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:151clinicname:test clinicUserName:Password:1c79c768fdda6fe5408d3287f16aa4f8fullname:Raghvendra Patelfirstname:lastname:mobile_number:8982793236Email:opulence.dev110@gmail.comemail_verify:1about:address:country:state:city:zip:UserRoleID:2Status:0fb_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:Wm7DPUcoupon_code:22qualified_physiotherapist:1terms_and_conditions:1Physio_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 linksthose 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 ?