触摸时拖动视图并返回触摸释放时的原始位置

Drag a view on touch and back to origin position on touch release

我想在整个屏幕上触摸拖动一个视图,然后在触摸释放时我想将视图带回原来的位置。

这是我的代码:

private float xCoOrdinate, yCoOrdinate;
private float top2AxOriginXCoordinate, top2AxOriginYCoordinate;

private void initControls(View view) {
    top2AX = view.findViewById(R.id.top2AX);
    top2AxOriginXCoordinate = top2AX.getX();
    top2AxOriginYCoordinate = top2AX.getY();
    addDragListener();
}

private void addDragListener() {
     myView.setOnTouchListener(new View.OnTouchListener() {

            @Override
            public boolean onTouch(View view, MotionEvent event) {
                switch (event.getActionMasked()) {
                    case MotionEvent.ACTION_DOWN:
                        xCoOrdinate = view.getX() - event.getRawX();
                        yCoOrdinate = view.getY() - event.getRawY();
                        break;
                    case MotionEvent.ACTION_MOVE:
                        view.animate().x(event.getRawX() + xCoOrdinate).y(event.getRawY() + yCoOrdinate).setDuration(0).start();
                        break;
                    case MotionEvent.ACTION_UP:
                        view.animate().x(top2AxOriginXCoordinate).y(top2AxOriginYCoordinate).setDuration(500).start();
                        break;
                    default:
                        return false;
                }
                return true;
            }
        });
}

触摸和拖动效果很好,但是当我松开触摸时它不会回到原来的位置。它位于屏幕的左上角,并留有少量的左边距。我在这里做错了什么?

 top2AxOriginXCoordinate = top2AX.getX();
top2AxOriginYCoordinate = top2AX.getY();

可能这两个变量是0.0,如果是的话告诉我我会帮你的