Question

我正在使用JavaFX中的一个非常简单的游戏。我有两个代表玩家的对象：

播放器 - 播放器类的实例
playerObj - Rectangle类的实例，用于在地图上表示玩家

现在我实现了玩家移动的基本功能，如goUp，goLeft等。这些都改变了玩家类中的属性X和Y（这些是DoubleProperty类型）。现在我想这样做，所以玩家的X和Y属性的每一个变化都会反映在playerObj中，所以我使用了这样的属性绑定：

 playerObj.yProperty().bind(player.yProperty());

但是当我调用goUp（）方法时，player对象中的“y属性”会改变，但是playerObj对象中的“y属性”不会改变。所以：

    pane.setOnKeyPressed(e -> {
        switch(e.getCode()) {
            case UP:
                player.goUp();
                System.out.println("player Y property: "+player.yProperty().getValue());
                System.out.println("playerObj Y property: "+playerObj.yProperty().getValue());
                break;
        }
    });

会导致这种情况（在调用goUp（）方法3次后）：

player Y property: 245.0
playerObj Y property: 250.0
player Y property: 240.0
playerObj Y property: 250.0
player Y property: 235.0
playerObj Y property: 250.0

当我将玩家Y属性绑定时，为什么playerObj Y属性也没有改变？

编辑：完整代码（删除不相关的部分）

Player.java

public class Player {

    private DoubleProperty x;
    private DoubleProperty y;
    private Scene scene;

    public Player(Scene scene) {
        this(DEFAULT_NAME, DEFAULT_COLOR, scene);
    }

    public Player(String name, Color color, Scene scene) {
        this.name = new SimpleStringProperty(name);
        this.color = color;
        this.scene = scene;
    }

    public Rectangle drawPlayer() {
        Rectangle player = new Rectangle(getX(),getY(),SIZE_OF_PLAYER,SIZE_OF_PLAYER);
        player.setFill(color);
        return player;
    }

    public Player goUp() {
        if(getY() != 0 && getY() != scene.getHeight()) {
           setY(yProperty().subtract(5).getValue());
       }
    }


    public DoubleProperty xProperty() {
        return x;
    }

    public DoubleProperty yProperty() {
        return y;
    }

    public double getX() {
        return x.getValue();
    }

    public void setX(double x) {
        this.x = new SimpleDoubleProperty(x);
    }

    public double getY() {
        return y.getValue();
    }

    public void setY(double y) {
        this.y = new SimpleDoubleProperty(y);
    }
}

Game.java

public class Game extends Application {

    private Scene scene;
    private Player player;
    private Rectangle playerObj;
    private Pane pane;

    @Override
    public void start(Stage primaryStage) {
        pane = new Pane();
        scene = new Scene(pane,500,500);
        player = new Player(scene);

        player.setX(scene.getWidth() / 2);
        player.setY(scene.getHeight() / 2);
        playerObj = player.drawPlayer();
        pane.getChildren().add(playerObj);
        playerObj.yProperty().bind(player.yProperty());

        pane.setOnKeyPressed(e -> {
            switch(e.getCode()) {
                case LEFT:
                    player.goLeft();
                    break;
                case UP:
                    player.goUp();
                    System.out.println("player Y property: "+player.yProperty().getValue());
                    System.out.println("playerObj Y property: "+playerObj.yProperty().getValue());
                    break;
                case RIGHT:
                    player.goRight();
                    break;
                case DOWN:
                    player.goDown();
            }
        });

        pane.requestFocus();

        primaryStage.setScene(scene);
        primaryStage.setTitle("Game");
        primaryStage.setResizable(false);
        primaryStage.show();

    }

    public static void main(String[] args) {
        Application.launch(args);
    }
}

Answer 1

x类中的y和Player属性未正确实施。

当您致电player.setY(...)时，您需要创建新 DoubleProperty：

public void setY(double y) {
    this.y = new SimpleDoubleProperty(y);
}

当然，这不是您的节点之前绑定的属性。因此，它的值不会反映在节点中。

你需要

public void setY(double y) {
    this.y.set(y);
}

，同样适用于setX(...)。

绑定属性不起作用

1 个答案: