gpt4 book ai didi

php - 具有两种身份验证方法的 FOSUserBundle

转载 作者:搜寻专家 更新时间:2023-10-31 21:23:02 25 4
gpt4 key购买 nike

我有一个 Symfony 2.8 项目,我在其中使用了 FOSUserBundleFOSUser 身份验证方法使用 fos_user 表来识别和验证凭据以及使用 sha512 加密的 key 。

是否可以修改或扩展一些类,以便在表 fos_user 中找不到用户时,在使用 加密 key 的用户表中查找它>md5?

根据madshvero的日落更新:

我创建了一个用户类:

namespace AppBundle\Security\User;

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

class WebserviceUser 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 WebserviceUser) {
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;
}
}

我还创建了用户提供程序:

namespace AppBundle\Security\User;

use AppBundle\Security\User\WebserviceUser;
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 WebserviceUserProvider implements UserProviderInterface
{
public function loadUserByUsername($username)
{
// make a call to your webservice here
$userData = true;
// pretend it returns an array on success, false if there is no user

if ($userData) {
$username = 'prueba';
$password = 'e10adc3949ba59abbe56e057f20f883e'; // md5('123456')
$salt = '';`enter code here`
$roles = [ROLE_SUPER_ADMIN];
// ...

return new WebserviceUser($username, $password, $salt, $roles);
}

throw new UsernameNotFoundException(
sprintf('Username "%s" does not exist.', $username)
);
}

public function refreshUser(UserInterface $user)
{
if (!$user instanceof WebserviceUser) {
throw new UnsupportedUserException(
sprintf('Instances of "%s" are not supported.', get_class($user))
);
}

return $this->loadUserByUsername($user->getUsername());
}

public function supportsClass($class)
{
return WebserviceUser::class === $class;
}
}

并修改security.yml:

security:
access_denied_url: /login
encoders:
FOS\UserBundle\Model\UserInterface: sha512
AppBundle\Security\User\WebserviceUser: md5

providers:
chain_provider:
chain:
providers: [fos_userbundle, webservice]

fos_userbundle:
id: fos_user.user_provider.username

webservice:
id: app.webservice_user_provider

firewalls:
main:
pattern: ^/
fr3d_ldap: ~
form_login:
provider: fos_userbundle
check_path: /login_check
login_path: /login
always_use_default_target_path: true
default_target_path: /
logout:
path: /logout
target: /login
anonymous: true
dev:
pattern: ^/(_(profiler|wdt)|css|images|js)/
security: false
login:
pattern: ^/login$
security: false

当然我也修改了services.yml添加服务app.webservice_user_provider: 服务: app.form.group: 类别:AppBundle\Form\GroupFormType 标签: - { 名称:form.type,别名:app_group_registration }

    app.form.user:
class: AppBundle\Form\ProfileFormType
tags:
- { name: form.type, alias: app_user_profile }

app.webservice_user_provider:
class: AppBundle\Security\User\WebserviceUserProvider

完成后,行为是系统允许 fos_user 提供者的用户访问,但不允许我的自定义提供者的用户访问。什么失败了?

这是日志:

[2017-02-16 11:37:08] request.INFO:匹配路由“fos_user_security_check”。 {"route_parameters":{"_controller":"AppBundle\Controller\SecurityController::checkAction","_route":"fos_user_security_check"},"request_uri":"http://127.0.0.1:8000/app_dev.php/login_check"} []

[2017-02-16 11:37:08] doctrine.DEBUG: SELECT t0.id_aspirante AS id_aspirante1, t0.correo AS correo2, t0.clave AS clave3, t0.status_cuenta AS status_cuenta4 FROM aspirantes2 t0 WHERE t0.correo = ? LIMIT 1 ["userFoo"] []

[2017-02-16 11:37:08] doctrine.DEBUG: SELECT t0.id_aspirante AS id_aspirante1, t0.correo AS correo2, t0.clave AS clave3, t0.status_cuenta AS status_cuenta4 FROM aspirantes2 t0 WHERE t0.correo = ? LIMIT 1 ["userFoo"] []

