解析通过 Symfony 组件的 JsonResponse 返回的 JSON 时出错
Error parsing JSON returned via Symfony Component's JsonResponse
我的 Ajax 电话:
$.ajax({
url : path,
type: 'POST',
dataType : 'json',
data: data,
success: function(memberExtra) {
console.log (memberExtra);
}
});
我的回复:
HTTP/1.0 201 Created
Cache-Control: no-cache
Content-Type: application/json
Date: Tue, 10 Feb 2015 23:49:09 GMT
{"memberExtras":{"label":"seller","dropdown":"abc"}}
我的PHP:
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\JsonResponse;
/**
* Update the pulldown menus.
*
* @Route("/classification", name="classification")
* @Template()
*/
public function classificationAction(Request $request)
{
$memberType = $request->request->get('classification');
$label = $memberType["user"]["memberType"];
$dropdown = "abc";
$response = new Response(json_encode(array('memberExtras' => array(
'label' => $label,
'dropdown' => $dropdown,
))), Response::HTTP_CREATED);
$response->headers->set('Content-Type', 'application/json');
return new Response($response);
}
console.log不输出任何东西。即使像 ("test").
这样的正则文本表达式
如果我删除 dataType : 'json' 声明并尝试通过 $.parseJSON(memberExtra ),我得到这个错误:
SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data
并不意外。基本上,解析器似乎被 Symfony class 返回的 header 绊倒了。我怎样才能避免这个 header 而只是到达 JSON?
谢谢!
将return new Response($response);
替换为return $response;
基本语法:
$response = new Response();
$response->setContent(json_encode(array(
'id' => $entity->getId(),
'other' => $entity->getOther(),
)));
$response->headers->set('Content-Type', 'application/json');
return $response;
简单尝试:
return $response;
而不是 return
new Response($response);
顺便说一句,我建议你简单地使用
return new JsonResponse($myarray)
并从您的方法中删除注解@Template。
希望对您有所帮助
我的 Ajax 电话:
$.ajax({
url : path,
type: 'POST',
dataType : 'json',
data: data,
success: function(memberExtra) {
console.log (memberExtra);
}
});
我的回复:
HTTP/1.0 201 Created
Cache-Control: no-cache
Content-Type: application/json
Date: Tue, 10 Feb 2015 23:49:09 GMT
{"memberExtras":{"label":"seller","dropdown":"abc"}}
我的PHP:
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\JsonResponse;
/**
* Update the pulldown menus.
*
* @Route("/classification", name="classification")
* @Template()
*/
public function classificationAction(Request $request)
{
$memberType = $request->request->get('classification');
$label = $memberType["user"]["memberType"];
$dropdown = "abc";
$response = new Response(json_encode(array('memberExtras' => array(
'label' => $label,
'dropdown' => $dropdown,
))), Response::HTTP_CREATED);
$response->headers->set('Content-Type', 'application/json');
return new Response($response);
}
console.log不输出任何东西。即使像 ("test").
这样的正则文本表达式如果我删除 dataType : 'json' 声明并尝试通过 $.parseJSON(memberExtra ),我得到这个错误:
SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data
并不意外。基本上,解析器似乎被 Symfony class 返回的 header 绊倒了。我怎样才能避免这个 header 而只是到达 JSON?
谢谢!
将return new Response($response);
替换为return $response;
基本语法:
$response = new Response();
$response->setContent(json_encode(array(
'id' => $entity->getId(),
'other' => $entity->getOther(),
)));
$response->headers->set('Content-Type', 'application/json');
return $response;
简单尝试:
return $response;
而不是 return
new Response($response);
顺便说一句,我建议你简单地使用
return new JsonResponse($myarray)
并从您的方法中删除注解@Template。
希望对您有所帮助