android – 单眼触摸检测在Ontouch方法的视图

我需要在自定义视图的onuch方法中进行单边触摸检测.我尝试在ACTION-DOWN和ACTION-UP中获取x和y值,而在ACTION-UP中我给出的条件是,如果ACTIONDOWN和ACTION-UP中的X和Y的值相等,则将其作为单次点击.

我的代码如下

@Override
public boolean onTouchEvent(MotionEvent ev) {
   if (!mSupportsZoom && !mSupportsPan) return false;
 
    mScaleDetector.onTouchEvent(ev);
 
    final int action = ev.getAction();
    switch (action & MotionEvent.ACTION_MASK) {
    case MotionEvent.ACTION_DOWN: {
        final float x = ev.getX();
        final float y = ev.getY();
 
        mLastTouchX = x;  //here i get x and y values in action down
        mLastTouchY = y;
        mActivePointerId = ev.getPointerId(0);
 
        break;
    }
 
    case MotionEvent.ACTION_MOVE: {
        final int pointerIndex = ev.findPointerIndex(mActivePointerId);
        final float x = ev.getX(pointerIndex);
        final float y = ev.getY(pointerIndex);
 
        if (mSupportsPan && !mScaleDetector.isInProgress()) {
            final float dx = x - mLastTouchX;
            final float dy = y - mLastTouchY;
 
            mPosX += dx;
            mPosY += dy;
            //mFocusX = mPosX;
            //mFocusY = mPosY;
 
            invalidate();
        }
 
        mLastTouchX = x;
        mLastTouchY = y;
 
        break;
    }
 
    case MotionEvent.ACTION_UP: {
 
        final float x = ev.getX();
        final float y = ev.getY();
 
        touchupX=x;   //here is get x and y values at action up
        touchupY=y; 
 
        if(mLastTouchX == touchupX && mLastTouchY == touchupY){  //my condition if both the x and y values are same .
 
            PinchZoomPanActivity2.tapped1(this.getContext(),100); //my method if the singletap is detected
 
        }
        else{
 
        }
 
        mActivePointerId = INVALID_POINTER_ID;
 
        break;
    }
 
    case MotionEvent.ACTION_CANCEL: {
        mActivePointerId = INVALID_POINTER_ID;
        break;
    }
 
    case MotionEvent.ACTION_POINTER_UP: {
        final int pointerIndex = (ev.getAction() & MotionEvent.ACTION_POINTER_INDEX_MASK) 
                >> MotionEvent.ACTION_POINTER_INDEX_SHIFT;
        final int pointerId = ev.getPointerId(pointerIndex);
        if (pointerId == mActivePointerId) {
 
            final int newPointerIndex = pointerIndex == 0 ? 1 : 0;
            mLastTouchX = ev.getX(newPointerIndex);
            mLastTouchY = ev.getY(newPointerIndex);
            mActivePointerId = ev.getPointerId(newPointerIndex);
        }
        break;
    }
    }
 
    return true;
}

但我不能完成.我的意思是在每一个动作,我的方法被称为.即使动作和动作的x和y值都不相同.我想我也需要把一些范围放在单边,我们用手指触摸屏幕.有人可以给我一些建议吗

解决方法

最近我也遇到了同样的问题,最终不得不实施去抖动来使其工作.这不是理想的,但它是非常可靠的,直到我能找到更好的东西.

View.onClickListener对我来说更可靠,但不幸的是,我需要OnTouchListener的MotionEvent.

编辑：删除导致其在此失败的多余代码

class CustomView extends View {
 
    private static long mDeBounce = 0;
 
    static OnTouchListener listenerMotionEvent = new OnTouchListener() {
        @Override
        public boolean onTouch(View view,MotionEvent motionEvent) {
            if ( Math.abs(mDeBounce - motionEvent.getEventTime()) < 250) {
                //Ignore if it's been less then 250ms since
                //the item was last clicked
                return true;
            }
 
            int intCurrentY = Math.round(motionEvent.getY());
            int intCurrentX = Math.round(motionEvent.getX());
            int intStartY = motionEvent.getHistorySize() > 0 ? Math.round(motionEvent.getHistoricalY(0)) : intCurrentY;
            int intStartX = motionEvent.getHistorySize() > 0 ? Math.round(motionEvent.getHistoricalX(0)) : intCurrentX;
 
            if ( (motionEvent.getAction() == MotionEvent.ACTION_UP) && (Math.abs(intCurrentX - intStartX) < 3) && (Math.abs(intCurrentY - intStartY) < 3) ) {
                if ( mDeBounce > motionEvent.getDownTime() ) {
                    //Still got occasional duplicates without this
                    return true;
                }
 
                //Handle the click
 
                mDeBounce = motionEvent.getEventTime();
                return true;
            }
            return false;
        }
    };
}

android – 单眼触摸检测在Ontouch方法的视图

解决方法

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