Kein Wert für ID? JSON php android java

Question

Im stumped jetzt mit diesem Fehler, wie ich versuche, Daten von mysql Server zu meinem Android-Studio über PHP aber erfolglos zu bekommen. Irgendeine Idee warum gibt es keinen Wert für ID?

08-07 17:37:23.994 20693-20693/edu.nyp.reportify D/HttpAsyncTask: {"results":[{"id":"36","userid":"91","image":"reportImage\/579edd1aeb961 .png","reportType":"Crime","address":"Republic Crescent","latitude":"1.4436883","longitude":"103.786","details":"Car and bus accident","dateTime":"Aug 1, 2016 1:24:02 PM","status":"pending"},{"id":"39","userid":"91","image":"reportImage\/57a6d32ae45ff .png","reportType":"Accident","address":"Republic Crescent","latitude":"1.4436883","longitude":"103.786","details":"test","dateTime":"Aug 7, 2016 2:20:00 PM","status":"pending"}]} 
08-07 17:37:23.995 20693-20693/edu.nyp.reportify W/System.err: org.json.JSONException: No value for id 


try { 
       JSONArray resultArray = jsonObject.getJSONArray("results"); 
       reports.clear(); 
       // Store all results into locations table layout 
       for (int i = 0; i < resultArray.length(); i++) { 

        //JSONObject resultObj = resultArray.getJSONObject(i); 
        String reportid = jsonObject.getString("id"); 
        String userid = jsonObject.getString("userid"); 
        String address = jsonObject.getString("address"); 
        String status = jsonObject.getString("status"); 
        JSONObject ReportObj = jsonObject.getJSONObject("resultObj"); 
        reports.add(new ReportObj(reportid, reportType, address, status)); 
       } 

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

Answer 1

Sie müssen das Objekt erhalten das Array bilden, und versuchen, die Attribute von dort

JSONObject tmpobject = resultArray.getJSONObject(i); 
String reportid = tmpobject.getString("id"); 
String userid = tmpobject.getString("userid"); 
String address = tmpobject.getString("address"); 
String status = tmpobject.getString("status"); 

zu erhalten