我无法将RectF转换为Geometry,然后在其中检测到一个点:

public static Geometry RectFtoGeometry(RectF r) {
    GeometricShapeFactory gsf = new GeometricShapeFactory();
    gsf.setBase(new Coordinate(r.left, r.bottom));
    gsf.setNumPoints(4);
    gsf.setWidth(r.width());
    gsf.setHeight(r.height());

    Geometry rect = gsf.createRectangle(),
        point = new GeometryFactory().createPoint(new Coordinate(r.centerX(), r.centerY()));

    if(!rect.contains(point))
        throw new IllegalArgumentException();//This gets thrown

    return gsf.createRectangle();
}


如何从“可以包含”其点的R​​ectF创建几何?

提前致谢!

最佳答案

GeometricShapeFactory.java中的代码使用以下方法创建信封:

public Envelope getEnvelope() {
  if (base != null) {
    return new Envelope(base.x, base.x + width, base.y, base.y + height);
  }
  ...
}


Android的原点(0,0)在左侧的TOP处;您必须将RectF的TOP与其高度相加才能得出RectF的底部。因此,基础必须是:

gsf.setBase(new Coordinate(r.left, r.top));

07-23 09:20