得到一个onTouch ACTION_MOVE事件的Andr​​oid速度 [英] Get speed of a onTouch ACTION_MOVE event in Android

问题描述

这应该是pretty的简单的事。用户将画面上他的手指和拖动它在屏幕上。有两个事件射击onTouch：

• MotionEvent.ACTION_DOWN
• MotionEvent.ACTION_MOVE

现在，我该怎么计算ACTION_MOVE手势的速度？用户拖动手指的手势时慢或更快，所以我想我需要计算速度之间的两个中间感动点：本lastTouchedPointX，lastTouchedPointY和event.getX（），event.getY（）

有没有人这样做过？

解决方案

  @覆盖
        公共布尔的onTouchEvent（MotionEvent事件，图形页面图形页面）{


            如果（event.getAction（）== MotionEvent.ACTION_DOWN）{

                oldX = event.getX（）;
                oldY = event.getY（）;
                    //启动定时器

            }否则，如果（event.getAction（）== MotionEvent.ACTION_UP）{

                //长timerTime =的getTime两个事件之间向下向上
                下一页末= event.getX（）;
                newY = event.getY（）;

                浮距离=的Math.sqrt（（下一页末-oldX）*（下一页末-oldX）+（newY-oldY）*（newY-oldY））;
                浮动速度=距离/ timerTime;

            }
}
 

This should be a pretty simple thing to do. The user puts his finger on the screen and drags it around the screen. There are two events firing on onTouch:

• MotionEvent.ACTION_DOWN
• MotionEvent.ACTION_MOVE

Now, how can I calculate the speed of the ACTION_MOVE gesture ? The user drags the finger slower or faster during a gesture, so I think I need to calculate the speed between two intermediate touched points: the lastTouchedPointX,lastTouchedPointY and the event.getX(),event.getY().

Has anyone done this before ?

解决方案

@Override
        public boolean onTouchEvent(MotionEvent event, MapView mapView) {


            if(event.getAction() == MotionEvent.ACTION_DOWN) {

                oldX = event.getX();
                oldY = event.getY();
                    //start timer

            } else if (event.getAction() == MotionEvent.ACTION_UP) {   

                //long timerTime = getTime between two event down to Up
                newX = event.getX();
                newY = event.getY();

                float distance = Math.sqrt((newX-oldX) * (newX-oldX) + (newY-oldY) * (newY-oldY));
                float speed = distance / timerTime;

            }
}

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