How to access JSON response and display it in new activity in android?

Question

I am getting the JSON response from PHP as {"success":'Hi Namrata...'} when the details are filled out in the editText field. I want to read the success value and display it in the new activity and in case of {"error":'something happened'} I want to be on the same activity.. How can I do that. Here is my MainActivity

public class MainActivity extends AppCompatActivity {

public Button blogin;TextView content;
public EditText uname, pass;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    uname = (EditText) findViewById(R.id.uname);
    pass = (EditText) findViewById(R.id.pass);
    content=   (TextView)findViewById( R.id.content );
    blogin = (Button) findViewById(R.id.blogin);

    blogin.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {

            new JsonTask().execute("https://www.aboutmyclinic.com/test.php",uname.getText().toString(),pass.getText().toString());

        }
    });



}

public class JsonTask extends AsyncTask<String, String, String> {

    @Override
    protected String doInBackground(String... params) {

        HttpURLConnection connection = null;
        BufferedReader reader = null;


        try {

            URL url = new URL(params[0]);
            connection = (HttpURLConnection) url.openConnection();
            connection.setReadTimeout(10000);
            connection.setConnectTimeout(15000);
            connection.setRequestMethod("POST");
            connection.setDoInput(true);
            connection.setDoOutput(true);
            Uri.Builder builder = new Uri.Builder()
                    .appendQueryParameter("firstParam", params[1])
                    .appendQueryParameter("secondParam", params[2]);
                    //.appendQueryParameter("type", params[3]);
            String query = builder.build().getEncodedQuery();
            OutputStream os = connection.getOutputStream();
            BufferedWriter writer = new BufferedWriter(
                    new OutputStreamWriter(os, "UTF-8"));
            writer.write(query);
            writer.flush();
            writer.close();
            os.close();


            connection.connect();



            InputStream inputStream = connection.getInputStream();
            StringBuffer buffer = new StringBuffer();

            reader = new BufferedReader(new InputStreamReader(inputStream));

            String line;
            while ((line = reader.readLine()) != null) {

                buffer.append(line + "\n");
            }
            return buffer.toString();

        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            if (connection != null) {
                connection.disconnect();
            }
            if (reader != null) {
                try {
                    reader.close();
                } catch (final IOException e) {

                }
            }

        }
        return null;
    }

    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);
        content.setText(result);

    }
  }
}

http://www.startingandroid.com/registration-and-login-with-back-end-php-and-mysql/ please check this. here you will find full detail about sending and receiving data to server and how to parse it as well. — Mustanser Iqbal, Jan 15 '16 at 06:28
Probably JSON String is `result ` then what problem u are getting in sending `result ` String using Intent to next Activity — ρяσѕρєя K, Jan 15 '16 at 06:30
parse JSON coming in response : link http://stackoverflow.com/questions/9605913/how-to-parse-json-in-android — Kapil Rajput, Jan 15 '16 at 06:30
Possible duplicate of [Sending and Parsing JSON Objects](http://stackoverflow.com/questions/2818697/sending-and-parsing-json-objects) — Gandalf the White, Jan 15 '16 at 06:30

raj · Accepted Answer · 2016-01-15T07:20:25.430

0

do this in your postexecute method

  @Override
        protected void onPostExecute(String result) {
            super.onPostExecute(result);
             JSONObject json=new JSONObject(result);
            String sucessvalue= json.getString("success");

          Intent i=new Intent(MainActivity.this,NewActivity.class);
          i.putExtra("sucessval",sucessvalue);
          startActivity(i);

        }

get in another activity

String valueofsucess = getIntent().getStringExtra("sucessval");

edited Jan 15 '16 at 07:20

answered Jan 15 '16 at 06:37

raj

2,088
14
23

I want this success value on a new activity not on the same activity. – Namrata Singh Jan 15 '16 at 07:15
i am going on the new activity but I am not getting the success value. – Namrata Singh Jan 15 '16 at 07:43
have you put the second line on second activity – raj Jan 16 '16 at 05:54
you have to put this line after setContentView in your second activity – raj Jan 16 '16 at 05:57
I did the same. Still getting the same result. – Namrata Singh Jan 16 '16 at 06:33
can i see on team viewer – raj Jan 16 '16 at 06:34
I added this in the second activity: TextView textView = new TextView(this); textView.setTextSize(40); textView.setText(valueofsucess); RelativeLayout layout = (RelativeLayout) findViewById(R.id.content1); layout.addView(textView);... NOw I am getting the success value on next page.. Thanks a lot!! :) – Namrata Singh Jan 16 '16 at 06:53

kevz · Answer 2 · 2016-01-15T08:56:03.117

0

Change below line -

new JsonTask().execute.....

to

new JsonTask(MainActivity.this).execute

Inside the class JsonTask modify -

public class JsonTask extends AsyncTask<String, String, String> {

    Context context;

