Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
110 views
in Technique[技术] by (71.8m points)

java - Get real position of a node in JavaFX

What is the best way to get the absolute position of a node in JavaFX?

Imagine we have a node in a Pane (Hbox, Stackpane, or any other pane) and that may have a parent itself.

I want to get the absolute position of that node and use it in another pane.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

It depends a little what you mean by "absolute". There is a coordinate system for the node, a coordinate system for its parent, one for its parent, and so on, and eventually a coordinate system for the Scene and one for the screen (which is potentially a collection of physical display devices).

You probably either want the coordinates relative to the Scene, in which case you could do

Bounds boundsInScene = node.localToScene(node.getBoundsInLocal());

or the coordinates relative to the screen:

Bounds boundsInScreen = node.localToScreen(node.getBoundsInLocal());

In either case the resulting Bounds object has getMinX(), getMinY(), getMaxX(), getMaxY(), getWidth() and getHeight() methods.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...