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Kevin
Kevin

Reputation: 5092

API Platform - How to Use a DTO for Posting?

I'm using API platform in my Symfony4 app to expose my resources. It's a great framework but it force you by default to have all your Business logic in the front-end side, because it expose all your Entities and not a Business Object.

I don't like that and I prefer to have my business logic in the back-end side.

I need to create users, but there are different type of users. So I have create a UserFactory in the back-end-side. So the front just need to push a Business object and the back-end take care of everything.

The front front can never persist a User Object directly in the DB. It is the role of the back-end

Following this tutorial to use DTO for Reading: https://api-platform.com/docs/core/dto/#how-to-use-a-dto-for-reading

I'm trying to do the same for posting. And it works. Here is my Controller code:

/**
 * @Route(
 *     path="/create/model",
 *     name="create-model",
 *     methods={"POST"},
 *     defaults={
 *          "_api_respond"=true,
 *          "_api_normalization_context"={"api_sub_level"=true},
 *          "_api_swagger_context"={
 *              "tags"={"User"},
 *              "summary"="Create a user Model",
 *              "parameters"={
 *                  
 *              },
 *              "responses"={
 *                  "201"={
 *                      "description"="User Model created",
 *                      "schema"={
 *                          "type"="object",
 *                          "properties"={
 *                              "firstName"={"type"="string"},
 *                              "lastName"={"type"="string"},
 *                              "email"={"type"="string"},
 *                          }
 *                      }
 *                  }
 *              }
 *          }
 *     }
 * )
 * @param Request $request
 * @return \App\Entity\User
 * @throws \App\Exception\ClassNotFoundException
 * @throws \App\Exception\InvalidUserException
 */
public function createModel(Request $request)
{
    $model = $this->serializer->deserialize($request->getContent(), Model::class, 'json');
    $user = $this->userFactory->create($model);
    $this->userRepository->save($user);

    return $user;
}

It works great, but I would love my new resource to work in the Swagger UI, so I can Create via POST method new resources directly in the web interface.

For that I think I need to complete the parameter section in my _api_swagger_context. But I don't fin any documentation about that.

How can I do that?

Upvotes: 0

Views: 4551

Answers (1)

Kevin
Kevin

Reputation: 5092

Found the answer here: https://github.com/api-platform/docs/issues/666

You can fill parameters like this :

 "parameters" = {
     {
        "name" = "data",
        "in" = "body",
        "required" = "true",
        "schema" = {
            "type" = "object",
            "properties" = {
                 "firstName"={"type"="string"},
                 "lastName"={"type"="string"},
                 "email" = {"type" = "string" }
             }
         },
     },
 },

More docs about parameters for swagger here : https://swagger.io/docs/specification/2-0/describing-parameters/

Upvotes: 1

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