Symfony 3.1發佈到Web Api

Question

我正在嘗試使用Symfony創建Web API，但POST方法不起作用。我的代碼不會給出任何錯誤。

實體/狀態：

<?php 

namespace RestBundle\Entity; 

use Doctrine\ORM\Mapping as ORM; 
use Symfony\Component\Validator\Constraints as Assert; 


/** 
* 
* 
* @ORM\Table(name="status") 
* @ORM\Entity 
*/ 
class Status 
{ 
    /** 
    * @var integer 
    * 
    * @ORM\Column(name="id", type="integer", nullable=false) 
    * @ORM\Id 
    * @ORM\GeneratedValue(strategy="IDENTITY") 
    */ 
    private $id; 

    /** 
    * @var string 
    * 
    * @ORM\Column(name="nome", type="string", length=100, nullable=true) 
    * 
    * @Assert\NotBlank() 
    */ 
    private $nome; 
} 
?> 

POST功能（控制器/ StatusController）：

public function postStatusAction(Request $request) 
{ 
    $user = new Status(); 
    $form = $this->createForm(StatusType::class, $user); 
    $form->handleRequest($request); 

    if ($form->isValid()) { 
     $em = $this->getDoctrine()->getManager(); 
     $em->persist($user); 
     $em->flush(); 

     return $user; 
    } 
    //var_dump($form->getErrors()); 
    throw new HttpException(400, "Invalid data"); 

} 

窗體/ StatusType：

<?php 
/** 
* Created by PhpStorm. 
* Date: 10/08/16 
* Time: 10:39 
*/ 

namespace RestBundle\Form; 


use Symfony\Component\Form\AbstractType; 
use Symfony\Component\Form\FormBuilderInterface; 
use Symfony\Component\OptionsResolver\OptionsResolver; 
use Symfony\Component\Form\Extension\Core\Type\TextType; 

class StatusType extends AbstractType 
{ 
    /** 
    * @param FormBuilderInterface $builder 
    * @param array $options 
    */ 
    public function buildForm(FormBuilderInterface $builder, array $options) 
    { 
     $builder->add('nome', TextType::class); 
    } 

    /** 
    * @param OptionsResolver $resolver 
    */ 
    public function configureOptions(OptionsResolver $resolver) 
    { 
     $resolver->setDefaults(array(
      'data_class' => 'RestBundle\Entity\Status', 
      'csrf_protection' => false, 
     )); 
    } 

    /** 
    * @return string 
    */ 
    public function getBlockPrefix() 
    { 
     return ''; 
    } 
} 

感謝。

Answer 1

控制器必須返回一個響應：

public function postStatusAction(Request $request) 
{ 
    $user = new Status(); 
    $form = $this->createForm(StatusType::class, $user); 
    $form->handleRequest($request); 

    if ($form->isValid()) { 
     $em = $this->getDoctrine()->getManager(); 
     $em->persist($user); 
     $em->flush(); 

     return new Response($user->getUsername()); 
    } 
    //var_dump($form->getErrors()); 
    throw new HttpException(400, "Invalid data"); 

}