Symfony 3.4 - Api 其余修改值

Question

我有一个简单的问题！我不知道如何为 table 命名个人资料中的 $email 设置值，并且该个人资料 table 链接到用户 table（多对一）。所以我成功地使用 POST 方法添加了一个用户：

{
"username": "TestTest10",
"password": "TestTest10",
}

之后，我自动创建了一个链接到用户的个人资料实体，但我手动输入了个人资料的信息，因此它创建了我想要的用户，但链接到该帐户的个人资料对于任何用户都是相同的（所有有电子邮件 ok@ok.com)（像这样：

$profile = new Profile();
$profile->setEmail('ok@ok.com');
$user->setIdProfile($profile);
$em = $this->getDoctrine()->getManager();
$em->persist($profile);
$em->flush();

但我想用 POST 方法编辑它：

{
"username": "TestTest10",
"password": "TestTest10",
"email": "imanoob@symfony.com"
}

那我该怎么做呢？

我的 PostUserFunction ：

/**
 *
 * @Rest\Post(
 *     path = "/users",
 *     name = "api_users_add"
 * )
 * @Rest\View(StatusCode=201, serializerGroups={"user_detail"})
 * @ParamConverter(
 *     "user",
 *     converter="fos_rest.request_body",
 *     options={"deserializationContent"={"groups"={"Deserialize"}}}
 * )
 */
public function postUserAction(Request $request, User $user, ConstraintViolationListInterface $violations)
{
    if (count($violations) > 0) {
        return new JsonResponse([
            'success' => "false",
        ]);
    }

    $profile = new Profile();

    $profile->setEmail('ok@ok.com');
    $profile->setLastConnexion(new \DateTime('now'));
    $profile->setCreatedAccount(new \DateTime('now'));
    $profile->setBirth(new \DateTime('now'));

    $user->setIdProfile($profile);

    $em = $this->getDoctrine()->getManager();
    $em->persist($profile);
    $em->flush();

    $this->encodePassword($user);
    $user->setRoles([User::ROLE_USER]);

    $this->persistUser($user);

    return new JsonResponse([
        'success' => "true",
        'data' => [
            'Id' => $user->getId(),
            'username' => $user->getUsername(),
            'email' => $profile->getEmail(),
        ]
    ]);
}

我也许可以在参数中接收配置文件：

    public function postUserAction(Request $request, User $user, Profile $profile, ConstraintViolationListInterface $violations)

但我不知道如何将两个 paramConverter 放入：

* @ParamConverter(
 *     "user",
 *     converter="fos_rest.request_body",
 *     options={"deserializationContent"={"groups"={"Deserialize"}}},
 * )

感谢所有尝试回答的人 :p

Answer 1

好的，伙计们，如果有人遇到同样的问题，响应真的很简单，但你现在需要：你需要像这样放置两个参数转换器：

* @Rest\View(StatusCode=201, serializerGroups={"user_detail"})
 * @ParamConverter(
 *     "user",
 *     converter="fos_rest.request_body",
 *     options={"deserializationContent"={"groups"={"Deserialize"}}},
 * )
 * @ParamConverter(
 *     "profile",
 *     converter="fos_rest.request_body",
 *     options={"deserializationContent"={"groups"={"Deserialize"}}},
 * )

整个代码如下所示：

/**
 *
 * @Rest\Post(
 *     path = "/users",
 *     name = "api_users_add"
 * )
 * @Rest\View(StatusCode=201, serializerGroups={"user_detail"})
 * @ParamConverter(
 *     "user",
 *     converter="fos_rest.request_body",
 *     options={"deserializationContent"={"groups"={"Deserialize"}}},
 * )
 * @ParamConverter(
 *     "profile",
 *     converter="fos_rest.request_body",
 *     options={"deserializationContent"={"groups"={"Deserialize"}}},
 * )
 */
public function postUserAction(Request $request, User $user, Profile $profile, ConstraintViolationListInterface $violations)
{
    if (count($violations) > 0) {
        return new JsonResponse([
            'success' => "false",
        ]);
    }
    $profile->setLastConnexion(new \DateTime('now'));
    $profile->setCreatedAccount(new \DateTime('now'));
    $profile->setBirth(new \DateTime('now'));

    $user->setIdProfile($profile);

    $em = $this->getDoctrine()->getManager();
    $em->persist($profile);
    $em->flush();

    $this->encodePassword($user);
    $user->setRoles([User::ROLE_USER]);

    $this->persistUser($user);

    return new JsonResponse([
        'success' => "true",
        'data' => [
            'Id' => $user->getId(),
            'username' => $user->getUsername(),
            'email' => $profile->getEmail(),
        ]
    ]);
}

Symfony 3.4 - Api 其余修改值

Symfony 3.4 - Api Rest modify value

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