InputListener.touchDragged

Question

我有一个演员，我想用触摸拖动。

class Tile extends Actor {
    Tile (char c) {
        addListener(new InputListener() {
            private float prevX, prevY;

            @Override
            public void touchDragged (InputEvent event, float x, float y, int pointer) {
                Gdx.app.log(TAG, "touchDrag: (" + x + "," + y);
                Tile cur = (Tile)event.getTarget();
                cur.setPosition(  //this call seems to cause the problem
                        cur.getX() + (x - prevX),
                        cur.getY() + (y - prevY) );
                prevX = x; prevY = y;
            }
        });
    }

    @Override
    public void draw(Batch batch, float alpha) {
        batch.draw(texture, getX(), getY());
    }

}

在被拖动时，瓷砖会颤抖，移动速度大约是触摸速度的一半。这由记录线确认，该记录线输出如下的坐标：

I/Tile: touchDrag: (101.99991,421.99994)
I/Tile: touchDrag: (112.99985,429.99994)
I/Tile: touchDrag: (101.99991,426.99994)
I/Tile: touchDrag: (112.99985,433.99994)
I/Tile: touchDrag: (101.99991,429.99994)
I/Tile: touchDrag: (112.99985,436.99994)

如果删除注释行（即不重置演员的位置），拖动输出看起来更合理：

I/Tile: touchDrag: (72.99997,78.99994)
I/Tile: touchDrag: (65.99997,70.99994)
I/Tile: touchDrag: (61.99997,64.99994)
I/Tile: touchDrag: (55.99997,58.99994)
I/Tile: touchDrag: (51.99997,52.99994)
I/Tile: touchDrag: (42.99997,45.99994)

有什么想法吗？谢谢你的期待！

Answer 1

InputListener方法中的坐标是根据Actor的位置给出的，因此如果您同时移动Actor，它们将无法与之前的值相比。

相反，存储原始位置并相对于此移动。数学计算可以适应你的动作：

    addListener(new InputListener() {
        private float startX, startY;

        @Override
        public boolean touchDown (InputEvent event, float x, float y, int pointer, int button) {
            startX = x;
            startY = y;
            return true;
        }

        @Override
        public void touchDragged (InputEvent event, float x, float y, int pointer) {
            Tile cur = (Tile)event.getTarget();
            cur.setPosition(
                    cur.getX() + (x - startX),
                    cur.getY() + (y - startY) );
        }
    });