Question

我是Symfony2和许多软件包的新手，但我正在研究原型REST API，我使用了FOSRestBundle，还使用了Symfony2生成的本机CRUD。但是，我发现虽然Symfony2 CRUD代码返回正确的JSON格式响应，但POST给了我一个错误，我正在寻找有关如何解决问题的解释和教程。查看实体的GET和POST代码，在这种情况下可以说地址：

/**
 * Lists all Address entities.
 *
 * @Route("/", name="address")
 * @Method("GET")
 * @Template()
 */
public function indexAction()
{
    $em = $this->getDoctrine()->getManager();

    $entities = $em->getRepository('MyWebServicesBundle:Address')->findAll();

    return array(
        'entities' => $entities,
    );
}
/**
 * Creates a new Address entity.
 *
 * @Route("/", name="address_create")
 * @Method("POST")
 * @Template("MyWebServicesBundle:Address:new.html.twig")
 */
public function createAction(Request $request)
{
    $entity = new Address();
    $form = $this->createCreateForm($entity);
    $form->handleRequest($request);

    // Inserted new code for deserialization
    $entity->setUseruid($request->request->get("useruid"));
    $entity->setCity($request->request->get("city"));
    $entity->setLatitude($request->request->get("latitude"));
    $entity->setLongitude($request->request->get("longitude"));

    $serializer = JMS\Serializer\SerializerBuilder::create()->build();
    $entity = $serializer->deserialize($request->request->all(), 'Name\BundleName\Entity\Address', 'json');

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

        return $this->redirect($this->generateUrl('address_show', array('id' => $entity->getId())));
    }

    return array(
        'entity' => $entity,
        'form'   => $form->createView(),
    );
}

假设我使用AJAX向API发送请求，这里分别是我的GET和POST代码：

        $.ajax({
            dataType: "json",
            type: "GET",
            url: "/symfony/web/app_dev.php/address/",
            success: function (responseText)
            {
                alert("Request was successful, data received: " + responseText); 
            },
            error: function (error) {
                alert(JSON.stringify(error));
            }
        });

        $.ajax({
            dataType: "json",
            type: "POST",
            data: {"id":1,"useruid":"Nothing","type":"Office in Space","latitude":"74.3","longitude":"33.2","displayed":true,"public":true,"verified":true,"street":"Something","city":"Something","country":"Space","region":"North America","created":"2009-03-07T00:00:00-0500","delete_status":"active"},
            url: "/symfony/web/app_dev.php/address/",
            success: function (responseText)
            {
                console.log("Request was successful, data received: " + JSON.stringify(responseText)); 

            },
            error: function (error) {
                alert(JSON.stringify(error));
            }
        });

当GET返回正确时，POST返回以下错误：{“code”：500，“message”：“警告：json_encode（）：在/ var / www / projects / symfony / vendor / jms / serializer中检测到递归/src/JMS/Serializer/JsonSerializationVisitor.php第29行“}。我需要做什么来解决错误？我可以返回单个实体或所有实体，但POST只是将数据发送到创建实体的路径，从而产生错误。我已将我的JMSSerializer软件包从0.12。*更新为dev-master，并检查以确保AJAX发送的数据中没有NULL值，但错误仍然存在。如何让我的POST控制器从POST中发送给它的有效JSON创建数据？

我也试过PUT，结果是一样的，它没有更新资源，不应该只更新表中的记录吗？如果我需要提供更多信息以找出此错误的来源，请告诉我们！ 代码已在上面进行了编辑。

Answer 1

$ this-＆gt; redirect返回不是您想要的RedirectResponse对象。如果您想要返回重定向网址，只需返回array('redirect_url' => $this->generate(...))，然后使用javascript重定向客户端。

Symfony2 CRUD，GET返回JSON，但POST创建EMPTY记录，PUT什么都不做

1 个答案: