Symfony4 + jwt-auth rescipe 总是返回 "code":401,"message":"Bad credentials"

Posted 2023-03-10

【中文标题】Symfony4 + jwt-auth rescipe 总是返回 "code":401,"message":"Bad credentials"

【英文标题】：Symfony4 + jwt-auth rescipe always return "code":401,"message":"Bad credentials"Symfony4 + jwt-auth rescipe 总是返回 "code":401,"message":"Bad credentials"

【发布时间】：2018-09-10 12:48:25

【问题描述】：

我有一个 Symfony4 安装，其中安装了一些常用的 Flex 包

制作 jwt-auth 注释 行为 phpunit 服务器

我有这个路由文件：

api_login_check:
    path: /api/login_check

我有这个配置：

security:

    encoders:
        App\Security\User: plaintext

    providers:
        app.provider:
            id: App\Security\UserProvider

    firewalls:

        login:
            pattern:  ^/api/login
            stateless: true
            anonymous: true
            provider: app.provider
            form_login:
                check_path: /api/login_check
                success_handler: lexik_jwt_authentication.handler.authentication_success
                failure_handler: lexik_jwt_authentication.handler.authentication_failure
                require_previous_session: false

        api:
            pattern:   ^/api
            stateless: true
            provider: app.provider
            guard:
                authenticators:
                    - lexik_jwt_authentication.jwt_token_authenticator

    access_control:
        -  path: ^/api/login, roles: IS_AUTHENTICATED_ANONYMOUSLY 
        -  path: ^/api,       roles: IS_AUTHENTICATED_FULLY 

还有这个用户：

<?php

namespace App\Security;

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

class User implements UserInterface, EquatableInterface

    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 isEqualTo(UserInterface $user)
    
        if (!$user instanceof User) 
            return false;
        

        if ($this->password !== $user->getPassword()) 
            return false;
        

        if ($this->salt !== $user->getSalt()) 
            return false;
        

        if ($this->username !== $user->getUsername()) 
            return false;
        

        return true;
    

最后是这个用户提供者：

<?php

namespace App\Security;

use App\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)
    
        $username = 'senso';
        $password = 'rario';
        $salt     = 'sale';
        $roles    = ['ROLE_ADMIN'];

        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 User::class === $class;
    

当我跑步时

curl -X POST http://localhost:8000/api/login_check -d _username=senso -d _password=rario "code":401,"message":"凭证错误"

我总是得到

"code":401,"message":"凭证错误"

我的问题是：

如何解决这个问题？ 为什么从未调用过 UserProvider::loadUserByUsername()？

【问题讨论】：

您可以发布您尝试过的正确curl 吗？我想检查一下：网址是否正确（而不是http://localhost:800ogin_check），选择的密码是rario，而不是test。

可以在login防火墙下添加provider: app.provider吗？

完成了，还是不行

我假设您在路由器配置中注册了/api/login_check，您能确认一下吗？

是的。我已经回答了问题

【参考方案1】：

如果您使用PlaintextPasswordEncoder 并为您的User 类提供salt，那么您的User 的getPassword 方法应该返回plain_passwordsalt。

在这种情况下，

$username = 'senso';
$password = 'rariosale';
$salt     = 'sale';
$roles    = ['ROLE_ADMIN'];

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

或

$username = 'senso';
$password = 'rario';
$salt     = '';
$roles    = ['ROLE_ADMIN'];

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

会很好用。