get JSONException：解析JSON响应时，无法将java.lang.String类型的值转换为JSONObject

我开发了一个Android应用程序，它请求服务器的地点坐标以JSON格式响应（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); } 

当应用获取坐标时，它会尝试以这种方式解析它们：

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

getplaces（）：

 public JSONObject getPlaces(){ // Building Parameters List params = new ArrayList(); 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 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; } } 

这是带有错误的JSON：

当App试图解析JSON响应时，它会崩溃并发送一个JSONException：java.lang.String类型的值无法转换为JSONObject我该如何解决这个问题？ 谢谢

Answer 1

您的Web服务未创建有效的JSON。 JSON字符串只能以{或[ 。 你的字符串"Array" 。

您可以在此处阅读其Wikipedia条目中的JSON格式。

Answer 2

刚刚开始删除“数组”，json形成不好