Question

我有扩展FosRestController的控制器。如何从序列化返回正常结果？

现在来自jsonp return

functionName([{},{},{},{},{},{},{},{},{},{},{},{}]);

现在来自json return

[{"id":1,"title":"\u053f\u0565\u0576\u057f\u0580\u0578\u0576"},{"id":11,"title":"\u0546\u0578\u0580\u0584-\u0544\u0561\u0580\u0561\u0577"}]

这是我的行动

/**
 * This function is used to get all Districts.
 *
 * @ApiDoc(
 *  resource=true,
 *  section="District",
 *  description="This function is used to get all Districts",
 *  statusCodes={
 *         200="Returned when successful",
 *         404="Returned when the Districts is empty"
 *     }
 * )
 *
 * @return mixed
 * @Rest\View(serializerGroups={"main"})
 */
public function cgetAction(Request $request)
{
    // for jsonP
    $callback = $this->get('request')->get('callback');

    $em = $this->getDoctrine()->getManager();
    $districts = $em->getRepository(self::ENTITY)->findAll();

    if(!is_null($request->getRequestFormat()) && !is_null($callback))
    {
        $response = new JsonResponse($districts, 200, array());
        $response->setCallback($callback);
        return $response;
    }
    else
    {
        return $districts;
    }
}

Answer 1

你必须通过以下方式构建json：

$response = new JsonResponse(array_map(function($obj) {
    return array('id' => $obj->getId(), 'title' => $obj->getTitle());
}, $districts), 200, array());

我如何从Symfony2休息控制器返回jsonp结果

1 个答案: