM\Column(type="integer")
*/
private $id;
/**
* @ORM\Column(type="string",length=255)
*/
private $nombre;
/**
* @ORM\Column(type="string",length=255)
*/
private $apellidos;
/**
* @ORM\Column(type="string",length=180,unique=true)
*/
private $email;
/**
* @ORM\Column(type="json")
*/
private $roles = [];
/**
* @var string The hashed password
* @ORM\Column(type="string")
*/
private $password;
/**
* @ORM\Column(type="string",length=255)
*/
private $prefijo;
/**
* @ORM\Column(type="string",length=255)
*/
private $telefono;
public function getId(): ?int
{
return $this->id;
}
public function getEmail(): ?string
{
return $this->email;
}
public function setEmail(string $email): self
{
$this->email = $email;
return $this;
}
/**
* A visual identifier that represents this user.
*
* @see UserInterface
*/
public function getUsername(): string
{
return (string) $this->email;
}
/**
* @see UserInterface
*/
public function getRoles(): array
{
$roles = $this->roles;
// guarantee every user at least has ROLE_USER
$roles[] = 'ROLE_USER';
return array_unique($roles);
}
public function setRoles(array $roles): self
{
$this->roles = $roles;
return $this;
}
/**
* @see UserInterface
*/
public function getPassword(): string
{
return $this->password;
}
public function setPassword(string $password): self
{
$this->password = $password;
return $this;
}
/**
* Returning a salt is only needed,if you are not using a modern
* hashing algorithm (e.g. bcrypt or sodium) in your security.yaml.
*
* @see UserInterface
*/
public function getSalt(): ?string
{
return null;
}
/**
* @see UserInterface
*/
public function eraseCredentials()
{
// If you store any temporary,sensitive data on the user,clear it here
// $this->plainPassword = null;
}
public function getNombre(): ?string
{
return $this->nombre;
}
public function setNombre(string $nombre): self
{
$this->nombre = $nombre;
return $this;
}
public function getApellidos(): ?string
{
return $this->apellidos;
}
public function setApellidos(string $apellidos): self
{
$this->apellidos = $apellidos;
return $this;
}
public function getPrefijo(): ?string
{
return $this->prefijo;
}
public function setPrefijo(string $prefijo): self
{
$this->prefijo = $prefijo;
return $this;
}
public function getTelefono(): ?string
{
return $this->telefono;
}
public function setTelefono(string $telefono): self
{
$this->telefono = $telefono;
return $this;
}
}
控制器
namespace App\Controller;
use App\Entity\RegistroUsuario;
use App\Form\RegistroUsuarioForm;
use App\Repository\PrefijoTfnoPaisesRepository;
use App\DataFixtures\UsuarioFixtures;
use App\Entity\Usuario;
use Symfony\Bundle\FrameworkBundle\Controller\AbstractController;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\Routing\Annotation\Route;
use Symfony\Component\PasswordHasher\Hasher\UserPasswordHasherInterface;
/**
* @Route("/usuario",name="usuario")
*/
class UsuarioController extends AbstractController
{
/**
* @Route("/",name="usuario")
*/
public function index(Request $request,PrefijoTfnoPaisesRepository $prefijoTfnoPaisesRepository,UserPasswordHasherInterface $passwordHasher)
{
$paramRegistroForm = new RegistroUsuario();
$registroFrom = $this->createForm(RegistroUsuarioForm::class,$paramRegistroForm);
$registroFrom->handleRequest($request);
$paisesPrefijo = $prefijoTfnoPaisesRepository->findAll();
if ($registroFrom->isSubmitted() && $registroFrom->isValid()){
$manager = $this->getDoctrine()->getManager();
// encode the plain password
$user = new Usuario();
$user
->setNombre("nombre")
->setApellidos("apellidos")
->setEmail("email@a.com")
->setRoles(["rol"])
->setPassword($passwordHasher->hashPassword(
$user,'contraseña'
))
->setPrefijo("+34")
->setTelefono("telefono")
;
$manager->persist($user);
$manager->flush();
dd($request->request->get('registro_usuario_form'));
}
return $this->render('usuario/index.html.twig',[
'registroFormulario' => $registroFrom->createView(),'paisesPrefijo' => $paisesPrefijo
]);
}
}
security.yaml
security:
password_hashers:
App\Entity\Usuario:
# Use native password hasher,which auto-selects the best
# possible hashing algorithm (starting from Symfony 5.3 this is "bcrypt")
algorithm: auto
解决方法
您 Usuario
实体应该实现 UserInterface
。
这是一个非常基本和简单的界面:
interface UserInterface
{
public function getRoles();
public function getPassword();
public function getSalt();
public function eraseCredentials();
public function getUsername();
}
您已经实现了必要的方法,但没有声明接口实现:
class Usuario implements UserInterface
{ /* class implementation */ }
为了向前兼容,您还应该实现 getIdentifier()
(getUsername()
将从 Symfony 6 的界面中删除,并替换为该方法)。
简单地说:
class Usuario implements UserInterface
{
public function getIdentifier():string
{
return $this->email;
}
// rest of the implementation follows
}
此外,由于您的 Usuario
实体还实现了 getPassword()
中的 UserInterface
,因此您还应该实现 PasswordAuthenticatedUserInterface
。这个新接口是在 5.3 中引入的,在 6.0 中 getPassword()
将从
UserInterface
:
class Usuario implements UserInterface,PasswordAuthenticatedUserInterface
{
}
到 6.0 时,您可能也想删除 getSalt()
,因为它将从 UserInterface
中删除(并移至 LegacyPasswordAuthenticatedUserInterface()
),但不太可能该方法可用于您的应用程序(因为任何相对现代的哈希方法都会产生自己的随机盐)。
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。