onActivityResult Intent在MainActivity上爲null

Question

我的主要活動啓動登錄活動。如果用戶成功登錄，則ApiResponseHandler對象調用activity.finish()。似乎所有操作都已正確完成。我看不到任何可能導致它成爲null的意圖。

public class LoginActivity extends Activity { 
    public static final Logger logger = Logger.getLogger(LoginActivity.class.getName()); 

    Button  loginButton; 
    EditText loginField; 
    EditText passwordField; 

    /** 
    * Called when the activity is first created. 
    */ 
    @Override 
    public void onCreate(Bundle savedInstanceState) { 
     super.onCreate(savedInstanceState); 
     setContentView(R.layout.login); 

     loginButton = (Button)findViewById(R.id.loginFormLabelButton_Login); 
     loginField = (EditText)findViewById(R.id.loginFormLogin); 
     passwordField = (EditText)findViewById(R.id.loginFormPassword); 
    } 

    /** 
    * Login a user when the button is clicked 
    * @param v 
    */ 
    public void logUserIn(View v) { 
     loginButton.setText(R.string.loginFormLabelButton_Login_Working); 

     ApiRequest request = new ApiRequest(); 

     request.setLogin(loginField.getText().toString()); 
     request.setPassword(passwordField.getText().toString()); 

     if (request.getLogin().length() == 0) { 
      showErrorDialog(getString(R.string.loginErrorDialog_LoginRequired)); 
      return; 
     } 
     if (request.getPassword().length() == 0) { 
      showErrorDialog(getString(R.string.loginErrorDialog_PasswordRequired)); 
      return; 
     } 

     //make request and handle results 
     ApiRequestHandler<User> apiHandler = new ApiRequestHandler<User>(User.class); 
     apiHandler.setUrl(getString(R.string.loginFormApiUrl)); 
     apiHandler.setApiRequest(request); 
     apiHandler.setResponseHandler(new ApiResponseHandler(this, getIntent())); 
     apiHandler.execute(); 

     loginButton.setText(R.string.loginFormLabelButton_Login); 
    } 

    .... 
} 

的new ApiResponseHandler(this, getIntent())看起來像這樣...

public class ApiResponseHandler implements com.Bible_Bowl_Management.Api.ApiResponseHandler<User> { 
    private Activity activity; 
    private Intent intent; 

    public ApiResponseHandler(Activity activity, Intent intent) { 
     this.activity = activity; 
     this.intent = intent; 
    } 

    @Override 
    public void ResponseSuccessful(User user) { 
     intent.putExtra("User", user); 
     activity.setResult(Activity.RESULT_OK); 
     activity.finish(); 
    } 

    @Override 
    public void ResponseNoContent() { 
     Toast.makeText(this.activity.getApplicationContext(), "No account found with these credentials", Toast.LENGTH_LONG).show(); 
    } 
} 

Answer 1

您：

錯誤的點由下面留言MainActivity

public class MainActivity extends Activity { 
     static final int LOGIN_INTENT_ID = 0; 

     @Override 
     public void onCreate(Bundle savedInstanceState) { 
      super.onCreate(savedInstanceState); 

      //launch login activity 
      startActivityForResult(new Intent(this, LoginActivity.class), LOGIN_INTENT_ID); 
     } 

     protected void onActivityResult(int requestCode, int resultCode, Intent intent) { 
      super.onActivityResult(requestCode, resultCode, intent); 

      //handle activity response 
      if (requestCode == LOGIN_INTENT_ID) { 
       if (resultCode == Activity.RESULT_OK) { 
//intent is null, so .getSerializableExtra() fails 
        User user = (User) intent.getSerializableExtra("User"); 
        Toast.makeText(getApplicationContext(), "Logged in as: " + user.getFirstName() + " " + user.getLastName(), Toast.LENGTH_SHORT).show(); 
       } 
      } 
     } 
    } 

我登錄活動中指出傳遞getIntent（）作爲意圖返回到活動

您應該爲此創建一個新的Intent對象。

例如：

Intent returnIntent = new Intent(); 
    returnIntent.putExtra("SelectedBook",book); 
    setResult(RESULT_OK,returnIntent);