Javascript 将对象从sql转换为json数据
我想将数据对象转换为json格式,这是我在symfony中的代码Javascript 将对象从sql转换为json数据,javascript,json,symfony,prepared-statement,Javascript,Json,Symfony,Prepared Statement,我想将数据对象转换为json格式,这是我在symfony中的代码 $connection = $em->getConnection(); $statement = $connection->prepare("select * from contact where id_user=$id"); $statement->execute(); $contacts = $statement->fetchAll(); 在渲染对象时,我这样做 return
$connection = $em->getConnection();
$statement = $connection->prepare("select * from contact where id_user=$id");
$statement->execute();
$contacts = $statement->fetchAll();
在渲染对象时,我这样做
return $this->render('...index.html.twig', array('form' => $form->createView(), 'errors' => $form->getErrors(), 'contacts'=>json_encode($contacts)));
但这不起作用,请尝试以下帮助:
$contacts= json_encode($contacts);
$Response = new Response(
'...myTemplate.html.twig',
['form' => $form->createView()],
['errors' => $form->getErrors()],
['contacts' => $contacts]
);
$Response->headers->set('Content-Type', 'application/json');
return $oResponse;
不要忘记使用:
use Symfony\Component\HttpFoundation\Response;
或者可以使用JsonResponse类。以下是文档:
试试这个:
$contacts= json_encode($contacts);
$Response = new Response(
'...myTemplate.html.twig',
['form' => $form->createView()],
['errors' => $form->getErrors()],
['contacts' => $contacts]
);
$Response->headers->set('Content-Type', 'application/json');
return $oResponse;
不要忘记使用:
use Symfony\Component\HttpFoundation\Response;
或者可以使用JsonResponse类。以下是文档: