如何在使用FXML设计的JavaFX中创建内部弹出窗口
我有一个FXML文件，我用它来允许用户在请求时输入。现在我只是将它放在一个新的阶段并执行 Stage.show（）。我希望它不会出现在一个新窗口中，表现得更像 ContextMenu 。
I have an FXML file that I'm using to allow user input when requested. Right now I just put it in a new stage and do Stage.show(). I would like to not have it appear in a new window and behave more like a ContextMenu.
看着 ContextMenu 类似乎我不能根据FXML文件设置内容。有没有办法用 ContextMenu 或 Popup 或其他一些我不知道的类？
Looking at ContextMenu class it doesn't appear that I can set the content based off an FXML file. Is there a way to do this either with ContextMenu or Popup or some other class I am unaware of?
虽然该库非常好，但我想要一些简单的东西，不需要第三方下载。我想出了这个：
Although that library is quite nice, I wanted something simple that didn't require 3rd party downloads. I came up with this:
Popup popup = new Popup();
CustomController controller = new CustomController();
FXMLLoader loader = new FXMLLoader(getClass().getResource(fxmlfile));
loader.setController(controller);
popup.getContent().add((Parent)loader.load());
问题是我没有意识到父$ c对于方法，$ c>可被视为 节点 Popup＃getContent #add
The problem was I didn't realize that a Parent could be considered a Node for the method Popup#getContent#add