Symfony 4 Validator - 如果无效则自动注销

Question

我想用 Symfony Validator 验证我的用户对象 ($user) 并在表单输入数据有效/无效时返回 JsonResponse ($response)。 但是我有一个问题，当无法验证数据时我会自动注销。我必须再次登录，当某些数据无效时，这不是我所期望的行为。我找到了一种解决方法（见下面的 cmets），但这不是很令人满意：/

这是我的控制器的方法：

 /**
 * Update user profile data
 *
 * @Route("/api/users/updateprofile")
 * @Security("is_granted('USERS_LIST')")
 */

public function apiProfileUpdate(ValidatorInterface $validator, FlashMessageBuilder $flashMessageBuilder)
{
    $request = Request::createFromGlobals();
    // Prepare Response
    $response = new JsonResponse();
    $response->setData([]);
    /** @var User $user */
    $user = $this->getUser();
    $oldName = $user->getName();
    $oldEmail = $user->getEmail();

    $user->setName($request->request->get('name'));
    $user->setEmail($request->request->get('email'));

    $errors = $validator->validate($user);
    if (count($errors) > 0) { // if this -> auto logout
        $user->setName($oldName); // if I set the both attributes back to the old value
        $user->setEmail($oldEmail); // then I don't get logged out automatically but this is just a workaround and not satisfying
        $entityManager = $this->getDoctrine()->getManager(); // forgot to remove this
        $entityManager->persist($user); // and this line, this is actually deleted in the real code

        foreach ($errors as $error) {
            $errorMessage = $error->getMessage();
            $errorField = $error->getPropertyPath();
            $flashMessageBuilder->addErrorMessage($errorMessage, $errorField);
        };

        return $response;
    }

    $entityManager = $this->getDoctrine()->getManager();
    $entityManager->persist($user);
    $entityManager->flush();

    $flashMessageBuilder->addSuccessMessage("Success!");
    return $response;
}

对不起，我的英语不好，提前谢谢你！

Answer 1

即使验证中出现错误，您也会保留用户对象，这可能会导致注销问题。

仅在没有验证错误的情况下尝试更新：

public function apiProfileUpdate(ValidatorInterface $validator, FlashMessageBuilder $flashMessageBuilder)
{
    $request = Request::createFromGlobals();

    // Prepare Response
    $response = new JsonResponse();
    $response->setData([]);

    /** @var User $user */
    $user = $this->getUser();

    $user->setName($request->request->get('name'));
    $user->setEmail($request->request->get('email'));

    $errors = $validator->validate($user);
    if (count($errors) == 0) {
        $entityManager = $this->getDoctrine()->getManager();
        $entityManager->persist($user);
        $entityManager->flush();

        $flashMessageBuilder->addSuccessMessage("Success!");
        return $response;
    }

    foreach ($errors as $error) {
        $errorMessage = $error->getMessage();
        $errorField = $error->getPropertyPath();
        $flashMessageBuilder->addErrorMessage($errorMessage, $errorField);
    };

    return $response;
}

但我认为您不应该将 API 调用和利用 FlashMessages 的经典表单方法混为一谈，而应返回正确的 JSON 结果。

因此考虑相应地更改代码：

public function apiProfileUpdate(ValidatorInterface $validator, FlashMessageBuilder $flashMessageBuilder)
{
    $request = Request::createFromGlobals();

    /** @var User $user */
    $user = $this->getUser();

    $user->setName($request->request->get('name'));
    $user->setEmail($request->request->get('email'));

    $errors = $validator->validate($user);
    if (count($errors) == 0) {
        $entityManager = $this->getDoctrine()->getManager();
        $entityManager->persist($user);
        $entityManager->flush();

        return new JsonResponse(['success' => true]);
    }

    $data = [];
    foreach ($errors as $error) {
        $data[$error->getPropertyPath()] = $error->getMessage();
    };

    return new JsonResponse(['success' => false, 'errors' => $data], 400);
}

现在您的调用代码可以处理 200 结果（成功）和状态码为 400 的错误情况，并显示结果正文中 errors 部分中所有失败字段的错误消息。