Question

我在我的Symfony2项目中使用FosUserBundle，我有登录工作，我现在想进一步自定义登录，包括检查id以及用户名，电子邮件和密码。我在Entity文件夹中有一个User实体，我在Repository文件夹中有一个UserRepository

如何自定义Fos UserBundle登录以在其查询中包含ID
我可以采取哪些不同的方式列出
欢迎任何其他改进代码的建议

-thankyou

用户实体：

<?php
    namespace Example\Bundle\ExampleBundle\Entity;

    use FOS\UserBundle\Entity\User as BaseUser;
    use Doctrine\ORM\Mapping as ORM;
    use Symfony\Component\Validator\Constraints as Assert;

 /**
  * User
  *
  * @ORM\Table(name="Example_user")
  * @ORM\Entity(repositoryClass="Example\Bundle\ExampleBundle\Repository\UserRepository")
  */
class User extends BaseUser {

/**
 * @var integer
 *
 * @ORM\Column(name="id", type="integer")
 * @ORM\Id
 * @ORM\GeneratedValue(strategy="AUTO")
 */
protected $id;

/*
*  some Class properties here 
*
*/

/**
 * Get id
 *
 * @return integer 
 */
public function getId() {
    return $this->id;
}

/*
* some setters and getters here
*/
}

用户存储库

<?php
  namespace Example\Bundle\ExampleBundle\Repository;

  use Doctrine\ORM\EntityRepository;
  use Symfony\Component\Security\Core\User\UserProviderInterface;

  use Example\Bundle\ExampleBundle\siteConfig;

  class UserRepository extends EntityRepository implements UserProviderInterface{

public $university = siteConfig::university_id;

/**
 * 
 * @param type $username
 */
public function FindUsernameOrEmailInUniversity($username, $universityId) {

    return  $this->createQueryBuilder('user')
            ->where('user.university_id = :universityId')
            ->andWhere('user.username = :username OR user.email = :email')
            ->setParameter('universityId', $universityId)
            ->setParameter('username', $username)
            ->setParameter('email', $username)
            ->getQuery()
            ->getOneOrNullResult();

}

/**
 * 
 * @param type $username
 */
public function loadUserByUsername($username) {

    $user = $this->FindUsernameOrEmailInUniversity($username, $this->university); //check order of parameters use type hinting

    if(!$user){
        throw new \Symfony\Component\Security\Core\Exception\UsernameNotFoundException('User Name '.$username.' Not Found');
    }

    return $user;
}

/**
 * 
 * @param \Symfony\Component\Security\Core\User\UserInterface $user
 */
public function refreshUser(\Symfony\Component\Security\Core\User\UserInterface $user) {

    return $user;
}

/**
 * 
 * @param type $class
 */
public function supportsClass($class) {

    return ;
}
}

Security.yml

# app/config/security.yml
security:
    encoders:
        FOS\UserBundle\Model\UserInterface: bcrypt

role_hierarchy:
    ROLE_ADMIN:       ROLE_USER
    ROLE_SUPER_ADMIN: ROLE_ADMIN

providers:
    fos_userbundle:
        entity: { class: Example\Bundle\ExampleBundle\Repository\UserRepository }

firewalls:
    main:
        pattern: ^/
        form_login:
            provider: fos_userbundle
            csrf_provider: security.csrf.token_manager # Use form.csrf_provider instead for Symfony <2.4
        logout:       true
        anonymous:    true

access_control:
    - { path: ^/login$, role: IS_AUTHENTICATED_ANONYMOUSLY }
    - { path: ^/register, role: IS_AUTHENTICATED_ANONYMOUSLY }
    - { path: ^/resetting, role: IS_AUTHENTICATED_ANONYMOUSLY }
    - { path: ^/admin/, role: ROLE_ADMIN }

以下是错误

The class 'Example\Bundle\ExampleBundle\Repository\UserRepository' was not found in the chain configured namespaces Example\Bundle\ExampleBundle\Entity, FOS\UserBundle\Entity

Answer 1

在扩展FOS\UserBundle\Doctrine\UserManager的实体文件夹中创建一个类（我假设有一个名为siteConfig.php的类，其中包含由db查询更新的静态University_id字段）

<?php
namespace Example\Bundle\ExampleBundle\Entity

use FOS\UserBundle\Doctrine\UserManager as BaseUserManager;
use Doctrine\ORM\EntityManager;
use Symfony\Component\Security\Core\Encoder\EncoderFactoryInterface;
use FOS\UserBundle\Util\CanonicalizerInterface;
use FOS\UserBundle\Model\UserInterface;
use Example\Bundle\ExampleBundle\siteConfig;

class UserManager extends BaseUserManager {
public function __construct(EncoderFactoryInterface $encoderFactory, CanonicalizerInterface $usernameCanonicalizer, 
                                    CanonicalizerInterface $emailCanonicalizer, EntityManager $em, $class) {

    parent::__construct($encoderFactory, $usernameCanonicalizer, $emailCanonicalizer, $em, $class);

}
/**
 * this overides the findUserByUsernameOrEmail in FOS\UserBundle\Doctrine\UserManager
 **/
public function findUserByUsernameOrEmail($usernameOrEmail) {
    if (filter_var($usernameOrEmail, FILTER_VALIDATE_EMAIL)) {
        return $this->findUserBy(array('emailCanonical' => $this->canonicalizeEmail($usernameOrEmail), 'university' => siteConfig::$university_id));
    }

    return $this->findUserBy(array('usernameCanonical' => $this->canonicalizeUsername($usernameOrEmail), 'university' => siteConfig::$university_id));
}
}

用户管理器的依赖注入，在services.yml内部

ExampleUserManager:
class: namespace Example\Bundle\ExampleBundle\Entity\UserManager
     arguments: [@security.encoder_factory, @fos_user.util.username_canonicalizer, @fos_user.util.email_canonicalizer, @fos_user.entity_manager, namespace Example\Bundle\ExampleBundle\Entity\User]

在config.yml内部，在fos_user

下添加以下配置

service:
    user_manager: ExampleUserManager

在提供商下的security.yml内添加以下内容

providers:
    fos_userbundle:
        id: fos_user.user_provider.username_email

Answer 2

首先是错误类型 - 在为实体行定义存储库类时需要额外的paranthese：

/**
 * User
 *
 * @ORM\Table(name="Example_user")
 * @ORM\Entity(repositoryClass="Example\Bundle\ExampleBundle\Repository\UserRepository")
 */
 class User extends BaseUser {

试试这个，我对你的方法做了一些修改：

/**
 * @param type $username
 */
 public function FindUsernameOrEmailInUniversity($username, University $university) {

      return  $this->createQueryBuilder('user') //It should be alias for your entity name, and it could be anything you write, like 'user' or just 'u'
          ->where('user.university = :university')
          ->andWhere('user.username = :username OR user.email = :email')
          ->setParameter('university', $university)
          ->setParameter('username', $username)
          ->setParameter('email', $username)
          ->getQuery()
          ->getOneOrNullResult();

}

你也有大学，电子邮件和用户名的getter，setter和属性吗？我假设您希望将大学作为与用户的关系，因此您需要设置正确的学说关系。

Symfony2 FosUserBundle自定义登录查询

2 个答案: