期望 "Symfony\Component\Security\Core\User\UserInterface" 的实例作为第一个参数

Question

我正在尝试为我的用户的密码添加散列，我遵循了 symfony 5.3 的 this 指南，当我使用

->setPassword($passwordHasher->hashPassword(
                $user,
                'contraseña'
            ))

在测试它是否有效时，出现错误：

Expected an instance of "Symfony\Component\Security\Core\User\UserInterface" as first argument, but got "App\Entity\Usuario".

我不明白，因为它是按照指南显示的字面意思写的。
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

Answer 1

您 Usuario 实体应实施 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，你还应该实现getIdentifier()（getUsername()将从Symfony 6的接口中删除，并用这种方法代替）。

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 UserInterface 的 getPassword()，您也应该实现 PasswordAuthenticatedUserInterface。这个新接口是在 5.3 中引入的，在 6.0 getPassword() 中将被删除 UserInterface:

class Usuario implements UserInterface, PasswordAuthenticatedUserInterface
{
}

到 6.0 时，您可能还想摆脱 getSalt()，因为它将从 UserInterface 中删除（并移至 LegacyPasswordAuthenticatedUserInterface()），但它该方法不太可能在您的应用程序中有任何用处（因为任何相对现代的哈希方法都会产生自己的随机盐）。

Answer 2

您应该通过替换

来修改您的 security.yaml 文件

password_hashers

和

encoders

期望 "Symfony\Component\Security\Core\User\UserInterface" 的实例作为第一个参数

Expected an instance of "Symfony\Component\Security\Core\User\UserInterface" as first argument

php

symfony

symfony5

symfony-security