Get real position of a node in JavaFX

java javafx javafx-2 javafx-8
maryam picture maryam · Jun 30, 2015 · Viewed 31.3k times · Source

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.

Answer

James_D picture James_D · Jul 1, 2015

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.

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