java - JavaFX8 中使用 FXML 的 Lambda 函数

Question

我正在使用 Scene Builder 在 JavaFX 中开发入侵者游戏(我必须用大炮射击飞机)，但我在 lambda 函数方面遇到了问题。

我确信代码是正确的，因为如果我通过场景生成器的选项(操作时)关联代码，它就可以工作，但是，当我尝试使用 lambda 函数时，我无法更改我的场景。我哪里错了？

public class SchermataGiocoController {

private Parent Menu, Avvio;
private TranslateTransition tt;
private Cannone cannone; //cannone = cannon
private Aereo aereo; //aereo = plane
private Proiettile proiettile; //proiettile = bullet
private RotateTransition rt;

@FXML
private Button down;

@FXML
private Circle circle;

@FXML
private Button su;

@FXML
private Button exit;

@FXML
private ImageView cannone_im;

@FXML
private AnchorPane SchermataGioco;

@FXML
private ImageView aereo_im;

@FXML
private Button menu;

@FXML
private Button home;

@FXML
void initialize() {
    assert exit != null : "fx:id=\"exit\" was not injected: check your FXML file 'SchermataGioco.fxml'.";
    assert cannone_im != null : "fx:id=\"cannone_im\" was not injected: check your FXML file 'SchermataGioco.fxml'.";
    assert su != null : "fx:id=\"su\" was not injected: check your FXML file 'SchermataGioco.fxml'.";
    assert SchermataGioco != null : "fx:id=\"SchermataGioco\" was not injected: check your FXML file 'SchermataGioco.fxml'.";
    assert aereo_im != null : "fx:id=\"aereo_im\" was not injected: check your FXML file 'SchermataGioco.fxml'.";
    assert menu != null : "fx:id=\"menu\" was not injected: check your FXML file 'SchermataGioco.fxml'.";
    assert down != null : "fx:id=\"down\" was not injected: check your FXML file 'SchermataGioco.fxml'.";
    assert home != null : "fx:id=\"home\" was not injected: check your FXML file 'SchermataGioco.fxml'.";
    cannone = new Cannone(141, 104, 14, 362);
    proiettile = new Proiettile(16, 16, 14, 362); 
    TranslateTransition();
}


 @FXML
 public void vaiMenu() {
    menu.setOnAction((ActionEvent event) -> {                      
        try {                
            FXMLLoader loader = new FXMLLoader();                
            loader.setLocation(Game.class.getResource("/game/view/Menu.fxml"));
            Menu = (Parent) loader.load();
            menu.getScene().getWindow().hide();
        } catch (IOException ioe) {
          ioe.getMessage();
        }            
        Stage stage = new Stage();
        stage.setScene(new Scene(Menu));
        stage.show();
    });
}

Answer 1

如果将 vaiMenu() 与 SceneBuilder 中的“On Action”相关联，那么(正如预期的那样)，例如当在 Button 上执行单击时，将执行 vaiMenu() 方法，该方法将执行以下操作:为以下操作事件分配一个监听器(通过 lambda) Button，因此它永远不会被执行，只是一次又一次添加。

如果您想通过 SceneBuilder 分配监听器，例如:

<Button fx:id="menu" onAction="#vaiMenu" />

您不能使用 lambda(匿名函数)，因为您需要在 FXML 文件(在 onAction 属性中)中引用命名函数。

如果您将当前的 lambda 插入例如 Controller 的 initialize() 方法，它将正常工作，没有任何问题，但在这种情况下，您不应在 FXML 文件中定义此属性。