javascript - 如何在 Symfony 3 中以 Json 形式从存储库返回对象

Question

我正在尝试将存储库对象作为 json 发送
我的 Controller 代码:

if($request->isXmlHttpRequest()){   
        $data = $request->request->get('id');
        $bedroom = $em->getRepository('EpitaHousingBundle:Bedroom')->findOneBy(array('id'=>$data));

        $this->container->get('logger')->addInfo('somesh');

        return $json = new Response(json_encode(array('bedroom' => $bedroom)));   
       } 

我的 JavaScript 代码:

$(document).ready(function(){            
     var id_select = $('#bedroom_view_bedroom').val();        
     $.ajax({
            type: 'POST', 
            url: '{{ (path('listingview', {'id': id})) }}',
            contentType: 'application/x-www-form-urlencoded',
            data: {id:id_select}, 
        success: function(result,status,xhr){
           var inst_arr = JSON.parse(result);
           console.log(inst_arr);
          },
        error: function(xhr, status, error) {     
                console.log(status);
            } 
        }); 
  });

这里我得到空对象作为响应。我如何使用 symfony3 将存储库对象作为 json 发送。帮助我任何人。提前致谢...

编辑:

我不知道哪里出了问题，我尝试了下面的代码，例如

if ($request->isXmlHttpRequest()) {
        $data = $request->request->get('id');
        $bedroom = $em->getRepository('EpitaHousingBundle:Bedroom')->findOneBy(array('id' => $data));
        if (null === $bedroom) {
            $bedroomJson = '';
        }
        $this->container->get('logger')->addInfo('somesh');  
        $serializer = $this->container->get('serializer');           
        $bedroomJson = $serializer->serialize($bedroom, 'json');

        return new Response($bedroomJson, Response::HTTP_OK, ['content-type' => 'application/json']);
    }

控制台显示

jquery.min.js:4 POST http://localhost/epitahousing/web/app_dev.php/provider/listing/view/8 500 (Internal Server Error)error

如果我尝试在 xmlHttpRequest 之前找到存储库对象，例如

$bedroom = $em->getRepository('EpitaHousingBundle:Bedroom')->findOneBy(array('id' => '12'));
    if ($request->isXmlHttpRequest()) {           
      //code
    }

这里显示了正确的对象，但我真的不知道 json 出了什么问题

编辑2:

$bedroom = $this->em->getRepository('EpitaHousingBundle:Bedroom')->findOneBy(array('id'=>$data));

        $output['rentamount'] = $bedroom->getRentamount();
        $output['rentcurrency'] = $bedroom->getRentcurrency()->getValue();
        $output['rentduration'] = $bedroom->getRentduration()->getValue();
        $output['bondamount'] = $bedroom->getBondamount();
        $output['bondcurrency'] = $bedroom->getBondcurrency()->getValue();
        $output['leaseminduration'] = $bedroom->getLeaseminduration()->getValue();
        $output['leasemaxduration'] = $bedroom->getLeasemaxduration()->getValue();
return $json = new Response(json_encode($output));

我这样做了，现在效果很好...... 感谢大家...如此有值(value)的回复

Answer 1

Symfony 附带 Serializer Component (选择您的 Sf 版本)，您应该使用它来将对象转换为 json。

app/config/config.yml

启用序列化器组件

serializer:
        enabled: true

您的 Controller

if($request->isXmlHttpRequest()){   
    $data = $request->request->get('id');
    $bedroom = $em->getRepository('EpitaHousingBundle:Bedroom')->findOneBy(array('id'=>$data));
    if(null === $bedroom) {
        // return some error response
    }
    $this->container->get('logger')->addInfo('somesh');
    $serializer = $this->container->get('serializer');
    $bedroomJson = $serializer->serialize($bedroom, 'json');

    return new Response($bedroomJson, Response::HTTP_OK, ['content-type' => 'application/json']);   
}

$bedroomJson 应该是代表实体对象的 json 字符串。如果有问题请告诉我:)希望有帮助!