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();  
}