使用FXML在窗格内创建ContextMenu

Question

我有一个工作示例,用于在JavaFX FXML中的窗格上定义ContextMenu,但我不确定它是否是最佳的.目前,只有JavaFX标准控件(例如Button,TextField)定义用于指定弹出ContextMenu的属性.然而,我想在一个窗格中出现一个弹出菜单,在我的例子中是一个VBox.

我采用扩展VBox的方法来支持上下文菜单.这是一个'笨重'的解决方案,但有效.有更好的方法吗？我错过了一些基本概念吗？

这是我的解决方案......

FXML:

<?xml version="1.0" encoding="UTF-8"?>

<?import javafx.scene.control.*?>
<?import javafx.scene.layout.*?>
<?import custommenu.view.ContextMenuPane?>

<AnchorPane xmlns:fx="http://javafx.com/fxml" fx:controller="custommenu.controller.CustomMenuController">
    <children>
        <VBox fx:id="vbox" onContextMenuRequested="#showMenu"
            onMousePressed="#hideMenu" prefHeight="200" prefWidth="200">
        </VBox>
        <ContextMenuPane>
            <contextMenu>
                <ContextMenu fx:id="menu">
                    <items>
                        <MenuItem text="add" onAction="#add" />
                    </items>
                </ContextMenu>
            </contextMenu>
        </ContextMenuPane>
    </children>
</AnchorPane>

CustomMenuPane ...

package custommenu.view;

import javafx.scene.control.ContextMenu;
import javafx.scene.layout.Pane;

public class ContextMenuPane extends Pane {

    private ContextMenu contextMenu;

    public void setContextMenu(ContextMenu contextMenu) {
        this.contextMenu = contextMenu;
    }

    public ContextMenu getContextMenu() {
        return contextMenu;
    }

}

控制器...

package custommenu.controller;

import javafx.fxml.FXML;
import javafx.scene.control.ContextMenu;
import javafx.scene.input.ContextMenuEvent;
import javafx.scene.layout.VBox;

public class CustomMenuController {

    @FXML private VBox vbox;
    @FXML private ContextMenu menu;

    @FXML public void add() {
        System.out.println("add");
    }

    @FXML
    public void showMenu(ContextMenuEvent event) {
        System.out.println("showMenu");

        menu.show(vbox, event.getScreenX(), event.getScreenY());
        event.consume();
    }

    @FXML public void hideMenu() {
        menu.hide();
    }
}

主要应用......

package custommenu;
import javafx.application.Application;
import javafx.fxml.FXMLLoader;
import javafx.scene.Scene;
import javafx.scene.layout.Pane;
import javafx.stage.Stage;

public class CustomMenuApplication extends Application {

    @Override
    public void start(Stage primaryStage) throws Exception {
        Pane myPane = (Pane)FXMLLoader.load(getClass().getResource("/custommenu/custom_menu_main.fxml"));
        Scene scene = new Scene(myPane);
        primaryStage.setScene(scene);
        primaryStage.show();
    }

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

Answer 1

尝试在弹出窗口中执行此操作？捕获 MouseButton.SECONDARY 的 MouseEvent 检查并在弹出窗口中显示 VBox。

private void showContextMenu() // showContextMenu(MouseEvent event)
{
    Popup popup = new Popup();

    VBox content = new VBox();
    Button b = new Button("Click Me!");
    b.setOnAction(new EventHandler<ActionEvent>() { });
    content.getChildren().addAll(b);

    popup.getContent().add(content);
    popup.setX(0); // or get mouse event x and y
    popup.setY(0); // event.getY()
    popup.setAutoHide(true);
    popup.show(your popup owner);
}