user_id
을 JSON을 통해 PHP에 보내려고하지만 Error parsing data org.json.JSONException: Value 2 of type java.lang.Integer cannot be converted to JSONObject
이 표시됩니다.Android : 데이터를 파싱하는 중 오류가 발생하여 JSONObject로 변환 할 수 없습니다.
어떻게하면됩니까? 미리 감사드립니다.
클래스 : 문제는 당신이 된 JSONObject로 정수 유형의 개체를 변환하려고
jsonobject = jsonarray.getJSONObject(i);
이 라인에 의해 발생
public class Home1 extends Activity {
JSONObject jsonobject;
JSONArray jsonarray;
ListView listview;
ListViewAdapter1 adapterrr;
SharedPreferences pref;
String uid;
String user_i,us;
String ques_i;
ArrayList<HashMap<String, String>> arraylist;
//static String BET_ID = "bet_id";
static String QUESTION = "question";
static String QUES_ID = "ques_id";
static String ANSWER = "answer";
ArrayList<HashMap<String, String>> data;
HashMap<String, String> resultp = new HashMap<String, String>();
Set<String> newset=new HashSet<String>();
@SuppressLint("NewApi")
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
pref = PreferenceManager.getDefaultSharedPreferences(this);
setContentView(R.layout.questionlist1);
uid = pref.getString("user_id",null);
Log.d("uid", ""+uid);
Intent i = getIntent();
ques_i = i.getStringExtra("qsid");
Log.d("qsid", ""+ques_i);
new DownloadJSON().execute();
}
private class DownloadJSON extends AsyncTask<Void, Void, Void> {
/*protected ArrayList<NameValuePair> parameters;
public DownloadJSON() {
parameters = new ArrayList<NameValuePair>();
NameValuePair us = new BasicNameValuePair("user_id", uid);
parameters.add(us);
}*/
@Override
protected Void doInBackground(Void... params) {
arraylist = new ArrayList<HashMap<String, String>>();
/*jsonobject = JSONfunctions
.getJSONfromURL("http://192.168.1.23/mutilatedphp/QuizGame/filtercheck.php");*/
try {
jsonobject = JSONfunctions
.getJSONfromURL("http://192.168.1.23/mutilatedphp/QuizGame/filtercheck.php?user_id="+uid);
jsonarray = jsonobject.getJSONArray("ques");
for (int i = 0; i < jsonarray.length(); i++) {
HashMap<String, String> map = new HashMap<String, String>();
jsonobject = jsonarray.getJSONObject(i);
map.put("ques_id", jsonobject.getString("ques_id"));
map.put("question", jsonobject.getString("question"));
map.put("answer", jsonobject.getString("answer"));
arraylist.add(map);
}
} catch (JSONException e) {
Log.e("Error", e.getMessage());
e.printStackTrace();
}
return null;
}
@Override
protected void onPostExecute(Void args) {
listview = (ListView) findViewById(R.id.listView2);
adapterrr = new ListViewAdapter1(Home1.this, arraylist);
listview.setAdapter(adapterrr);
}
서버의 json_response는 무엇입니까? – nitesh