问题描述
现在我达到了结果:
他带有以下代码:
namespace App\Controller;
use App\Entity\LogStatus;
use Doctrine\ORM\EntityManagerInterface;
use Symfony\Bundle\FrameworkBundle\Controller\AbstractController;
use Symfony\Component\HttpFoundation\JsonResponse;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\Routing\Annotation\Route;
use Symfony\Component\Serializer\SerializerInterface;
class LogController extends AbstractController
{
/**
* @Route("/logs/statuses",name="log_statuses",methods={"POST"})
*/
public function index(EntityManagerInterface $entityManager): Response
{
$request = new Request($_POST);
$data = $request->getContent();
$data = json_decode($data);
$data = (array)$data;
$deviceid = $data['filter'];
$logStatuses = $entityManager->getRepository(LogStatus::class)
->findBy(['deviceid' => $deviceid],['createdat' => 'DESC']);
return new JsonResponse([
'rows' => (array)$logStatuses[0]
]
);
}
}
问题:如何对$logStatuses
进行编码,以在不迭代每个对象的情况下获得预期的Json格式(或进行扩展操作)。
解决方法
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