maryam maryam - 10 months ago 63
Java Question

get real position of a node in javaFX

What is the best way to get absolute position of a node in JavaFX ?
Suppose we have a node in a a pane(Hbox, Stackpane , ...or any other pane) and that may have parent itself.
I want to get abosolute position of that node and use it in another pane ?

Answer Source

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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