Java 如何使节点在滚动时始终可见?

Java 如何使节点在滚动时始终可见?,java,user-interface,scroll,javafx-8,Java,User Interface,Scroll,Javafx 8,我有一个滚动窗格里面一个锚窗格。在这个锚窗格中,我有一个标签。我需要此标签在滚动时始终可见。我需要移动,以便用户始终看到它,就好像它没有改变一样。有人能帮忙吗 @FXML private ScrollPane s; public void initialize(URL location, ResourceBundle resources) { AnchorPane p = new AnchorPane()

我有一个滚动窗格里面一个锚窗格。在这个锚窗格中,我有一个标签。我需要此标签在滚动时始终可见。我需要移动,以便用户始终看到它,就好像它没有改变一样。有人能帮忙吗

@FXML
 private ScrollPane s;


public void initialize(URL location, ResourceBundle resources) {                           
                AnchorPane p = new AnchorPane();
                VBox v = new VBox();
                p.getChildren().add(v);
                s.setContent(p);}

实现这一点的最简单方法不是将
标签
放在内容中,而是放在同时包含
滚动窗格
标签的
堆栈窗格
中:

@Override
public void start(Stage primaryStage) {
    Region content = new Region();
    content.setPrefSize(2000, 2000);
    content.setBackground(new Background(new BackgroundFill(
            new LinearGradient(0, 0, 1, 1, true, CycleMethod.NO_CYCLE, new Stop(0, Color.RED), new Stop(1, Color.BLUE)),
            CornerRadii.EMPTY,
            Insets.EMPTY)));

    Label label = new Label("Hello World!");
    label.setTextFill(Color.WHITE);
    StackPane.setAlignment(label, Pos.TOP_LEFT);
    StackPane.setMargin(label, new Insets(10));

    ScrollPane scrollPane = new ScrollPane(content);

    StackPane root = new StackPane();
    root.getChildren().addAll(scrollPane, label);

    Scene scene = new Scene(root, 200, 200);

    primaryStage.setScene(scene);
    primaryStage.show();
}
或者,也可以使用“平移”属性调整位置:

@Override
public void start(Stage primaryStage) {
    Label label = new Label("Hello World!");
    label.setTextFill(Color.WHITE);

    Pane content = new Pane(label);
    content.setPrefSize(2000, 2000);
    content.setBackground(new Background(new BackgroundFill(
            new LinearGradient(0, 0, 1, 1, true, CycleMethod.NO_CYCLE, new Stop(0, Color.RED), new Stop(1, Color.BLUE)),
            CornerRadii.EMPTY,
            Insets.EMPTY)));

    ScrollPane scrollPane = new ScrollPane(content);

    double targetX = 10;
    double targetY = 10;

    InvalidationListener listener = o -> {
        Bounds viewportBounds = scrollPane.getViewportBounds();
        Bounds contentBounds = content.getBoundsInLocal();
        Bounds labelBounds = label.getBoundsInLocal();

        double factorX = Math.max(contentBounds.getWidth() - viewportBounds.getWidth(), 0);
        double factorY = Math.max(contentBounds.getHeight() - viewportBounds.getHeight(), 0);

        label.setTranslateX(targetX + scrollPane.getHvalue() * factorX - labelBounds.getMinX());
        label.setTranslateY(targetY + scrollPane.getVvalue() * factorY - labelBounds.getMinY());
    };

    scrollPane.viewportBoundsProperty().addListener(listener);
    scrollPane.hvalueProperty().addListener(listener);
    scrollPane.vvalueProperty().addListener(listener);
    label.boundsInLocalProperty().addListener(listener);

    Scene scene = new Scene(scrollPane, 200, 200);

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

    listener.invalidated(null);
}

你应该展示一些代码,这将有助于得到一些答案,节点(标签)的位置也必须是动态的@BoHalim我如何修复@BoHalim Nope的可能重复,一个是关于滚动以使某个节点可见,这一个是关于保持一个节点可见而不受滚动位置的影响。