Why ajax returns code 500 [ symfony3 ]

Question

Good day to all. I have problem with ajax request in symfony3. In this case it returns code 200 and everything ok. This is "dirty" code example, but with "good" code i am getting respond with code 500. I know that this syntax is more appropriate. So question is how to fix syntax which is applies to symfony documentation.

 /**
 * @Route("/ajax", name="ajax_handler")
 */
public function ajaxAction(Request $request) {

    if ($request->isXMLHttpRequest()) {
        if(isset($_POST['secret']))
        echo json_encode(array('ok'=>1));
        exit;
    }
}

Why if i will write appropriate to "all known" syntax code it will returns code 500?

  /**
     * @Route("/ajax", name="ajax_handler")
     */
    public function ajaxAction(Request $request) {

        if ($request->isXMLHttpRequest()) {
//            echo json_encode(array('ok' => 1));
//            exit;

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

            return $response;
        }
    }

JavaScript:

$(document).ready(function () {

                $(".add-one-email").click(function () {

                    var data = {"test": 1};

                    data = $.param(data);

                    $.ajax({
                    type:"POST",
                            dataType:"JSON",
                            url:"/ajax",
                            data: data,
                            success:function (s) {
                                alert(s['ok']);
                                }
                    });
                });

            });

if ($request->isXMLHttpRequest()) {
        $response = new Response('sadasdasd');
        return $response;
    }

In upper case it returns 500 StuckTrace:The controller must return a response | in vendor/symfony/symfony/src/Symfony/Component/HttpKernel/HttpKernel.php at line 171 |  if (null === $response) {
                    $msg .= ' Did you forget to add a return statement somewhere in your controller?';
                }
                throw new \LogicException($msg);

Basic rule of thumb: If you get a 500, your FIRST stop is the server's error logs to get details about the 500. Until you have that, there is no point in fiddling with your code or asking why it's not working - there is just NO way to tell what caused the 500. maybe the code above is 100% perfect but you have a syntax error in a .htaccess file. — Marc B, Aug 26 '16 at 15:24
/enterBundle/Controller/DefaultController.php at line 273 ( line 273 is ` $response = new Response(json_encode(array('name' => $name)));` ) — Yevhenii Shashkov, Aug 26 '16 at 15:30
In you question description the var `$name` does not exist. Please check this. — yceruto, Aug 26 '16 at 15:46
Are you really sure about this? Where does that value come from? Also the `JsonResponse` makes everything easier, you should check it out. — Emanuel Oster, Aug 26 '16 at 15:46
Cannot access protected property Symfony\Component\BrowserKit\Response::$headers (500 Internal Server Error) — Yevhenii Shashkov, Aug 26 '16 at 16:20
`BrowserKit\Response`? Unless it's changed for Symfony3, I'd be expecting the `HttpFoundation\Response` type... Looks like you might have imported the wrong namespace. — Jonnix, Aug 26 '16 at 16:31
use enterBundle\Entity\Users; use Symfony\Bundle\FrameworkBundle\Controller\Controller; use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route; — Yevhenii Shashkov, Aug 26 '16 at 16:38
use Symfony\Component\HttpFoundation\Request; use Symfony\Component\Form\Extension\Core\Type\PasswordType; use Symfony\Component\Form\Extension\Core\Type\RepeatedType; use Symfony\Component\Form\Extension\Core\Type\TextType; use Symfony\Component\Form\Extension\Core\Type\SubmitType; use PHPMailer; use enterBundle\Repository\usersRepository; use Doctrine\ORM\Query; use Sensio\Bundle\FrameworkExtraBundle\Configuration\Method; use Symfony\Component\BrowserKit\Response; — Yevhenii Shashkov, Aug 26 '16 at 16:39
So `Symfony\Component\BrowserKit\Response; ` is probably wrong. Probably wants to be `Symfony\Component\HttpFoundation\Response` — Jonnix, Aug 26 '16 at 16:42
anyway `echo json_encode(array('ok'=>1));` show 200 in browser i can use it until i solve BrowserKit — Yevhenii Shashkov, Aug 26 '16 at 16:44
@fonjeekay That only works because you `exit` afterwards, otherwise you'd have another error. Try fixing the problems instead of just hacking around them — Jonnix, Aug 26 '16 at 16:45
use Symfony\Component\HttpFoundation\JsonResponse; return new JsonResponse(array('ok' => 1)); works good — Yevhenii Shashkov, Aug 26 '16 at 17:56
From Symfony 3.1 use: `return $this->json(array('ok' => 1));` — malcolm, Aug 27 '16 at 09:01
/ajax looks like it is probably going to a production version, are you sure cache is cleared for production if that is the case? I've got a 500 error occasionally because of this. — Scott Flack, Aug 29 '16 at 05:53

Yevhenii Shashkov · Answer 1 · 2016-08-27T04:40:58.807

0

This syntax works good

use Symfony\Component\HttpFoundation\JsonResponse;

  return new JsonResponse(array('ok' => 1));

Problem was that i was using old syntax and looks like it is not work in symfony3, but this code works good.

edited Aug 27 '16 at 04:40

answered Aug 26 '16 at 17:56

Yevhenii Shashkov

464
1
8
19

Why ajax returns code 500 [ symfony3 ]

1 Answers1