    JsonTask(Context context){
        this.context = context;
    }

rest all same

Modify your onPostExecute() as below -

@Override
    protected void onPostExecute(String s) {
        super.onPostExecute(s);

        String result = "error";
        try {
            JSONObject jObj = new JSONObject(s);
            if(jObj.has("success")){

                result = jObj.getString("success");
                startActivity(new Intent(context, NextActivity.class)
                .putExtra("result", result));
            }else{
                content.setText(result);
            }
        } catch (JSONException e) {
            e.printStackTrace();
        }
    }

edited Jan 15 '16 at 08:56

answered Jan 15 '16 at 06:44

kevz

2,727
14
39

@Namrata Singh: context = MainActivity.this that u passed to ur AsyncTask – kevz Jan 15 '16 at 07:12
It says "Unfortunately, MyApplication has been stopped". – Namrata Singh Jan 15 '16 at 07:19
@Namrata Singh: well u didn't passed the context to Asynck. I've edit my answer check it out I'm sure it'll help :) – kevz Jan 15 '16 at 07:21
@NamrataSingh: plz lets talk in chat http://chat.stackoverflow.com/rooms/100739/how-to-access-json-response-and-display-it-in-new-activity-in-android – kevz Jan 15 '16 at 07:31
I don't have enough reputation to chat. I am new here. – Namrata Singh Jan 15 '16 at 07:42
@Namrata Singh: where u stuck at? – kevz Jan 15 '16 at 08:45
Error is - java.lang.NullPointerException: Attempt to invoke virtual method 'int java.lang.String.length()' on a null object reference – Namrata Singh Jan 15 '16 at 08:53
@Namrata Singh: I've moved the String result declaration outside if block can u plz check it now. – kevz Jan 15 '16 at 08:57
@NamrataSingh can you please try retrofit tutorials for beginners – silentsudo Jan 15 '16 at 09:06
@NamrataSingh: on which line r u getting the error? – kevz Jan 15 '16 at 09:07
@NamrataSingh you webservice define is wrong it does not stands properly for JSON format please check with you service developer :D – silentsudo Jan 15 '16 at 09:17
@kevz Now I'm getting a different error - java.lang.RuntimeException: Unable to start activity ComponentInfo{com.example.namrata.clinic/com.example.namrata.clinic.Login}: java.lang.IllegalStateException: This Activity already has an action bar supplied by the window decor. Do not request Window.FEATURE_ACTION_BAR and set windowActionBar to false in your theme to use a Toolbar instead. – Namrata Singh Jan 16 '16 at 05:48
@sector11 I'm not getting you and I'll try retrofit later. – Namrata Singh Jan 16 '16 at 05:48

silentsudo · Answer 3 · 2016-01-15T09:16:09.037

I assume you are starting project right away can you start moving to some networking libraries. Like Retrofit. Because it will help you for longer run in project.

    public class RetrofitActivity extends AppCompatActivity implements Callback<ResponseBody> {

        private ProgressDialog dialog;
        @Override
        protected void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.activity_retrofit);
            makeRequest("ashish", "ashish");
        }


        private void makeRequest(String userName, String password) {
            Retrofit retrofit = new Retrofit.Builder()
                    .baseUrl("https://www.aboutmyclinic.com/")
                    .build();
            Service service = retrofit.create(Service.class);
            User user = new User(userName, password);
            Call<ResponseBody> loginRequest = service.login(user.getFirstParam(), user.getSecondParam());
            dialog = new ProgressDialog(this);
            dialog.setMessage("Logging in");
            dialog.show();
            loginRequest.enqueue(this);
        }

        @Override
        public void onResponse(Response<ResponseBody> response) {
            Log.i("Response", "" + response);
 Log.i("Response", "" + response.body().toString());
        try {
            Log.i("Response", "" + response.body().string());
        } catch (IOException e) {
            e.printStackTrace();
        }
            dialog.cancel();
        }

        @Override
        public void onFailure(Throwable t) {
            Log.e("Error", "Unexpected response");
            dialog.cancel();
        }

        public class User {
            String firstParam;
            String secondParam;

            public User(String userName, String password) {
                this.firstParam = userName;
                this.secondParam = password;
            }

            public String getFirstParam() {
                return firstParam;
            }

            public void setFirstParam(String firstParam) {
                this.firstParam = firstParam;
            }

            public String getSecondParam() {
                return secondParam;
            }

            public void setSecondParam(String secondParam) {
                this.secondParam = secondParam;
            }
        }
    }



    public interface Service {

        @POST("/test.php")
        public Call<ResponseBody> login(@Query("firstParam") String firstParam, @Query("secondParam")String secondParam);
    }

How to access JSON response and display it in new activity in android?

3 Answers3