how to return json response on ajax call in symfony3?

Question

I am using symfony3:

I am calling a method via ajax call:

Controller

/**
     * @Route("personnel/domainlist/{id}", name="ajax_method")
     * @Method("GET")
     */
    public function domainlist(Request $request,$id){
       $repository = $this->getDoctrine()->getRepository('AppBundle:ENTITYNAME');          
       $res=$repository->findBy(array('COLNAME' => $id));          
      // create a JSON-response with a 200 status code

      $response = new Response(json_encode($res));    
      $response->headers->set('Content-Type', 'application/json');
      return $response;
      die;
    }

form above code i am getting following result: print_r($res);

Array
(
    [0] =&gt; AppBundle\Entity\ENTITYNAME Object
        (
            [domain_id:AppBundle\Entity\ENTITYNAME:private] =&gt; 15
            [domain_title:protected] =&gt; ABC
            [domain_group_id:protected] =&gt; 1
        )
)

AJax code:

 $.ajax({
        url: "domainlist/" + pdoamin_id,      
        type: 'POST',
        dataType: 'json',        
         success: function(result) {
            alert(result);
             }
          });

  });

Anyone can help me how can i return json to ajax method in symfony3

zalex · Accepted Answer · 2016-04-05 11:46:14Z

1

You have to serialize your answer before push it to Response. There are two ways to do it (at least I know just two ways)

The Serializer Component
Implement interface \JsonSerializable (PHP version >=5.4 )

In both variants your json_encode function will be work as you expect.

I prefer second way for For its simplicity

answered Apr 5, 2016 at 11:46

zalex

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tlarson · Accepted Answer · 2019-01-04 16:43:14Z

1

Not sure if Symfony 3 has the class, but in Symfony 4, you can use the JsonResponse class.

import the class into your controller

use Symfony\Component\HttpFoundation\JsonResponse;

Then use one of its constructors in your return statement:

return new JsonResponse($dataToReturn);

That worked for me.

answered Jan 4, 2019 at 16:43

tlarson

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Ashish Kumar Saxena · Accepted Answer · 2016-04-05 13:04:43Z

0

$repository = $this->getDoctrine()->getRepository('AppBundle:ENTITYNAME');          
       $res=$repository->findBy(array('COL_NAME' => $id));          
       $normalizer = new ObjectNormalizer();      
       $encoder = new JsonEncoder();
       $serializer = new Serializer(array($normalizer), array($encoder));
       $response=$serializer->serialize($res, 'json');

answered Apr 5, 2016 at 13:04

Ashish Kumar Saxena

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