Symfony2 JSON响应返回奇怪的UTF字符

时间:2013-02-10 14:02:44

标签: json symfony unicode

这是我的控制器方法:

  public function sendjsonAction()
  {

    $message = $this->getDoctrine()
    ->getRepository('AcmeStoreBundle:Message')
    ->findAll();
    $serializer = new Serializer(array(new GetSetMethodNormalizer()), array('message' => new 
JsonEncoder()));
    $message = $serializer->serialize($message, 'json');    
    return new JsonResponse($message);

  }

这是我的路由器:

acme_store_sendjson:
    pattern:  /sendjson/
    defaults: { _controller: AcmeStoreBundle:Default:sendjson}

这是我去/ sendjson /时所得到的:

"[{\u0022id\u0022:1,\u0022iam\u0022:1,\u0022youare\u0022:2,\u0022lat\u0022:50.8275853,\u0022lng\u0022:4.3809764,\u0022date\u0022:{\u0022lastErrors\u0022:{\u0022warning_count\u0022:0,\u0022warnings\u0022:[],\u0022error_count\u0022:0,\u0022errors\u0022:[]},\u0022timezone\u0022:{\u0022name\u0022:\u0022UTC\u0022,\u0022location\u0022:{\u0022country_code\u0022:\u0022??

(同样地继续)

我尝试使用以下内容进行ajax调用(使用jQuery):

$.getJSON('/app_dev.php/sendjson', function(data) {
  var items = [];

  $.each(data, function(key, val) {
    items.push('<li id="' + key + '">' + val + '</li>');
  });

  $('<ul/>', {
    'class': 'my-new-list',
    html: items.join('')
  }).appendTo('body');
});

我得到了

Uncaught TypeError: Cannot use 'in' operator to search for '1549' in [{"id":1,...

当我更改Symfony2的响应类型时,我得到一个

列表

[对象] [对象] [对象] [对象] [对象] [对象] ...

我做错了什么?我应该解析将\ u0022转换为“或者从一开始我的反应是否错误?”

修改

我还尝试将控制器更改为:

  public function sendjsonAction()
  {
$encoders = array(new XmlEncoder(), new JsonEncoder());
$normalizers = array(new GetSetMethodNormalizer());
$serializer = new Serializer($normalizers, $encoders);

    $message = $this->getDoctrine()
    ->getRepository('AcmeStoreBundle:Message')
    ->findAll();
$serializer = $serializer->serialize($message, 'json');
    return new Response($serializer);
}

这次我获得了VALID JSON,(根据Jsonlint)buuttt标题不是application / json ...(有道理因为我发送了一个Response而不是JsonResponse ......)(但那就是我在尝试避免因为JsonResponse似乎正在改变添加奇怪的字符)

[{"id":1,"iam":1,"youare":2,"lat":50.8275853,"lng":4.3809764,"msgbody":"I saw you over there what's up!"},{"id":2,"iam":1,"youare":2,"lat":50.8275853,"lng":4.3809764,"msgbody":"I saw you over there what's up!"},{"id":3,"iam":1,"youare":2,"lat":50.8275853,"lng":4.3809764,"msgbody":"I saw you over there what's up!"},{"id":4,"iam":1,"youare":2,"lat":50.8275853,"lng":4.3809764,"msgbody":"I saw you over there what's up!"},{"id":5,"iam":1,"youare":2,"lat":50.8275853,"lng":4.3809764,"msgbody":"I saw you over there what's up!"},{"id":6,"iam":1,"youare":2,"lat":50.8275853,"lng":4.3809764,"msgbody":"I saw you over there what's up!"}]

6 个答案:

答案 0 :(得分:4)

我找到了答案。

1)只要JSON是有效的,内容类型不是application / json而是text / html并不“真正重要”。我的JS没有玩的原因是我要求val而不是val的属性,例如val.msgbody。 :

所以我的Javascript应该是

$.getJSON('/app_dev.php/sendjson', function(data) {
  var items = [];

  $.each(data, function(key, val) {
    items.push('<li id="' + key + '">' + val.msgbody + '</li>');
  });

  $('<ul/>', {
    'class': 'my-new-list',
    html: items.join('')
  }).appendTo('body');
});

如果要求Content-Type,那么控制器可能是这样的:

 public function sendjsonAction()
  {
    $encoders = array(new JsonEncoder());
    $normalizers = array(new GetSetMethodNormalizer());
    $serializer = new Serializer($normalizers, $encoders);
    $message = $this->getDoctrine()
      ->getRepository('AcmeStoreBundle:Message')
      ->findAll();
    $response = new Response($serializer->serialize($message, 'json')); 
    $response->headers->set('Content-Type', 'application/json');
    return $response;
  }

答案 1 :(得分:2)

序列化是规范化的过程 - 创建一个表示对象的数组 - 并对该表示进行编码(即对JSON或XML)。 JsonResponse为您处理编码部分(查看类的名称),因此您无法传递“序列化对象”,否则将再次编码。 因此,解决方案只是规范化对象并将其传递给JsonResponse:

public function indexAction($id)
{
    $repository = $this->getDoctrine()->getRepository('MyBundle:Product');
    $product = $repository->find($id);

    $normalizer = new GetSetMethodNormalizer();

    $jsonResponse = new JsonResponse($normalizer->normalize($product));
    return $jsonResponse;
}

答案 2 :(得分:0)

您可以仅使用规范化器而不使用序列化器来解决此问题:

    $normalizer = new ObjectNormalizer();
    $array = $normalizer->normalize($newEntry);
    $entryJSONFile = json_encode($array, JSON_UNESCAPED_UNICODE);

答案 3 :(得分:0)

https://www.php.net/manual/en/json.constants.php

https://www.php.net/manual/en/function.json-encode.php

要输出JSON字符串的对象\Symfony\Component\HttpFoundation\JsonResponse使用函数json_encode,通过对象setEncodingOptions的方法,您可以按JSON_UNESCAPED_UNICODE之类的按位常量设置输出JSON字符串选项:

  

按字面意义编码多字节Unicode字符(默认为转义为\ uXXXX)。自PHP 5.4.0起可用。

$jsonResponse = new \Symfony\Component\HttpFoundation\JsonResponse($arrayForJSON);
$jsonResponse->setEncodingOptions($this->getEncodingOptions()|JSON_UNESCAPED_UNICODE|JSON_NUMERIC_CHECK|JSON_PRETTY_PRINT);
$jsonResponse->send();

答案 4 :(得分:0)

您的代码两次在json中编码。当您使用序列化程序自己进行json编码时,请使用Response类。

将返回新的JsonResponse($ message)替换为返回新的Response($ message)

答案 5 :(得分:-2)

问题是您是将字符串传递给JsonResponse而不是数组。

您的控制器代码是:

...
return new JsonResponse($message)
...

您的控制器代码必须是:

...
return new JsonResponse(json_decode($message, true))
...