2

我需要获取画布内图像的边界以检测是否已被触摸，但如果我旋转画布 getbounds() 保持相同的值，如何在画布旋转后获得正确的值？

//CODE ON VIEW CLASS:

    canvas.save();  
    canvas.rotate(rotation, Xpos, Ypos);

    secHandImg.setBounds(left,top,right,bottom);    
    secHandImg.setAntiAlias(true);
    secHandImg.draw(canvas);

    circleHandImg.setBounds(left,top,right,bottom); 
    circleHandImg.setAntiAlias(true);
    circleHandImg.draw(canvas);

    canvas.restore();


//CODE ON FRAGMENT CLASS:

        public boolean onTouch(View view, MotionEvent event) {

                Rect imageBounds = MyClass.secHandImg.getBounds();          

                int action = event.getAction();

                final int x = (int)event.getX();
                final int y = (int)event.getY();


                    if(action == MotionEvent.ACTION_DOWN)   {

    if(x >= imageBounds.left && x < (imageBounds.left + imageBounds.width())
    && y >= imageBounds.top && y < (imageBounds.top + imageBounds.height())){

//THIS DON´T WORK IF CANVAS ROTATES 
                imageBoundsTouch = true;


                }

        }

这里有一张图片可以更好地解释我的问题：

http://s11.postimg.org/z7j0xgwmb/Imagen_2.png

4

2 回答 2

1

您可以使用三角函数来获取旋转矩形的边界：

function BBoxDimensions(width,height,radianAngle){
  var c = Math.abs(Math.cos(radianAngle));
  var s = Math.abs(Math.sin(radianAngle));
  return({  width: height * s + width * c,  height: height * c + width * s });
}

您可以使用 trig 获取旋转边界框的 XY：

// where cx/cy is the center of rotation of the target
// and startingX/startingY is the starting xy of the unrotated target

var x1 = startingX - cx;
var y1 = startingY - cy;
newX =cx+ x1*Math.cos(angle) - y1*Math.sin(angle);
newY =cy+ y1*Math.cos(angle) + x1*Math.sin(angle);
于 2013-04-24T21:17:37.393 回答
0

重新计算你的边界值 onOrientationChanged()

于 2013-04-24T21:03:39.793 回答