Java (Android) HTTP 发布到 PHP 页面

Question

这是一个 Android 应用程序，我的移动开发人员试图将 json 数据发布到我的 PHP 页面。

下面是正在使用的函数:

public static String postData(String url, String postData) {
  // Create a new HttpClient and Post Header
     InputStream is = null;
     StringBuilder sb = null;
     String result = "";

//   StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
//   StrictMode.setThreadPolicy(policy);
     HttpClient httpclient = new DefaultHttpClient();

     HttpPost httppost = new HttpPost(url);

     try {
         httppost.setEntity(new StringEntity(postData));
         httppost.setHeader("Accept", "application/json");
         httppost.setHeader("Content-type", "application/json");
         // Execute HTTP Post Request
         HttpResponse response = httpclient.execute(httppost);
         HttpEntity entity = response.getEntity();
         is = entity.getContent();

     } catch (Exception e) {
         Log.e("log_tag", "Error in http connection" + e.toString());
         //throw new CustomException("Could not establish network connection");
     }
        try {
                BufferedReader reader = new BufferedReader(new InputStreamReader(
                        is, "utf-8"), 8);

                sb = new StringBuilder();
                sb.append(reader.readLine() + "\n");
                String line = "0";

                while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
                }
                is.close();
                result = sb.toString();

        } catch (Exception e) {
            Log.e("log_tag", "Error converting result " + e.toString());
            //throw new CustomException("Error parsing the response");
        }
  return result;

 }

其中 url 是指向我的 php 网页的链接。在我的 php 页面上，我只是想打印出发布的数据:

print_r($_POST);

但它显示的是一个空数组。我什至尝试使用 Firefox 的 REST 插件并做同样的事情，但它只显示一个空白数组。

如果有人能指出我遗漏了什么，那就太好了。

谢谢。

Answer 1

你需要获取如下的json内容:

if(isset($_POST))
    {
     $json = file_get_contents('php://input');
     $jsonObj = json_decode($json);
     echo $jsonObj;
    }