如何以编程方式登录/验证用户？

Question

我想在注册过程后立即让用户登录，而不通过登录表单。

这可能吗 ？我找到了 FOSUserBundle 的解决方案，但我没有在我实际从事的项目中使用它。

这是我的 security.yml，我正在使用两个防火墙。纯文本编码器仅用于测试。

 security:
    encoders:
        Symfony\Component\Security\Core\User\User: plaintext
        Ray\CentralBundle\Entity\Client: md5

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

    providers:
        in_memory:
            users:
                admin: { password: admin, roles: [ 'ROLE_ADMIN' ] }
        entity:
            entity: { class: Ray\CentralBundle\Entity\Client, property: email }

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

        user_login:
            pattern:    ^/user/login$
            anonymous:  ~

        admin_login:
            pattern:    ^/admin/login$
            anonymous:  ~

        admin:
            pattern:    ^/admin
            provider: in_memory
            form_login:
                check_path: /admin/login/process
                login_path: /admin/login
                default_target_path: /admin/dashboard
            logout:
                path:   /admin/logout
                target: /

        site:
            pattern:    ^/
            provider: entity
            anonymous:  ~
            form_login:
                check_path: /user/login/process
                login_path: /user/login
                default_target_path: /user
            logout:
                path:   /user/logout
                target: /

    access_control:
        - { path: ^/user/login, roles: IS_AUTHENTICATED_ANONYMOUSLY }
        - { path: ^/admin/login, roles: IS_AUTHENTICATED_ANONYMOUSLY }
        - { path: ^/user, roles: ROLE_USER }
        - { path: ^/admin, roles: ROLE_ADMIN }
        - { path: ^/, roles: IS_AUTHENTICATED_ANONYMOUSLY }

原文由 rayfranco 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

是的，您可以通过类似于以下内容的方式执行此操作：

 use Symfony\Component\EventDispatcher\EventDispatcher,
    Symfony\Component\Security\Core\Authentication\Token\UsernamePasswordToken,
    Symfony\Component\Security\Http\Event\InteractiveLoginEvent;

public function registerAction()
{
    // ...
    if ($this->get("request")->getMethod() == "POST")
    {
        // ... Do any password setting here etc

        $em->persist($user);
        $em->flush();

        // Here, "public" is the name of the firewall in your security.yml
        $token = new UsernamePasswordToken($user, $user->getPassword(), "public", $user->getRoles());

        // For older versions of Symfony, use security.context here
        $this->get("security.token_storage")->setToken($token);

        // Fire the login event
        // Logging the user in above the way we do it doesn't do this automatically
        $event = new InteractiveLoginEvent($request, $token);
        $this->get("event_dispatcher")->dispatch("security.interactive_login", $event);

        // maybe redirect out here
    }
}

当您将令牌设置到上下文中时，最后触发的事件不会自动完成，而通常在使用登录表单或类似表单时会自动完成。因此将它包括在这里的原因。您可能需要根据您的用例调整使用的令牌类型 - 上面显示的 UsernamePasswordToken 是核心令牌，但如果需要，您可以使用其他令牌。

编辑：根据下面 Franco 的评论，调整以上代码以解释“public”参数，并将用户角色添加到令牌创建中。

原文由 richsage 发布，翻译遵循 CC BY-SA 3.0 许可协议

Answer 2

对于 symfony 6，以下工作：

 public function login(User $user, Request $request, UserCheckerInterface $checker, UserAuthenticatorInterface $userAuthenticator, FormLoginAuthenticator $formLoginAuthenticator): void
{
    $checker->checkPreAuth($user);
    $userAuthenticator->authenticateUser($user, $formLoginAuthenticator, $request);
}

您可以选择将此功能移动到服务中，以便依赖注入更容易：

 # config/services.yaml

services:
    App\Service\LoginService:
        arguments:
            $formLoginAuthenticator: '@security.authenticator.form_login.main'

 # src/Service/LoginService.php

namespace App\Service;

use App\Entity\User;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\Security\Core\User\UserCheckerInterface;
use Symfony\Component\Security\Http\Authentication\UserAuthenticatorInterface;
use Symfony\Component\Security\Http\Authenticator\FormLoginAuthenticator;

class LoginService
{
    private UserCheckerInterface $checker;
    private UserAuthenticatorInterface $userAuthenticator;
    private FormLoginAuthenticator $formLoginAuthenticator;

    /**
     * @param UserCheckerInterface $checker
     * @param UserAuthenticatorInterface $userAuthenticator
     * @param FormLoginAuthenticator $formLoginAuthenticator
     */
    public function __construct(UserCheckerInterface $checker, UserAuthenticatorInterface $userAuthenticator, FormLoginAuthenticator $formLoginAuthenticator)
    {
        $this->checker = $checker;
        $this->userAuthenticator = $userAuthenticator;
        $this->formLoginAuthenticator = $formLoginAuthenticator;
    }

    public function login(User $user, Request $request): void
    {
        $this->checker->checkPreAuth($user);
        $this->userAuthenticator->authenticateUser($user, $this->formLoginAuthenticator, $request);
    }
}

Source 是一个 RFC ，请求一种更简单的编程登录方式。这已经 实现，并将与 symfony 6.2 一起发布。

原文由 Florian Moser 发布，翻译遵循 CC BY-SA 4.0 许可协议