从Android发送JSON HTTP POST请求

Question

我正在使用下面的代码发送http POST请求，该请求将对象发送到WCF服务。可以，但是如果我的WCF服务还需要其他参数怎么办？如何从Android客户端发送它们？这是我到目前为止编写的代码：StringBuilder sb = new StringBuilder();  String http = "http://android.schoolportal.gr/Service.svc/SaveValues";  HttpURLConnection urlConnection=null;  try {      URL url = new URL(http);      urlConnection = (HttpURLConnection) url.openConnection();    urlConnection.setDoOutput(true);       urlConnection.setRequestMethod("POST");      urlConnection.setUseCaches(false);      urlConnection.setConnectTimeout(10000);      urlConnection.setReadTimeout(10000);      urlConnection.setRequestProperty("Content-Type","application/json");       urlConnection.setRequestProperty("Host", "android.schoolportal.gr");    urlConnection.connect();      //Create JSONObject here    JSONObject jsonParam = new JSONObject();    jsonParam.put("ID", "25");    jsonParam.put("description", "Real");    jsonParam.put("enable", "true");    OutputStreamWriter out = new   OutputStreamWriter(urlConnection.getOutputStream());    out.write(jsonParam.toString());    out.close();      int HttpResult =urlConnection.getResponseCode();      if(HttpResult ==HttpURLConnection.HTTP_OK){          BufferedReader br = new BufferedReader(new InputStreamReader(              urlConnection.getInputStream(),"utf-8"));          String line = null;          while ((line = br.readLine()) != null) {              sb.append(line + "\n");          }          br.close();          System.out.println(""+sb.toString());      }else{              System.out.println(urlConnection.getResponseMessage());      }  } catch (MalformedURLException e) {           e.printStackTrace();  }  catch (IOException e) {      e.printStackTrace();      } catch (JSONException e) {    // TODO Auto-generated catch block    e.printStackTrace();}finally{      if(urlConnection!=null)      urlConnection.disconnect();  }  

Answer 1

在服务器端，参数顺序无关紧要。似乎您将json放入有效负载中...在此otherParametersUrServiceNeed = "param1=a&param2=b&param3=c";放置服务器所需的参数名称，这仅是示例..您应首先确定请求中服务器所需的参数String otherParametersUrService如我的示例所示，添加这些参数并在其中添加值。