Question

我有几条以/experiment/{id}/...开头的路线，我厌倦了重写相同的逻辑来检索已签名的用户实验。我想我可以重构我的代码，但我猜@ParamConverter将是一个更好的解决方案。

如何重写以下代码以利用Symfony的@ParamConverter功能？

/**
 * Displays details about an Experiment entity, including stats.
 *
 * @Route("/experiment/{id}/report", requirements={"id" = "\d+"}, name="experiment_report")
 * @Method("GET")
 * @Template()
 * @Security("has_role('ROLE_USER')")
 */
public function reportAction(Request $request, $id)
{
    $em = $this->getDoctrine()->getManager();

    $experiment = $em->getRepository('AppBundle:Experiment')
        ->findOneBy(array(
            'id'   => $id,
            'user' => $this->getUser(),
        ));

    if (!$experiment) {
        throw $this->createNotFoundException('Unable to find Experiment entity.');
    }

    // ...
}

实验实体的composite primary keys如下：

class Experiment
{
    /**
     * @var integer
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     */
    protected $id;

    /**
     * @var integer
     *
     * @ORM\Column(name="user_id", type="integer")
     * @ORM\Id
     */
    protected $userId;

    /**
     * @ORM\ManyToOne(targetEntity="UserBundle\Entity\User", inversedBy="experiments", cascade={"persist"})
     * @ORM\JoinColumn(name="user_id", referencedColumnName="id")
     */
    protected $user;

    // ..

}

我想使用他们的用户id和路线中的实验id来检索已登录用户的实验。

Answer 1

您可以使用custom ParamConverter来实现这一目标。例如：

namespace AppBundle\Request\ParamConverter;

use Sensio\Bundle\FrameworkExtraBundle\Configuration\ParamConverter;
use Sensio\Bundle\FrameworkExtraBundle\Request\ParamConverter\ParamConverterInterface;
use Doctrine\ORM\EntityManager;
use Symfony\Component\Security\Core\Authentication\Token\Storage\TokenStorage;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpKernel\Exception\NotFoundHttpException;
use AppBundle\Entity\Experiment;

class ExperimentConverter implements ParamConverterInterface
{
    protected $em;
    protected $user;

    public function __construct(EntityManager $em, TokenStorage $tokenStorage)
    {
        $this->em = $em;
        $this->user = $tokenStorage->getToken()->getUser();
    }

    public function apply(Request $request, ParamConverter $configuration)
    {
        $object = $this->em->getRepository(Experiment::class)->findOneBy([
            'id'   => $request->attributes->get('id'),
            'user' => $this->user
        ]);

        if (null === $object) {
            throw new NotFoundHttpException(
                sprintf('%s object not found.', $configuration->getClass())
            );
        }

        $request->attributes->set($configuration->getName(), $object);

        return true;
    }

    public function supports(ParamConverter $configuration)
    {
        return Experiment::class === $configuration->getClass();
    }
}

您需要注册转换器服务并为其添加标签：

# app/config/config.yml
services:
    experiment_converter:
        class: AppBundle\Request\ParamConverter\ExperimentConverter
        arguments:
            - "@doctrine.orm.default_entity_manager"
            - "@security.token_storage"
        tags:
            - { name: request.param_converter, priority: 1, converter: experiment_converter }

Answer 2

不幸的是，您无法将当前登录的用户ID注入param转换器，除非您实际将其作为url中的参数传递。

您可以创建自己的转换器，但我认为您最好的选择是创建一个用于获取实验的受保护方法。它将像注释一样易于使用和维护：

protected function getCurrentUsersExperiment($experimentId)
{
    return $this->getDoctrine()->getManager()->getRepository('AppBundle:Experiment')
        ->findOneBy(array(
            'id' => $experimentId,
            'user' => $this->getUser()
        ));
}

public function reportAction(Request $request, $id)
{
    $experiment = $this->getCurrentUsersExperiment($id);

    ...
}

正如symfony best practices中所述：在适当时使用ParamFetcher，但不要过度思考。

Symfony @ParamConverter通过用户关系获取实体

2 个答案: