在Android时钟中移动分钟

时间:2011-09-28 04:45:34

标签: java android math

我正在尝试编写一个简单的概念验证应用程序,允许用户旋转时钟的分针。我很难为OnTouchEvent提供正确的逻辑。

到目前为止,我有以下代码:

    public boolean onTouchEvent(MotionEvent e) {

       float x = e.getX();
       float y = e.getY();

    switch (e.getAction()) {
    case MotionEvent.ACTION_MOVE:
        //find an approximate angle between them.

        float dx = x-cx;
        float dy = y-cy;
        double a=Math.atan2(dy,dx);

        this.degree = Math.toDegrees(a);
        this.invalidate();
    }
    return true;
    }


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

    boolean changed = mChanged;
    if (changed) {
        mChanged = false;
    }

    int availableWidth = getRight() - getLeft();
    int availableHeight = getBottom() - getTop();

    int x = availableWidth / 2;
    int y = availableHeight / 2;

    cx = x;
    cy = y;

    final Drawable dial = mDial;

    int w = dial.getIntrinsicWidth() + 100;
    int h = dial.getIntrinsicHeight() + 100;

    boolean scaled = false;

    if (availableWidth < w || availableHeight < h) {
        scaled = true;

        float scale = Math.min((float) availableWidth / (float) w, (float) availableHeight / (float) h);

        canvas.save();
        canvas.scale(scale, scale, x, y);
    }

    if (changed)
    {
        dial.setBounds(x - (w / 2), y - (h / 2), x + (w / 2), y + (h / 2));
    }

    dial.draw(canvas);

    canvas.save();
    float hour = mHour / 12.0f * 360.0f;

    canvas.rotate(hour, x, y);

    final Drawable hourHand = mHourHand;

    if (changed) {

        w = hourHand.getIntrinsicWidth() + 30;
        h = hourHand.getIntrinsicHeight() + 30;

        hourHand.setBounds(x - (w / 2), y - (h / 2), x + (w / 2), y  + (h / 2));
    }

    hourHand.draw(canvas);
    canvas.restore();

    canvas.save();
    float minute = mMinutes / 60.0f * 360.0f;

    if (bearing == 0)
    {
        canvas.rotate(minute, x, y);
    }
    else
    {
        canvas.rotate((float)bearing, x, y);
    }

    final Drawable minuteHand = mMinuteHand;

    if (changed) {

        w = minuteHand.getIntrinsicWidth() + 30;
        h = minuteHand.getIntrinsicHeight() + 30;

        minuteHand.setBounds(x - w, y - h, x + w, y  + h);
    }

    minuteHand.draw(canvas);
    canvas.restore();

    if (scaled) {
        canvas.restore();
    }
}

然后基于此,我的OnDraw方法将分针旋转到指定的“this.degree”(只调用canvas.rotate)。我假设我的数学不在这里。我试着按照这里的例子:Calculate angle for rotation in Pie Chart,但仍然没有正确旋转分针。任何帮助,将不胜感激。

2 个答案:

答案 0 :(得分:2)

数学看起来正确。你的计算应该给你触摸事件的角度,在中心点的正确右侧的触摸应该给你0度。

需要注意的一些事项

  • 确保您正朝着正确的方向旋转。很难保持这种直线,因此容易搞砸
  • 确保您考虑到值为0表示分针应指向右侧。例如,如果你从一个指向上方的分针开始,你必须在计算结果上加/减90度(取决于旋转方向 - 不确定哪个是正确的)
  • 确保(cx,cy)是您想要计算角度的中心点
  • 旋转时,您需要使用3 arg Canvas.rotate(float, float, float)方法,或单独添加其他翻译,以确保绕正确的点旋转。没有任何翻译,它将围绕(0,0)(视图的左上角)旋转

更多关于轮换: 旋转总是发生在“当前”(0,0)点附近。 “当前”是指当前矩阵应用后的(0,0)点。首次输入onDraw时,(0,0)点应该是视图的左上角。每当您应用平移/缩放/等时,您可能会相对于视图更改(0,0)点的位置。

我认为在设置正确的旋转中心方面,以下内容应该有效:

//first we save the initial matrix, so we can easily get
//back to this state after we're done rotating
canvas.save();
//I *think* you need to negate the center offsets here,
//because you are conceptually moving the canvas, rather
//than moving the center directly
canvas.translate(-cx, -cy);
//<perform the rotation and draw the clock hand>
//...

//and now restore the matrix back to the initial state   
canvas.restore();

答案 1 :(得分:0)

  

您的计算适用于测量分钟指针的角度   在模拟时钟的相应象限中旋转......这里很少   位变化可以使分钟或小时的手在旋转   触摸位置....在onTouch()方法中调用以下方法进行动作移动

public float getRotationAngle(float x, float y) {
        float dx = x - cx;
        float dy = y - cy;
        double a = Math.atan2(dy, dx);

        double degree = Math.toDegrees(a)+90;
        if(angle<0){
            degree=degree+360;
        }
        return (float) degree;
}

我有这种方法用于计算角度的矢量概念,但如果你想要的话,如果你想要的话,我会给出那个逻辑....