我已经开发了一个Android应用程序请求的地方，从一个服务器，响应JSON格式（暂时它只是发送两个地方）坐标：

I have developed an Android application that requests places coordinates from a server which responds in JSON format(for the moment it just sends two places):

这是从服务器上的PHP code：

this is the php code from the server:

$place = $db->getCoordinates($name);
if ($place != false) {

     $response[1]["success"] = 1;
     $response[1]["place"]["H"] = $place[1]["H"];
     $response[1]["place"]["V"] = $place[1]["V"];
     $response[1]["place"]["placeid"] = $place[1]["placeid"];
     $response[1]["place"]["name"] = $place[1]["name"];
     $response[1]["place"]["type"] = $place[1]["type"];
     $response[1]["place"]["note"] = $place[1]["note"];
     // place found
     // echo json with success = 1
     $response[2]["success"] = 1;
     $response[2]["place"]["H"] = $place[2]["H"];
     $response[2]["place"]["V"] = $place[2]["V"];
     $response[2]["place"]["placeid"] = $place[2]["placeid"];
     $response[2]["place"]["name"] = $place[2]["name"];
     $response[2]["place"]["type"] = $place[2]["type"];
     $response[2]["place"]["note"] = $place[2]["note"];
     echo json_encode($response);
} 

当应用程序获取坐标它试图分析它们是这样的：

When the app get the coordinates it try to parse them this way:

            JSONObject json_places = userFunction.getPlaces();
            JSONObject  places = json_places.getJSONObject("1");
            JSONObject  coord = places.getJSONObject("place");

getplaces（）：

getplaces():

public JSONObject getPlaces(){
    // Building Parameters
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("tag", allcoordinates_tag));
    JSONObject json = jsonParser.getJSONFromUrl("http://"+ip+placesURL, params);
    // return json
    Log.i("JSON", json.toString());
    return json;
}

JSONParser：

JSONParser:

public class JSONParser {

    static InputStream is = null;
    static JSONObject jObj = null;
    static String json = "";

    // constructor
    public JSONParser() {

    }

    public JSONObject getJSONFromUrl(String url, List<NameValuePair> params) {

        // Making HTTP request
        try {
            // defaultHttpClient
            HttpPost httpPost = new HttpPost(url);
            HttpParams httpParameters = new BasicHttpParams();
            int timeoutConnection = 9000;
            HttpConnectionParams.setConnectionTimeout(httpParameters, timeoutConnection);
            int timeoutSocket = 9000;
            HttpConnectionParams.setSoTimeout(httpParameters, timeoutSocket);
            DefaultHttpClient httpClient = new DefaultHttpClient(httpParameters);

            httpPost.setEntity(new UrlEncodedFormEntity(params));

            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();

        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        }
        catch (ConnectTimeoutException e) {

            e.printStackTrace();
        }
        catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (Exception e) {
            e.printStackTrace();
        } 


        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();
            json = sb.toString();
            Log.e("JSON", json);
        } catch (Exception e) {
            Log.e("Buffer Error", "Error converting result " + e.toString());
        }

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

        // return JSON String
        return jObj;

    }
}

和这里是有错误的JSON：

and here is the JSON with the error:

当应用程序试图解析JSON响应崩溃并发送JSONException：java.lang.String类型的值不能转换到的JSONObject 我该如何解决呢？ 谢谢

when the App tries to parse the JSON response it crashes and sends a JSONException: Value of type java.lang.String cannot be converted to JSONObject how can i solve this? THANKS

推荐答案

您的Web服务是不是建立有效的JSON。 JSON字符串只能先从 {或 [。你开始与字符串阵列。

You web service is not creating valid JSON. A JSON string can only start with { or [. Yours starts with the String "Array".

您可以阅读有关JSON格式的维基百科条目这里。

You can read about the JSON format in its Wikipedia entry here.