在 JavaFX 中获取节点的真实位置 [英] Get real position of a node in JavaFX
问题描述
在 JavaFX 中获取节点绝对位置的最佳方法是什么?
What is the best way to get the absolute position of a node in JavaFX?
想象一下,我们在窗格(Hbox、Stackpane 或任何其他窗格)中有一个节点,并且它本身可能有一个父节点.
Imagine we have a node in a Pane (Hbox, Stackpane, or any other pane) and that may have a parent itself.
我想获取该节点的绝对位置并在另一个窗格中使用它.
I want to get the absolute position of that node and use it in another pane.
推荐答案
这取决于您所说的绝对"是什么意思.节点有一个坐标系,其父级有一个坐标系,父级有一个坐标系,依此类推,最终有一个用于 Scene
的坐标系和一个用于屏幕的坐标系(这可能是一个物理显示设备的集合).
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).
你可能想要相对于 Scene
的坐标,在这种情况下你可以做
You probably either want the coordinates relative to the Scene
, in which case you could do
Bounds boundsInScene = node.localToScene(node.getBoundsInLocal());
或相对于屏幕的坐标:
Bounds boundsInScreen = node.localToScreen(node.getBoundsInLocal());
无论哪种情况,结果都是 Bounds
对象有 getMinX()
, getMinY()
, getMaxX()
, getMaxY()
code>、getWidth()
和 getHeight()
方法.
In either case the resulting Bounds
object has getMinX()
, getMinY()
, getMaxX()
, getMaxY()
, getWidth()
and getHeight()
methods.
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