[2017-02-16 11:37:08] security.INFO: Authentication request failed. {"exception":"[object] (Symfony\\Component\\Security\\Core\\Exception\\AuthenticationServiceException(code: 0): The user provider must return a UserInterface object. at /home/userx/projects/myproj/vendor/symfony/symfony/src/Symfony/Component/Security/Core/Authentication/Provider/DaoAuthenticationProvider.php:94, Symfony\\Component\\Security\\Core\\Exception\\AuthenticationServiceException(code: 0): The user provider must return a UserInterface object. at /home/userx/projects/myproj/vendor/symfony/symfony/src/Symfony/Component/Security/Core/Authentication/Provider/DaoAuthenticationProvider.php:86)"} []

[2017-02-16 11:37:08] security.DEBUG: Authentication failure, redirect triggered. {"failure_path":"/login"} []

[2017-02-16 11:37:08] request.INFO: Matched route "fos_user_security_login". {"route_parameters":{"_controller":"AppBundle\\Controller\\SecurityController::loginAction","_route":"fos_user_security_login"},"request_uri":"http://127.0.0.1:8000/app_dev.php/login"} []

[2017-02-16 11:37:08] security.INFO: Populated the TokenStorage with an anonymous Token. [] []

[2017-02-16 11:37:08] request.INFO: Matched route "_wdt". {"route_parameters":{"_controller":"web_profiler.controller.profiler:toolbarAction","token":"c368df","_route":"_wdt"},"request_uri":"http://127.0.0.1:8000/app_dev.php/_wdt/c368df"} []

[2017-02-16 11:37:08] security.INFO: Populated the TokenStorage with an anonymous Token. [] []

[2017-02-16 11:37:08] security.DEBUG: Access denied, the user is not fully authenticated; redirecting to authentication entry point. {"exception":"[object] (Symfony\\Component\\Security\\Core\\Exception\\AccessDeniedException(code: 403): Access Denied. at /home/userx/projects/myproj/vendor/symfony/symfony/src/Symfony/Component/Security/Http/Firewall/AccessListener.php:70)"} []

[2017-02-16 11:37:08] security.DEBUG: Calling Authentication entry point. [] []

[2017-02-16 11:37:08] request.INFO: Matched route "fos_user_security_login". {"route_parameters":{"_controller":"AppBundle\\Controller\\SecurityController::loginAction","_route":"fos_user_security_login"},"request_uri":"http://127.0.0.1:8000/app_dev.php/login"} []

[2017-02-16 11:37:08] security.INFO: Populated the TokenStorage with an anonymous Token. [] []

最佳答案

阅读文档后,我能够理解身份验证方法的逻辑,并且发现将数据库中的身份验证提供程序用于我的项目更加方便和容易。真的是一个非常简单的解决方案,记录在:How to Load Security Users from the Database (the Entity Provider)

在我使用 FOSUserBundle 的情况下,有两个注意事项:

  1. 它们必须存在两种身份验证方法:一种由 FOSUserBundle 提供,另一种由 MyBundle 提供。
  2. 身份验证过程应尝试以两种方法对用户进行身份验证。

为此,除了 How to Load Security Users from the Database (the Entity Provider) 中给出的建议外,您必须修改 security.yml 的某些部分,使它们看起来像这样::

encoders:
// The database method of FOSUserBundle
FOS\UserBundle\Model\UserInterface:
algorithm: sha512
// The data base method of mine
MyBundle\Entity\MyEntity:
//This values depends on how the keys were encrypted in the database
algorithm: md5
encode_as_base64: false
iterations: 0

providers:
chain_provider:
chain:
providers: [fos_userbundle, aspirante_db]

fos_userbundle:
id: fos_user.user_provider.username

myentity_db:
entity: { class: MyBundle\Entity\MyEntity, property: username }

firewalls:
main:
pattern: ^/
fr3d_ldap: ~
form_login:
provider: chain_provider //This is the important change
check_path: /login_check
login_path: /login
always_use_default_target_path: true
default_target_path: /
logout:
path: /logout
target: /login
anonymous: true
dev:
pattern: ^/(_(profiler|wdt)|css|images|js)/
security: false
login:
pattern: ^/login$
security: false

就是这样。我希望将来有人可以利用这篇文章。非常感谢 madshvero 的指导。

关于php - 具有两种身份验证方法的 FOSUserBundle,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42251200/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com