FOSRestBundle return 对象错误

Question

我使用 FOERestBundle 和 class 视图。当我验证实体时，我有这样的对象错误，这是：

[
 {
   "property_path": "main_skill",
   "message": "This value should not be blank."
 },
 {
   "property_path": "type",
   "message": "This value should not be blank."
 },
 {
   "property_path": "description",
   "message": "This value should not be blank."
 }
]

当用户不是这样的有效安全令牌时，我需要 return 对象错误

[
 {
   "property_path": "main_skill",
   "message": "This value should not be blank."
 },
]

现在我有了纯文本。这是我的终点

    /**
 * Update existing Bit from the submitted data.
 *
 * @ApiDoc(
 * resource = true,
 * description = "Update single Bit",
 *  parameters={
 *      {"name"="status", "dataType"="string", "required"=false, "description"="status for bit"},
 *      {"name"="text", "dataType"="string", "required"=true, "description"="text for rejected"},
 *      {"name"="token", "dataType"="string", "required"=true, "description"="is equally md5('email'.secret_word)"}
 *  },
 * statusCodes = {
 *      200 = "Bit successful update",
 *      400 = "Secret token is not valid"
 * },
 *  section="Bit"
 * )
 * @RestView()
 *
 * @param Request $request
 * @param string  $id
 *
 * @return View
 */
public function putBitAction(Request $request, $id)
{
    $manager = $this->getDoctrine()->getManager();
    $token = $this->get('request')->request->get('token');
    $user = $this->getDoctrine()->getRepository('MyBundle:Users')->findOneBySecuritytoken($token);
    $bit = $manager->getRepository('MyBundle:Bit')->find($id);
    $view = View::create();

    if (!empty($user) && !empty($bit) && !empty($token)) {

            *some logic
            $view = $this->view($bit, 200);

            return $this->handleView($view);
        }
    } else {
        $view = $this->view('Secret token is not valid', 400);

        return $this->handleView($view);
    }
}

现在我有了纯文本

Response Body [Raw]
"Secret token is not valid"

这是 return 对象错误验证，这没问题

[
 {
   "property_path": "main_skill",
   "message": "This value should not be blank."
 },
 {
   "property_path": "type",
   "message": "This value should not be blank."
 },
 {
   "property_path": "description",
   "message": "This value should not be blank."
 }
]

如何return自定义错误，比如对象不是纯文本？

Answer 1

只需像数组一样传递您的数据，并告诉视图将其呈现为 json 应该会生成您想要的输出

$view = $this->view(
              array(
                'property_path'  => 'main_skill',
                'message' => "error"
                //whatever your object/array structure is
              ),
              500 //error code for the error
            );

$view->setFormat('json');    
return $this->handleView($view);

Answer 2

您可以使用 Symfony 的 HTTPExceptions，因为它们将由 FOSRestBundle 处理。

参见：http://symfony.com/doc/current/bundles/FOSRestBundle/4-exception-controller-support.html

public function putBitAction(Request $request, $id)
{
    $token = $request->get('token');
    if (null === $token) {
        throw new BadRequestHttpException('Provide a secret token');
    }

    $manager = $this->getDoctrine()->getManager();
    $user = $manager->getRepository('MyBundle:Users')->findOneBySecuritytoken($token);
    if (null === $user) {
        throw new BadRequestHttpException('Secret token is not valid');
    }        

    $bit = $manager->getRepository('MyBundle:Bit')->find($id);
    if (null === $token) {
        throw new NotFoundHttpException('Bid not found');
    }

    $view = $this->view($bit, 200);
    return $this->handleView($view);
}

这是一个怎样的 PUT 请求？您应该重命名为 getBidAction.

FOSRestBundle return 对象错误

FOSRestBundle return object error

php

validation

symfony

fosrestbundle