Question

如果用户在一段时间内处于非活动状态，我想关闭我的JavaFX应用程序。我在Swing中有这个代码，我想在JavaFX中做同样的事情。如果在指定的时间内没有事件发生，则此类将用户重定向到登录面板。

import javax.swing.Timer;

public class AutoClose {

    private Timer timer;

    public AutoClose(JFrame centralpanel) {
        EventQueue.invokeLater(new Runnable() {
            @Override
            public void run() {

                Toolkit.getDefaultToolkit().addAWTEventListener(new AWTEventListener() {

                    @Override
                    public void eventDispatched(AWTEvent event) {
                        Object source = event.getSource();
                        if (source instanceof Component) {
                            Component comp = (Component) source;
                            Window win = null;
                            if (comp instanceof Window) {
                                win = (Window) comp;
                            } else {
                                win = SwingUtilities.windowForComponent(comp);
                            }
                            if (win == centralpanel) {
                                timer.restart();
                            }
                        }
                    }
                }, AWTEvent.KEY_EVENT_MASK | AWTEvent.MOUSE_EVENT_MASK | AWTEvent.MOUSE_MOTION_EVENT_MASK | AWTEvent.MOUSE_WHEEL_EVENT_MASK);

                timer = new Timer(3600000, new ActionListener() {
                    @Override
                    public void actionPerformed(ActionEvent e) {

                        centralpanel.dispose();

                        //Redirect to the login panel.
                        Login login = new Login();
                        login.setVisible(true);
                        timer.stop();
                        JOptionPane.showMessageDialog(null,"Connection closed due to inactivity");
                    }
                });
                timer.start();
            }
        });
    }
}

Answer 1

创建PauseTransition以触发延迟退出，并为重启过渡的所有场景添加InputEvent的事件过滤器：

@Override
public void start(Stage primaryStage) throws IOException {
    Button button = new Button("abcde");
    StackPane root = new StackPane(button);
    Scene scene = new Scene(root, 400, 400);

    // create transition for logout
    Duration delay = Duration.seconds(10);
    PauseTransition transition = new PauseTransition(delay);
    transition.setOnFinished(evt -> logout());

    // restart transition on user interaction
    scene.addEventFilter(InputEvent.ANY, evt -> transition.playFromStart());

    primaryStage.setScene(scene);
    primaryStage.show();

    transition.play();
}

private void logout() {
    // TODO: replace with logout code
    Platform.exit();
}

注意：使用事件过滤器非常重要，因为事件可以在到达事件处理程序 之前消耗。

Answer 2

我已完成此代码，现在它可以正常使用。

public class AutoClose {
    private Timeline timer;

    public AutoClose(VBox mainPanel) {

        timer = new Timeline(new KeyFrame(Duration.seconds(3600), new EventHandler<ActionEvent>() {

            @Override
            public void handle(ActionEvent evt) {
                Alert alert = new Alert(Alert.AlertType.INFORMATION);
                alert.setTitle("Inactivity");
                alert.setHeaderText("Connection closed due to inactivity");
                alert.show();

                try {
                    Stage mainWindowStage = Login.getPrimaryStage();

                    Parent root = FXMLLoader.load(getClass().getResource("/view/Login.fxml"));

                    Scene scene = new Scene(root);
                    mainWindowStage.setScene(scene);
                    mainWindowStage.show();

                    timer.stop();
                } catch (IOException ex) { 
                }
            }
        }));
        timer.setCycleCount(Timeline.INDEFINITE);
        timer.play();

        mainPanel.addEventFilter(MouseEvent.ANY, new EventHandler<Event>() {
            @Override
            public void handle(Event event) {
                timer.playFromStart();
            }
        });
    }
}

由于无效而自动关闭JavaFX应用程序

2 个答案: