Android

Question

i want create login page with display user detail in next activity.but does not display another information ...how to solve this problem plz help me..

Rajeev Punhani · Answer

Hi Aradhna, In above code it should be JSONArray jarr = new JSONArray( result );//error in this line Instead of "user" it should be result. If the above solution is helpful then, accept the answer.

Aradhna pandey · Answer

Which jar file added...

Aradhna pandey · Answer

This my android code... public class RegisterActivity extends Activity { TextView output; // All xml labels JSONArray user = null; JSONObject json; @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.second); output = (TextView) findViewById(R.id.textView3); new MyTask().execute(); } class MyTask extends AsyncTask { String result = ""; @Override protected String doInBackground(String... arg0) { String url ="http://physiodesk.com/api/userdetail.php?UserID=?"; Log.e("Mytask url---", url); String data =""; HttpClient cl = new DefaultHttpClient(); HttpPost post = new HttpPost(url); try { HttpResponse res = cl.execute(post); InputStream is = res.getEntity().getContent(); BufferedReader br = new BufferedReader(new InputStreamReader(is)); StringBuilder sb = new StringBuilder(); while(true) { String str = br.readLine(); if(str==null) { break; } data +=str; } System.out.println("*****************"+data); result = sb.toString(); } catch (ClientProtocolException e) { e.printStackTrace(); } catch (IOException e) { e.printStackTrace(); } return data; } @Override protected void onPostExecute(String result) { String data1= ""; Log.e("Group JSON", result); try { JSONArray jarr = new JSONArray("user");//error in this line for(int x=0; x { String json = jarr.getString(x); JSONObject job = new JSONObject(json); int userID = Integer.parseInt(job.optString("userID")); String fullname = job.optString("fullname").toString(); String mobile_number= job.getString("mobile_number").toString(); data1+= "Node"+x+" : 
 UserName= "+ fullname +" 
 mo_number= "+ mobile_number +" 
 "; } output.setText(data1); } catch (JSONException e) {e.printStackTrace();} } } } And My json data... {"user":[{"userID":"151","clinicname":"test clinic","UserName":"","Password":"1c79c768fdda6fe5408d3287f16aa4f8","fullname":"Raghvendra Patel","firstname":"","lastname":"","mobile_number":"8982793236","Email":"opulence.dev110@gmail.com","email_verify":"1","about":"","address":"","country":"","state":"","city":"","zip":"","UserRoleID":"2","Status":"0","fb_url":"","twt_url":"","photo":"","clinic_name":"","clinic_slogan":"","about_clinic":"","clinic_email":"","clinic_phone":"","clinic_fb":"","clinic_tw":"","clinic_address":"","clinic_city":"","clinic_state":"","clinic_zip":"","tokenno":"Wm7DPU","coupon_code":"22","qualified_physiotherapist":"1","terms_and_conditions":"1","Physio_ip":"182.77.71.106"}]}null

Rajeev Punhani · Answer

Hi Aradhna, The array clearly states that you are trying to store string in array.Have you added necessary jar file.Please provide the code snippet .

Aradhna pandey · Answer

hello, i was apply this code according to link.but does not solve problem...

Rajeev Punhani · Answer

Hi Aradhana, Please see the below link for the solution. http://stackoverflow.com/questions/18174064/value-of-type-java-lang-string-cannot-be-converted-to-jsonarray If the above solution is helpful then, accept the answer.

Aradhna pandey · Answer

This error show on logcat am not understand this error. org.json.JSONException: Value user of type java.lang.String cannot be converted to JSONArray.plz help

Rajeev Punhani · Answer

Hi Aradhna, Please see these links,those links will help http://www.mkyong.com/android/android-activity-from-one-screen-to-another-screen/ http://www.androidhive.info/2011/08/how-to-switch-between-activities-in-android/ If the above solution is helpful then, accept the answer.

Donovan IT · Answer

you will be able to define your problem , what's the problem ?

Android

Answers (9)