[英]How do you inject a Symfony repository into a custom user provider?
I have successfully injected repositories into services before, but can't get injection to work in a custom user provider. 我之前已经成功地将存储库注入了服务,但是无法在自定义用户提供程序中工作。 It seems like the __construct() method is never getting called, so the repository is never made available.
似乎从未调用过__construct()方法,因此该存储库从未可用。 The way you are supposed to make a user provider is to implement the interface UserProviderInterface, but the UserProvider class you create doesn't extend another class that would have a constructor.
应该使用户提供程序的方式是实现接口UserProviderInterface,但是您创建的UserProvider类不会扩展具有构造函数的另一个类。 When I try to access the repository using $this->personRepository->findStatusByUsername($username), I get the error:
当我尝试使用$ this-> personRepository-> findStatusByUsername($ username)访问存储库时,出现错误:
Using $this when not in object context
不在对象上下文中时使用$ this
One reason I think the constructor is never getting called is that it has two assignments to $this, but I'm not getting an error message about them. 我认为构造函数从未被调用的原因之一是它有两个给$ this的赋值,但是我没有得到关于它们的错误消息。
I need the Person repository to be injected so that I can check whether or not the person trying to authenticate is already approved in the app (status in the Person table = approved.) 我需要注入人员资料库,以便可以检查应用程序中是否已批准尝试进行身份验证的人员(人员表中的状态=批准)。
My services.yml file has these settings: 我的services.yml文件具有以下设置:
parameters:
ginsberg_transportation.user.class: Ginsberg\TransportationBundle\Services\User
user_provider.class: Ginsberg\TransportationBundle\Security\User\UserProvider
services:
ginsberg_user:
class: "%ginsberg_transportation.user.class%"
user_provider:
class: "%user_provider.class%"
arguments:
["@ginsberg_person.person_repository", "@logger"]
ginsberg_person.person_repository:
class: Doctrine\ORM\EntityRepository
factory_service: doctrine.orm.default_entity_manager
factory_method: getRepository
arguments:
- Ginsberg\TransportationBundle\Entity\Person
My UserProvider.php class starts like this: 我的UserProvider.php类开始如下:
namespace Ginsberg\TransportationBundle\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;
use Ginsberg\TransportationBundle\Entity\Person;
use Doctrine\ORM\EntityRepository;
class UserProvider implements UserProviderInterface
{
private $personRepository;
private $logger;
public function __construct(\Ginsberg\TransportationBundle\Entity\PersonRepository $personRepository, \Monolog\Logger $logger) {
$this->personRepository = $personRepository;
$this->logger = $logger;
}
public function loadUserByUsername($uniqname)
{
$password = "admin";
$salt = "";
$roles = array();
if (self::is_authenticated() && self::is_approved()) {
if (self::is_superuser()) {
$roles[] = 'ROLE_SUPER_ADMIN';
} elseif (self::is_admin()) {
$roles[] = 'ROLE_ADMIN';
} elseif (self::is_eligible() && self::is_approved()) {
$roles[] = 'ROLE_USER';
}
return new User($uniqname, $password, $salt, $roles);
}
throw new UsernameNotFoundException(
sprintf('Username "%s" does not exist.', $uniqname));
}
...
public static function is_approved()
{
$uniqname = self::get_uniqname();
$status = $this->personRepository->findStatusByUsername($username);
return($status == 'approved') ? TRUE : FALSE;
}
I also tried using property injection too, but that didn't work either. 我也尝试过使用属性注入,但这也不起作用。
Should I just make my UserProvider class extend another appropriate class that would make use of a constructor? 我是否应该让UserProvider类扩展使用构造函数的另一个合适的类? If so, any thoughts on what class would be fitting?
如果是这样,对什么课程合适的想法?
Thanks for any suggestions. 感谢您的任何建议。
The problem is that the method is_approved()
is static
. 问题是方法
is_approved()
是static
。 You cannot use $this
in a static context. 您不能在静态上下文中使用
$this
。 You need to declare your method non-static. 您需要声明您的方法是非静态的。
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