定位path.lineTo（）

Question

我目前正在Android Studio上学习Canvas，但不确定使用正确的标题，但是我在使用Seekbar拖动红点时如何控制三角形的位置上一无所知。下面是详细信息

下面是我成功绘制三角形的代码。

        @Override
        protected void onDraw(Canvas canvas) {
        super.onDraw(canvas);

        paint.setColor(Color.RED);
        canvas.drawRoundRect(100, 300, 900, 700, 50, 50,paint);

        Point pointTop = new Point(250, 400);
        Point pointLeft = new Point(150, 150);
        Point pointRight = new Point(350, 150);


        paint.setColor(Color.YELLOW);
        path.setFillType(Path.FillType.EVEN_ODD);
        path.lineTo(pointTop.x, pointTop.y);
        path.lineTo(pointLeft.x, pointLeft.y);
        path.lineTo(pointRight.x, pointRight.y);
        path.lineTo(pointTop.x, pointTop.y);
        path.close();

        canvas.drawPath(path, paint);
        }

图片结果：

预期结果

注意：只有1个红点和1个三角形。

非常感谢=）

Answer 1

先看Node类，然后再看setTranslateX。

Answer 2

在阅读了以下链接的文档并添加代码后，我已经解决了问题

https://developer.android.com/reference/android/graphics/Path

享受：）

        @Override
        protected void onDraw(Canvas canvas) {
        super.onDraw(canvas);

        paint.setColor(Color.RED);
        canvas.drawRoundRect(100, 300, 900, 700, 50, 50,paint);

        Point pointTop = new Point(250, 400);
        Point pointLeft = new Point(150, 150);
        Point pointRight = new Point(350, 150);


        paint.setColor(Color.YELLOW);
--------------------- added -------------------------------
        path.moveTo(pointLeft.x, pointLeft.y);
        path.moveTo(pointRight.x, pointRight.y);
        path.moveTo(pointTop.x, pointTop.y);
        path.reset();
--------------------- added -------------------------------
        path.setFillType(Path.FillType.EVEN_ODD);
        path.lineTo(pointTop.x, pointTop.y);
        path.lineTo(pointLeft.x, pointLeft.y);
        path.lineTo(pointRight.x, pointRight.y);
        path.lineTo(pointTop.x, pointTop.y);
        path.close();

        canvas.drawPath(path, paint);
        }