Question

这是我的控制器

public function index2Action($name)
    {   
  $em = $this->getDoctrine()->getEntityManager();
  $test = $em->getRepository('RestWebServiceBundle:Test')->findall();
  return new Response(json_encode(array('locations' => $test)));

    }

当我转到URL时，我得到了：

{"locations":[{}]}

但是当我使用时：

public function index2Action($name)
    {   
$name ="Adam";
  return new Response(json_encode(array('locations' => $name)));

    }

我得到了JSON。

我做错了什么？我想在第一个场景中获得JSON。

更新：我已经验证$ test变量确实不是空的，当我对它执行print_r时，它会显示以下内容：

Array
(
    [0] => Rest\WebServiceBundle\Entity\Test Object
        (
            [id:protected] => 1
            [title:protected] => test title
            [author:protected] => test author
            [blog:protected] => this is the blog
            [tags:protected] => 
            [comments:protected] => 
            [created:protected] => DateTime Object
                (
                    [date] => 2012-05-13 00:00:00
                    [timezone_type] => 3
                    [timezone] => America/Chicago
                )

            [updated:protected] => DateTime Object
                (
                    [date] => 2012-05-13 00:00:00
                    [timezone_type] => 3
                    [timezone] => America/Chicago
                )

        )

)
null

Answer 1

我强烈建议您将序列化程序用于返回实体。看看序列化程序组件或jmsserializerbundle。

Answer 2

        $obj = $test;

        $serializer = new Serializer(
            array(new GetSetMethodNormalizer()),
            array('json' => new JsonEncoder())
        );
        $json = $serializer->serialize($obj, 'json');

        $response = new \Symfony\Component\HttpFoundation\Response($json);
        $response->headers->set('Content-Type', 'application/json');
        return $response;

Answer 3

我在我的存储库类＆amp;中使用了getArrayResult（）而不是getResult（）来尝试这个。它的工作原理

无法在symfony 2中返回json

3 个答案: