Question

I am using symfony 1.4, to create my project with propel as ORM. i want to get the response in JSON format, when i call a url. I have set the headers to "application/json" but it is not working, i am getting the response back in HTML format..which i am not able to decode. How can we set content-type in symfony?? example code: Action-

 public function executeIndex(sfWebRequest $request)
 {
     $data_array=array("name" => "harry", "mobile" => "9876543210");
     $data_json=json_encode($data_array);
     $this->output=$data_json;  
 }

查看-

<?php
  header( Content-type: application/json );
  echo $output;
?>

Answer 1

ok i got where i was going wrong..yhe code should have been.. Action-

public function executeIndex(sfWebRequest $request)
{
 $this->getResponse()->setContentType( application/json );
 $data_array=array("name" => "harry", "mobile" => "9876543210");
 $data_json=json_encode($data_array);
 return $this->renderText($data_json); 
}

这个代码对我有效，如果你有更好的解决方案，请发布。。

Answer 2

您也可以在模块的view.yml（apps/｛app_name｝/modules/｛module_name｝/config/view.yml）文件中对此进行定义。

indexSuccess:
  has_layout: false
  http_metas:
    content-type: application/json

Answer 3

哦作为一种替代方案，我只想说我一直在使用路由系统，它提供了一种非常巧妙的方法来完成这项工作：

->；在你的路线上。yml

json_test:
  url: /test
  class: sfRequestRoute
  param: { module: test, action: index, sf_format: json }

然后，框架将自动获取您必须创建的index.json.php视图。如上所述，您可以使用json_encode在操作中生成内容，尽管有一些参数可以将其放入视图中。

Now... Ok, if you are interested in picking some more about this, have a look at the "Practical symfony" tutorial: Day 15: Web Services There s some good stuff down there !

Answer 4

在REST风格中，只需在URI中添加.format，创建相对模板，Symfony Routing系统就可以为我们完成其余的工作。

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