Question

我尝试过很多版本的解析来自我的.net网络服务和我当前的JSON数据在我的JSON数据前添加了一个{“d”：.....我不知道为什么。

手动调用我的webservice时，数据格式正确。但是当我使用这个解析器时，我已经检查过.names（）以列出JSONObjects的名称，它只显示[“d”]

是否有任何推荐的代码来解析java中的JSON Web服务？

继承我解析数据的代码

        //initialize
                InputStream is = null;
                String result = "";
                JSONObject jArray = null;
                String url = "http://10.0.2.2:1672/Eventurous/WsEventurousMobile.asmx/getEventsList";
                //http post
                try{

                    HttpClient httpclient = new DefaultHttpClient();
                    HttpPost httppost = new HttpPost(url);

                    JSONObject obj = new JSONObject();
                    obj.put("category",category);
                    httppost.setEntity(new StringEntity(obj.toString(),"UTF-8"));

                    httppost.setHeader("Accept","application/json");
                    httppost.setHeader("Content-type","application/json");

                    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());
                }

                //convert response to string
                try{
                    BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
                    StringBuilder sb = new StringBuilder();
                    String line = null;
                    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());
                }

                //try parse the string to a JSON object
                try{
                        jArray = new JSONObject(result);
                }catch(JSONException e){
                    Log.e("log_tag", "Error parsing data "+e.toString());
                }

                try{

                        JSONArray catalogObj = jArray.getJSONArray("Table");
   //Im using the newtonsoft json dataset converter and so the name of the JSONArray
       //is by default Table and i have no idea how to change it either

                        JSONObject event = catalogObj.getJSONObject(0);
                        return event.getString("EventName");        

                }
                catch(Exception e)
                {
                    return e.toString();
                }

Answer 1

我建议使用gson lib，http://code.google.com/p/google-gson/。

是否有任何推荐的代码来解析java中的JSON Web服务？

在不知道json数据的情况下，很难给出任何具体的建议。但无论如何，我建议使用gson lib，这在Android开发者中很受欢迎。

如何正确解析JSON .net Web服务到Android？

1 个答案: