【发布时间】:2015-05-30 07:55:33
【问题描述】:
我不想使用 FOSRestBundle 从 api 返回 json 响应,但 FOSRestBundle 想要返回 twig 而不是 json。
我的控制器方法如下所示
/**
* @Rest\Post("/producers-from-points",name="get-producers-from-points",options={"expose"=true})
* @Rest\View(serializerGroups={"list"}, statusCode=200)
*
* @return \FOS\RestBundle\View\View
* @throws \Exception
*/
public function getProducersFromPointsAction()
{
$json = parent::getJsonFromRequest();
if($json===false){
throw new \Exception("");
}
/** @var EntityManager $em */
$em = $this->getDoctrine()->getManager();
$poss = $em->getRepository('AppBundle:PointOfSale')->getPointOfSalesByAroundPoints($json->points);
return ['pointsOfSales'=>$poss];
}
config.yml
fos_rest:
routing_loader:
default_format: json
include_format: false
param_fetcher_listener: true
body_listener: true
format_listener: false
view:
view_response_listener: 'force'
exception:
messages:
'Exception': true
【问题讨论】:
标签: json symfony fosrestbundle