单元测试 - 如何模拟监听器

Question

我正在为基于 mvp 模式的应用程序编写模拟测试。以下是登录活动的代码

public class LoginActivity extends AppCompatActivity implements LoginContract.ILoginView{

presenter = new LoginPresenter(this);

@Override
    public void startMoviesActivity() {
       startActivity(new Intent(this, MoviesActivity.class));
    }
}



 public class LoginContract {
     public interface ILoginView {
            ...
            void startMoviesActivity();
        }

     public interface ILoginService {

            interface OnLoginFinishedListener {
                void onLoginSuccess();
                void onLoginFailure();
            }

            void login(String username, String password, OnLoginFinishedListener 
                 listener);
        }
}


public class LoginPresenter implements LoginContract.ILoginService.OnLoginFinishedListener {

    private LoginContract.ILoginView view;
    private LoginContract.ILoginService service;

    public LoginPresenter(LoginContract.ILoginView view) {
        this.view = view;
        service = new LoginService();
    }


    public void validateCredentials(String email, String password) {
        if(email.isEmpty()){
            view.setEmailEmptyError(R.string.error_email_empty);
            return;
        }
        if(password.isEmpty()){
            view.setPasswordEmptyError(R.string.error_password_empty);
            return;
        }

        view.showProgressDialog();
        service.login(email,password,this);
    }

    @Override
    public void onLoginSuccess() {
        view.hideProgressDialog();
        view.startMoviesActivity();
    }

    @Override
    public void onLoginFailure() {
        view.hideProgressDialog();
        view.showUnsuccessfulLoginError(R.string.error_login_unsuccessful);
    }
}

public class LoginService implements LoginContract.ILoginService {

    @Override
    public void login(String username, String password, final 
         OnLoginFinishedListener listener) {
   // Contains REST API which calls listener.onLoginSuccess() on valid credentials and listener.onLoginFailure on invalid.
    }
}

由于我使用接口的方式，我在编写用于检查 LoginPresenter 类中的 onLoginSuccess() 和 onLoginFailure() 的测试时遇到问题。当调用service.login(email,password,this);时，它应该返回对LoginPresenter中的那些方法的调用，这反过来又调用LoginActivity中的startMoviesActivity()方法。我是mockito的初学者......如果你能帮忙的话。谢谢。

Answer 1

在下面找到一些使用 mockito 为您的代码编写的测试，这些案例是我在我的机器上测试过的。但是，对于测试 android，我更喜欢使用 Robolectric(http://robolectric.org/)，它基本上是为 Android UI 单元测试而设计的，并且从代码覆盖率的角度来看它是好的。与 mockito 相比，覆盖率更高。

---

import org.junit.Before;
import org.junit.Test;
import org.powermock.core.classloader.annotations.PrepareForTest;

import static org.mockito.Mockito.doCallRealMethod;
import static org.mockito.Mockito.mock;

@PrepareForTest(LoginActivity.class)
public class LoginActivityTest {

    private LoginActivity loginActivity;
    @Before
    public void setUp()
    {
        loginActivity = mock(LoginActivity.class);
        doCallRealMethod().when(loginActivity).startMoviesActivity();
    }
    @Test
    public void testStartMovies()
    {
        loginActivity.startMoviesActivity();
    }
}


import org.junit.Before;
import org.junit.Test;
import org.junit.runner.RunWith;
import org.powermock.core.classloader.annotations.PrepareForTest;
import org.powermock.modules.junit4.PowerMockRunner;

    import static org.mockito.Mockito.doCallRealMethod;
    import static org.mockito.Mockito.mock;
    import static org.mockito.Mockito.when;

    @PrepareForTest(LoginPresenter.class)
    public class LoginPresenterTest {

        private LoginPresenter loginPresenter;

        @Before
        public void setUp() {
            loginPresenter = mock(LoginPresenter.class);
            doCallRealMethod().when(loginPresenter).onLoginSuccess();
            doCallRealMethod().when(loginPresenter).onLoginFailure();
        }

        @Test
        public void testLoginSuccess()
        {
            loginPresenter.onLoginSuccess();
            loginPresenter.onLoginSuccess();
        }
        @Test
        public void testLoginFailure()
        {
            loginPresenter.onLoginFailure();
        }
    }