CODEWITHSUNDEEP
phpsymfony
YeisonM
YeisonM

Reputation: 597

Api Platform - Symfony - return json in a persist fuction

I needd to return a JsonResponse inside persist fuction.

this is an example of my DataPersister class, the goal it's return a JsonResponse, when i try, i get the error: The controller must return a \"Symfony\Component\HttpFoundation\Response\" object but it returned an object of type App\Entity\VerificationCodes.

<?php
// api/src/DataPersister/UsersDataPersister.php

namespace App\DataPersister;

use ApiPlatform\Core\DataPersister\DataPersisterInterface;
use App\Entity\Users;
use Doctrine\Common\Persistence\ManagerRegistry;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\HttpFoundation\JsonResponse;

final class UsersDataPersister implements DataPersisterInterface 
{    
    private $managerRegistry;

    public function __construct(ManagerRegistry $managerRegistry)
    {
      $this->managerRegistry = $managerRegistry;
    }


    public function supports($data): bool
    {
        return $data instanceof Users;
    }

    public function persist($data){
        $em = $this->managerRegistry->getManagerForClass(Users::class);
        $user = new Users();
        //Persist User with encode password
        return $user;
        return new JsonResponse(['response'=>'yes']);
    }


    public function remove($data)
    {
      throw new \RuntimeException('"remove" is not supported');
    }
}

please help me or tell me what i can do, than you

Upvotes: 1

Views: 1228

Answers (1)

sensorario
sensorario

Reputation: 21698

If you want that App\Entity\Users class will be converted into a JSON you can implement JsonSerializable inferface (is PHP native interface). This interface force you to implement the method jsonSerialized that is automatically called when the object is converted to json (for example when you use json_encode($user);.

Finally you can try with

class Users implements \JsonSerializable
{
    public function jsonSerialize()
    {
        return []; // an array representation of your class
    }
}

then you can ...

return new JsonResponse(json_encode($user));

or (maybe) ...

return new JsonResponse($user);

Upvotes: 2

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