Question

只是想知道为什么这不起作用。

当在单独的方法中放置JSON返回时，我在控制器中收到以下错误：

The controller must return a response (null given). Did you forget to add a return statement somewhere in your controller? (500 Internal Server Error)

设置如下：

控制器

namespace UsedBundle\Controller;

use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Symfony\Component\HttpFoundation\Request;
...
use Symfony\Component\HttpFoundation\JsonResponse;


class AccountController extends Controller
{
private $status;
private $message;
private $data;

/**
 * @Route("/mon-compte", name="account_page")
 */

public function showAccount(Request $request){
    $factory = $this->get('security.encoder_factory');
    if (!$this->get('security.authorization_checker')->isGranted('IS_AUTHENTICATED_REMEMBERED')) {
        throw $this->createAccessDeniedException();
    }
    $user = $this->getUser();
    $session = $request->getSession();
    $email = $session->get('email');

    $em = $this->getDoctrine('doctrine')->getManager('used');
    $this->user_info = $em->getRepository('UsedBundle:User')
    ->UserAccountInfoAction( $email ); 

    $form = $this->createForm(UserType::class, $user);
    if ($request->isMethod('POST')) {
        if($_POST['action'] == 'update_user'){
            $this->updateProfile($request, $user, $form, $em);
        }elseif($_POST['action'] == 'delete_user'){
            $this->deleteUser( $user, $em );
        }elseif($_POST['action'] == 'update_password'){
            $this->updatePassword( $user, $em, $factory);
        }
               // \Doctrine\Common\Util\Debug::dump($this->data);
        $this->returnJson();//***** this is generating the error*****
    }else{
        // populate change profile form
        return $this->render('account/account.html.twig', [
            'user_info' => $this->user_info,
            'form' => $form->createView(),
        ]);
    }
}

然后，进一步在该类上我有returnJson（）方法：

    public function returnJson(){
    return new JsonResponse(array(
        'status' => $this->status, 
        'message' => $this->message,
        'data' => $this->data,
        )
    );         
}

如果我将该代码替换为showAccount（）上的$ this-＆gt; returnJson（），它将正常工作。

为什么退货不能作为单独的方法放置？或者我错过了什么。

由于

Answer 1

您的success: function (json) { $('#hasil').append("<div class='list-group-item row'>"); for (var i = 0; i < json.data.length; i++) { $('#hasil').append("<div class='col-md-4' id='col"+i+"'>"+json.data[i]+"</div>"); } $('#hasil').append("</div>");函数会将returnJson返回JsonResponse函数，而不是showAccount函数。

这应该有效：

return $this->returnJson();

symfony 3 - 控制器返回单独的函数会产生错误

1 个答案: