Question

我有一个应用程序，我需要根据当前聚焦的子视图调整视图位置（它是一个具有可聚焦项目的列表，当前聚焦的项目必须位于屏幕的中心 - 用于从电视遥控器控制的电视应用程序）。
必须使用动画调整位置我只使用了一个问题：如果用户在动画完成之前更改焦点（快速点击“向上”按钮两次），下一个动画以“跳转”开始 - 它从与第一个相同的位置开始。

所以我尝试做的是取消之前的动画并启动另一个动画，然后从第一个动画开始的点开始新动画，这样用户就可以在动画中看到一个非常明显的跳跃，看起来非常糟糕。 / p>

以下是代码：

@Override
public void requestChildFocus(final View child, final View focused) {
    super.requestChildFocus(child, focused);

    //this test code included for explanation
    Rect r = new Rect();
    child.getDrawingRect(r); //this will return view's position ignoring animation state
    Rect r2 = new Rect();
    child.getGlobalVisibleRect(r2); //as will this one too
    Log.d("Top: " + child.getTop() + "; Drawing rect: " +  r.toString() + "; global visible rect: " + r2.toString()); 
    //all of this methods will ignore changes that were made 
    //by animation object - they'll return numbers from LayoutParam

    //calculate current position inside view and position to move to
    //cursorOffset - is the "center" of the screen
    final int currentPosition = child.getTop();
    final int requaredPosition = cursorOffset - focused.getTop();

    //cancel current running animation - layout params will not change
    //if i do change layout params when cancelling animation, it looks even worse
    //because of jumping back list jumps forward
    if (currentAnimation != null) {
        Animation animation = currentAnimation;
        currentAnimation = null;
        animation.cancel();
    }

    //just a regular translate animation
    TranslateAnimation animation = new TranslateAnimation(0, 0, 0, requaredPosition - currentPosition);
    animation.setDuration(300);
    animation.setFillEnabled(true);
    animation.setFillBefore(true);
    animation.setAnimationListener(new AnimationListener() {

        @Override
        public void onAnimationStart(Animation animation) {
            currentAnimation = animation;
        }

        @Override
        public void onAnimationRepeat(Animation animation) {}

        @Override
        public void onAnimationEnd(Animation animation) {
            if (animation == currentAnimation) {
                //change layout params if animation finished running (wasn't cancelled)
                RelativeLayout.LayoutParams params = (LayoutParams) child.getLayoutParams();
                params.setMargins(0, requaredPosition, 0, 0);
                child.setLayoutParams(params);
            }
        }
    });
    child.startAnimation(animation);
}

所以必须提出的问题是：如何从以前的翻译动画留下的位置（假设它被取消）开始翻译动画？或者，用简单的话说，我如何确定视图的当前可见矩形？

Answer 1

显然你无法获得当前的观看位置，但你可以获得当前的动画状态所以你可以通过这样做得到当前的y偏移：

Transformation transformation = new Transformation();
float[] matrix = new float[9];
currentAnimation.getTransformation(AnimationUtils.currentAnimationTimeMillis(), transformation);
transformation.getMatrix().getValues(matrix);
float y = matrix[Matrix.MTRANS_Y];

这就是我如何能够取消一个动画并从我离开的位置开始另一个动画。如果有人关心，那就是完整的代码：

private Animation currentAnimation;
private float[] matrix = new float[9];
private Transformation transformation = new Transformation();

@Override
public void requestChildFocus(final View child, final View focused) {
    super.requestChildFocus(child, focused);

    final int currentPosition;

    if (currentAnimation != null) {
        currentAnimation.getTransformation(AnimationUtils.currentAnimationTimeMillis(), transformation);
        transformation.getMatrix().getValues(matrix);
        float y = matrix[Matrix.MTRANS_Y];

        RelativeLayout.LayoutParams params = (LayoutParams) child.getLayoutParams();
        params.topMargin += y;
        //child.getTop() will return wrong position until layout actually happens, 
        //so I use params.topMargin as a current position in case I need to cancel
        currentPosition = params.topMargin;
        child.requestLayout();

        currentAnimation.setAnimationListener(null);
        currentAnimation.cancel();
        currentAnimation = null;
    } else {
        currentPosition = child.getTop();
    }

    final int requaredPosition = cursorOffset - focused.getTop();

    TranslateAnimation animation = new TranslateAnimation(0, 0, 0, requaredPosition - currentPosition);
    animation.setDuration(300);
    animation.setFillEnabled(true);
    animation.setFillBefore(true);
    animation.setAnimationListener(new AnimationListener() {

        @Override
        public void onAnimationStart(Animation animation) {
            currentAnimation = animation;
        }

        @Override
        public void onAnimationRepeat(Animation animation) {}

        @Override
        public void onAnimationEnd(Animation animation) {
            if (animation == currentAnimation) {
                RelativeLayout.LayoutParams params = (LayoutParams) child.getLayoutParams();
                params.setMargins(0, requaredPosition, 0, 0);
                child.requestLayout();
            }
            currentAnimation = null;
        }
    });
    child.startAnimation(animation);
}

希望有人觉得这很有用。

Answer 2

为每个寻求帮助的人创造一个像我这样的时钟或速度表（寻找RotationAnimation＆lt; API lvl 9），我发现，设置AnimationListener并保存是个好主意布尔变量中的动画状态。在开始新动画之前，请检查状态。因此，您的动画将一直运行到最后并且看起来非常流畅。我很抱歉我的英语。

        if(isAnimationActive){
        //[...]
        rotate = new RotateAnimation(oldRotateDegree, rotateDegree, Animation.RELATIVE_TO_SELF, 0.5f,
                            Animation.RELATIVE_TO_SELF, 0.5f);
                    rotate.setDuration(rotateDuration);
                    rotate.setFillAfter(true);
                    rotate.setAnimationListener(this);
       }        
       //[...]

        @Override
        public void onAnimationEnd(Animation animation)
        {
            this.isAnimationActive = false;

        }
        @Override
        public void onAnimationStart(Animation animation)
        {
            this.isAnimationActive = true;  
        }

中断翻译动画并启动另一个

2 个答案: