如何在JavaFX中获取Node的绝对旋转 [英] How to get the absolute rotation of a Node in JavaFX

问题描述

在JavaFX中，Node的rotateProperty为我提供了相对于其父级的旋转度。有没有办法获得相对于场景或屏幕的度数绝对旋转？

In JavaFX the rotateProperty of a Node provides me with its rotation in degree relative to its parent. Is there a way to get the absolute rotation in degree, either relative to the Scene or the Screen?

例如，有没有办法从我可以从getLocalToSceneTransform（）获得的Node的Transform对象计算角度？

For example, is there a way to calculate the angle from the Transform object of a Node i can get from getLocalToSceneTransform()?

推荐答案

所以，我自己做了数学计算，对于我的情况，我要么通过以下方式获得Radians的轮换：

So, i did the math myself, and for my case i either get the rotation in Radians via:

double xx = myNode.getLocalToSceneTransform().getMxx();
double xy = myNode.getLocalToSceneTransform().getMxy();
double angle = Math.atan2(-xy, xx);

或

double yx = myNode.getLocalToSceneTransform().getMyx();
double yy = myNode.getLocalToSceneTransform().getMyy();
double angle = Math.atan2(yx, yy);

在这两种情况下都可以转换为360度：

In both cases this can be converted to 360-degrees:

angle = Math.toDegrees(angle);
angle = angle < 0 ? angle + 360 : angle;

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