Symfony - Open API 3 - Set JSON response as example in controlller

Question

Im using Open Api 3 (exactly NelmioApiDocBundle) in Symfony to document APIs and I have the following code:

* @Route("/login", name="user_login", methods={"POST"})
*
* @OA\Response(
*  response=200, 
*  @OA\JsonContent(
*      type="object",
*      @OA\Property(property="code", type="number"),
*      @OA\Property(property="error", type="boolean"),
*      @OA\Property(property="message", type="string")
*  )
* )

This works and shows me the following:

What I want to know is if there is an easier way to define these responses so I don't have to write so many lines at each endpoint.

I wish there was a way to define a JSON file as an interface and have it read from there. I've tried a thousand things but nothing works for me and I don't know how to do it and I've already given up.

Thanks a lot!

I have tried to use @Model, @Schema, modify YAML files... etc... I have read the documentation, but I have not seen any valid example for my case in this version of Open Api.

score 0 · Answer 1 · answered Oct 27 '22 at 07:46

If you use models, you can put in the following

@OA\JsonContent(ref=@Model(type=LoginResponse::class))

Your response class should be something like this:

use OpenApi\Annotations as OA;

class LoginResponse
{
    /** @OA\Property(type="number") */
    public $code;
    /** @OA\Property(type="boolean") */
    public $error;
    /** @OA\Property(type="string") */
    public $message;

}

Make sure the properties are publicly accessible and you have symfony/property-info installed. You can also use jms/serializer, but that's a bit more work

More info at: https://symfony.com/bundles/NelmioApiDocBundle/current/index.html#use-models

Symfony - Open API 3 - Set JSON response as example in controlller

1 Answers1