按下按钮时javafx循环

我有麻烦让这个循环时，按钮被按下，并停止循环时，按钮被释放

btnUp.pressedProperty().addListener((observable, wasPressed, pressed) -> {
    System.out.println("changed");
    if (pressed) {
        System.out.println("pressed");
        while(btnUp.isArmed()){ 
            try        
            {
                Thread.sleep(1000);
            } 
            catch(InterruptedException ex) 
            {
                Thread.currentThread().interrupt();
            }
            //moveflag = false;
            System.out.println("pressed");
        }
    } else {
        System.out.println("released");
    }
});

不要在UI操作中使用Thread.sleep（），它会阻止所有UI绘图

在changelistener中不要使用，只需要有一个单独的循环来检查按钮的状态

对于这样的循环，时间线是最好的

public void start(Stage stage) {
 Button btn = new Button("Press me");

 Timeline timeline = new Timeline(new KeyFrame(Duration.seconds(1), (ActionEvent event) -> {
    // this code will be called every second
    System.out.println(btn.isPressed() ? "pressed" : "released");
 }));
 timeline.setCycleCount(Timeline.INDEFINITE);
 timeline.play();


 StackPane root = new StackPane();
 root.getChildren().add(btn);

 Scene scene = new Scene(new StackPane(btn), 300, 250);
 stage.setTitle("Hello World!");
 stage.setScene(scene);
 stage.show();

}

不要在UI操作中使用Thread.sleep（），它会阻止所有UI绘图

在changelistener中不要使用，只需要有一个单独的循环来检查按钮的状态

对于这样的循环，时间线是最好的

public void start(Stage stage) {
 Button btn = new Button("Press me");

 Timeline timeline = new Timeline(new KeyFrame(Duration.seconds(1), (ActionEvent event) -> {
    // this code will be called every second
    System.out.println(btn.isPressed() ? "pressed" : "released");
 }));
 timeline.setCycleCount(Timeline.INDEFINITE);
 timeline.play();


 StackPane root = new StackPane();
 root.getChildren().add(btn);

 Scene scene = new Scene(new StackPane(btn), 300, 250);
 stage.setTitle("Hello World!");
 stage.setScene(scene);
 stage.show();

}

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@Folieadeux：您可以通过单击左侧的来接受答案。@Folieadeux：您可以通过单击左侧的来接受答案。