json - JMSSerializerBundle的响应内容必须是字符串或实现__toString()的对象"

Question

我遇到了 JMSSerializerBundle 的问题。

我在那里有我的实​​体 AGVote：

<?php

namespace K\AGBundle\Entity;

use JMS\SerializerBundle\Annotation\Type;
use JMS\SerializerBundle\Annotation\Accessor;
use JMS\SerializerBundle\Annotation\AccessType;
use JMS\SerializerBundle\Annotation\Exclude;
use JMS\SerializerBundle\Annotation\ExclusionPolicy;
use Doctrine\ORM\Mapping as ORM;


/**
 * K\AGBundle\Entity\AGVote
 * @ORM\Entity
 * @ORM\HasLifecycleCallbacks
 * 
 */


/*
 * 
/** @AccessType("public_method") */

class AGVote
{


   /**
* @Type("integer") 
* @Accessor(getter="getId") 
*/

 /**
 * @ORM\Id
 * @ORM\Column(type="integer")
 * @ORM\GeneratedValue(strategy="AUTO")
 */
public $id;

 /**
 * @ORM\Column(type="text")
  * @Accessor(getter="getQuestion")
  * @Type("text")
 */
public $question;

/**
 * @ORM\Column(type="smallint")
 * @Type("integer")
 * @Accessor(getter="getActif")
 */
public $actif;

 /**
 * @ORM\ManyToOne(targetEntity="\K\KBundle\Entity\Users", cascade={"all"})
  * @Exclude
 */
protected $users;



/**
 * Get id
 *
 * @return integer 
 */
public function getId()
{
    return $this->id;
}

/**
 * Set question
 * Nb : Only AG admin can set a question
 * @param text $question
 */
public function setQuestion($question)
{
    $this->question = $question;
}

/**
 * Get question
 *
 * @return text 
 */
public function getquestion()
{
    return $this->question;
}

/**
 * Set actif
 *
 * @param smallint $actif
 */
public function setActif($actif)
{
    $this->actif = $actif;
}

/**
 * Get actif
 *
 * @return smallint
 */
public function getActif()
{
    return $this->actif;
}

 /**
 * Set Users
 *
 * @param K\KBundle\Entity\Province $Users
 */
public function setUsers(\K\KBundle\Entity\Users $users)
{
    $this->users = $users;
}

/**
 * Get Users
 *
 * @return K\KBundle\Entity\Users
 */
public function getUsers()
{
    return $this->users;
}

  public function __toString()
{
   return $this->getquestion();
}

}

我制作了一个控制器，它在 Json 中向我返回了一个 AGVote 实体：

   public function jsonvoteAction($id) {
  $em = $this->getDoctrine()->getEntityManager();
  $entity = $em->getRepository('KAGBundle:AGVote')->findOneById($id);

  if ($entity->getActif() == 1) {
     $serializer = $this->container->get('serializer');
     $serializer->serialize($entity, 'json');
     $response = new Response($serializer);

     return $reponse;
     }
     }

我在 Json 中有一个响应，但它是一个错误说：

[{"message":"响应内容必须是实现__toString()的字符串或对象，\"object\"给出。","class":"UnexpectedValueException","trace":

事实上，我已经在我的所有实体中实现了一个 __toString() 方法。

有人有想法吗？

谢谢 ：）

Answer 1

When you call the serialize method on the $serializer, it returns the serialized data (a string). The problem is that you do not use this returned value, and create the response with the $serializer itself, which makes no sense.

First, store the serialized $entity:

$serializedEntity = $serializer->serialize($entity, 'json');

Then, you can return a new response using with string:

return new Response($serializedEntity, 200, array('Content-Type' => 'application/json'));