获取JSONException:类型为java的值。解析JSON响应时，无法将lang.String转换为JSONObject

Question

我开发了一个Android应用程序，请求从响应JSON格式的服务器的位置坐标（目前它只发送两个地方）：

这是来自服务器的php代码：

$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时，它会尝试以这种方式解析坐标：

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

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：

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;

    }
}

下面是带有错误的JSON：

当应用程序试图解析JSON响应时，它会崩溃，并发送一个JSON异常：类型java.lang.字符串的值不能转换为JSONObject如何解决这个问题？谢了

Answer 1

刚开始删除“Array”，json的格式就不好了

Answer 2

您的Web服务没有创建有效的JSON。JSON字符串只能以{或[开头。你的从String"Array"开始。

您可以在这里的Wikipedia条目中阅读JSON格式。