How can I send JSON response in symfony2 controller

Question

I am using jQuery to edit my form which is built in Symfony.

I am showing the form in jQuery dialog and then submitting it.

Data is entering correctly in database.

But I don't know whether I need to send some JSON back to jQuery. Actually I am bit confused with JSON thing.

Suppose I have added a row in my table with ``jQuery and when I submit the form then after data is submitted I want to send back those row data so that I can dynamically add the table row to show the data added.

I am confused how can get that data back.

This is my current code:

$editForm = $this->createForm(new StepsType(), $entity);

$request = $this->getRequest();

$editForm->bindRequest($request);

if ($editForm->isValid()) {
    $em->persist($entity);
    $em->flush();

    return $this->render('::success.html.twig');               
}

This is just the template with success message.

Answer 1

Symfony 2.1

$response = new Response(json_encode(array('name' => $name)));
$response->headers->set('Content-Type', 'application/json');

return $response;

---

Symfony 2.2 and higher

You have special JsonResponse class, which serialises array to JSON:

return new JsonResponse(array('name' => $name));

---

But if your problem is How to serialize entity then you should have a look at JMSSerializerBundle

Assuming that you have it installed, you'll have simply to do

$serializedEntity = $this->container->get('serializer')->serialize($entity, 'json');

return new Response($serializedEntity);

You should also check for similar problems on StackOverflow:

• How to encode Doctrine entities to JSON in Symfony 2.0 AJAX application?
• Symfony 2 Doctrine export to JSON

Answer 2

Symfony 2.1 has a JsonResponse class.

return new JsonResponse(array('name' => $name));

The passed in array will be JSON encoded the status code will default to 200 and the content type will be set to application/json.

There is also a handy setCallback function for JSONP.

Answer 3

Since Symfony 3.1 you can use JSON Helper http://symfony.com/doc/current/book/controller.html#json-helper

public function indexAction()
{
// returns '{"username":"jane.doe"}' and sets the proper Content-Type header
return $this->json(array('username' => 'jane.doe'));

// the shortcut defines three optional arguments
// return $this->json($data, $status = 200, $headers = array(), $context = array());
}

Answer 4

To complete @thecatontheflat answer I would recommend to also wrap your action inside of a try … catch block. This will prevent your JSON endpoint from breaking on exceptions. Here's the skeleton I use:

public function someAction()
{
    try {

        // Your logic here...

        return new JsonResponse([
            'success' => true,
            'data'    => [] // Your data here
        ]);

    } catch (\Exception $exception) {

        return new JsonResponse([
            'success' => false,
            'code'    => $exception->getCode(),
            'message' => $exception->getMessage(),
        ]);

    }
}

This way your endpoint will behave consistently even in case of errors and you will be able to treat them right on a client side.

Answer 5

If your data is already serialized:

a) send a JSON response

public function someAction()
{
    $response = new Response();
    $response->setContent(file_get_contents('path/to/file'));
    $response->headers->set('Content-Type', 'application/json');
    return $response;
}

b) send a JSONP response (with callback)

public function someAction()
{
    $response = new Response();
    $response->setContent('/**/FUNCTION_CALLBACK_NAME(' . file_get_contents('path/to/file') . ');');
    $response->headers->set('Content-Type', 'text/javascript');
    return $response;
}

If your data needs be serialized:

c) send a JSON response

public function someAction()
{
    $response = new JsonResponse();
    $response->setData([some array]);
    return $response;
}

d) send a JSONP response (with callback)

public function someAction()
{
    $response = new JsonResponse();
    $response->setData([some array]);
    $response->setCallback('FUNCTION_CALLBACK_NAME');
    return $response;
}

e) use groups in Symfony 3.x.x

Create groups inside your Entities

<?php

namespace Mindlahus;

use Symfony\Component\Serializer\Annotation\Groups;

/**
 * Some Super Class Name
 *
 * @ORM    able("table_name")
 * @ORM\Entity(repositoryClass="SomeSuperClassNameRepository")
 * @UniqueEntity(
 *  fields={"foo", "boo"},
 *  ignoreNull=false
 * )
 */
class SomeSuperClassName
{
    /**
     * @Groups({"group1", "group2"})
     */
    public $foo;
    /**
     * @Groups({"group1"})
     */
    public $date;

    /**
     * @Groups({"group3"})
     */
    public function getBar() // is* methods are also supported
    {
        return $this->bar;
    }

    // ...
}

Normalize your Doctrine Object inside the logic of your application

<?php

use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\Serializer\Mapping\Factory\ClassMetadataFactory;
// For annotations
use Doctrine\Common\Annotations\AnnotationReader;
use Symfony\Component\Serializer\Mapping\Loader\AnnotationLoader;
use Symfony\Component\Serializer\Serializer;
use Symfony\Component\Serializer\Normalizer\ObjectNormalizer;
use Symfony\Component\Serializer\Encoder\JsonEncoder;

...

$repository = $this->getDoctrine()->getRepository('Mindlahus:SomeSuperClassName');
$SomeSuperObject = $repository->findOneById($id);

$classMetadataFactory = new ClassMetadataFactory(new AnnotationLoader(new AnnotationReader()));
$encoder = new JsonEncoder();
$normalizer = new ObjectNormalizer($classMetadataFactory);
$callback = function ($dateTime) {
    return $dateTime instanceof \DateTime
        ? $dateTime->format('m-d-Y')
        : '';
};
$normalizer->setCallbacks(array('date' => $callback));
$serializer = new Serializer(array($normalizer), array($encoder));
$data = $serializer->normalize($SomeSuperObject, null, array('groups' => array('group1')));

$response = new Response();
$response->setContent($serializer->serialize($data, 'json'));
$response->headers->set('Content-Type', 'application/json');
return $response;

Answer 6

If you are using symfony 5 propably it will works.

$data = $serializer->serialize(array('key' => 'value'), JsonEncoder::FORMAT);

return new JsonResponse($data, Response::HTTP_OK, [], true);