Handle success request return with Android Async Http

Question

I'm new to android, coming from iOS I don't know much about Java and all of its features. I'm trying to build an application where the user need to log at launch. I'm using a private API which I use like : https://apiUrl.com/login?login=login&password=password It returns me a JSon Object :

{
token: "qqdpo9i7qo3m8lldksin6cq714"
}

So what I'm doing in my code is simple :

MainActivity.java :

Button button = (Button) findViewById(R.id.loginButton);

        button.setOnClickListener(new View.OnClickListener() {
            public void onClick(View v) {
                String login = (String) ((EditText) findViewById (R.id.userName)).getText().toString();
                String password = (String) ((EditText) findViewById (R.id.password)).getText().toString();

                if (login != "" && password != "")
                {
                    HashMap<String, String> postElements = new HashMap<String, String>();
                    postElements.put("login", login);
                    try {
                        postElements.put("password", URLEncoder.encode(password, "utf-8"));
                    } catch (UnsupportedEncodingException e) {
                        e.printStackTrace();
                    }
                    Button button = (Button) findViewById(R.id.loginButton);
                    button.setText("Login in ...");

                    String queryLogin = "https://apiUrl.com/login?";

                    String urlString = "";
                    try {
                        urlString = "login=";
                        urlString += URLEncoder.encode(login, "UTF-8");
                        urlString += "&password=";
                        urlString += URLEncoder.encode(password, "UTF-8");
                    } catch (UnsupportedEncodingException e) {

                        // if this fails for some reason, let the user know why
                        e.printStackTrace();
                        Toast.makeText(getApplicationContext(), "Error: " + e.getMessage(), Toast.LENGTH_LONG).show();
                    }

                    apiQuery.loginQuery(queryLogin, urlString);
}

apiQuery is of type APIQuery :

public void loginQuery(String url, String urlString) {

  // Prepare your search string to be put in a URL
  // It might have reserved characters or something
  // Create a client to perform networking
  AsyncHttpClient client = new AsyncHttpClient();

  // Have the client get a JSONArray of data
  // and define how to respond
  client.get(url + urlString,
        new JsonHttpResponseHandler() {
              @Override
              public void onSuccess(JSONObject jsonObject) {
                    String token = "";

                    if (jsonObject.has("token")) {
                    /*Toast.makeText(_mainContext, "Login Success!", Toast.LENGTH_LONG).show();*/
                    token = jsonObject.optString("token");
                    // 8. For now, just log results
                    Log.d("APIQuery Success", jsonObject.toString());
                     }
              }

              @Override
              public void onFailure(int statusCode, Throwable throwable, JSONObject error) {
                     // Display a "Toast" message
                     // to announce the failure
                     Toast.makeText(_mainContext, "Error: " + statusCode + " " + throwable.getMessage(), Toast.LENGTH_LONG).show();

                      // Log error message
                      // to help solve any problems
                      Log.e("APIQuery Failure", statusCode + " " + throwable.getMessage());
              }
        });
  }

My implementation works fine, I have a ToastMessage which appears on the screen with "Login Success" (or "Login Error" when it fails of course)

But I don't know how to handle that success in order to pass to the other activity I created .

I would like to do something like this :

if (apiQuery.loginQuery(...)) 
   show(activityLogged); // Where activityLogged is another activity

UPDATE

I added these lines :

if (jsonObject.has("token")) 
{
    /*Toast.makeText(_mainContext, "Login Success!",     Toast.LENGTH_LONG).show();*/
    token = jsonObject.optString("token");
    // 8. For now, just log results
    Log.d("APIQuery Success", jsonObject.toString());
    Intent i = new Intent(_mainContext, MainActivityLogged.class);
    _mainContext.startActivity(i);
}

And my manifest file looks like :

            <category android:name="android.intent.category.DEFAULT" />
            <category android:name="android.intent.category.BROWSABLE" />
            <!-- ATTENTION: This data URL was auto-generated. We recommend that you use the HTTP scheme.
              TODO: Change the host or pathPrefix as necessary. -->
            <data
                android:host="epidroid.charvoz.example.com"
                android:pathPrefix="/mainactivitylogged"
                android:scheme="http" />
        </intent-filters>

Answer 1

You could simply write an Intent to move to the next activity inside your onSuccess callback

 @Override
              public void onSuccess(JSONObject jsonObject) {
                    String token = "";

                    if (jsonObject.has("token")) {
                    /*Toast.makeText(_mainContext, "Login Success!", Toast.LENGTH_LONG).show();*/
                    token = jsonObject.optString("token");
                    Intent i = new Intent(context,LoggedActivity.class);
                    context.startActivity(i);
                     }
              }

In the above code

  Intent i = new Intent(context,LoggedActivity.class);
                    startActivity(i);

This is used to navigate to next page.Also Make sure you declare the activity inside the manifest file.

Answer 2

You could also pass some data to the next activity through the intent like so:

@Override
public void onSuccess(JSONObject jsonObject) {
    String token = "";
    if (jsonObject.has("token")) {
        /*Toast.makeText(_mainContext, "Login Success!", Toast.LENGTH_LONG).show();*/
        token = jsonObject.optString("token");
        Intent i = new Intent(context,LoggedActivity.class);
        i.putExtra("token", token);
        startActivity(i);
    }
}

And then retrieve the token from the next activity (inside the onCreate()) like so:

String token = getIntent().getStringExtra("token");