将postDelayed添加到视图中，这样getTop()就不会返回0。

binding.scrollViewLogin.postDelayed({
            val scrollTo = binding.textInputLayoutFirstName.top
            binding.scrollViewLogin.isSmoothScrollingEnabled = true
            binding.scrollViewLogin.smoothScrollTo(0, scrollTo)
     }, 400
) 

还要确保视图是scrollView的直接子视图，否则你会得到getTop()为零。示例:嵌入在TextInputLayout中的edittext的getTop()将返回0。在这种情况下，我们需要计算TextInputLayout的getTop()它是ScrollView的直接子。

<ScrollView>
    <TextInputLayout>
        <EditText/>
    </TextInputLayout>
</ScrollView>

参考资料:https://stackoverflow.com/a/6438240/2624806

接下来的工作要好得多。

mObservableScrollView.post(new Runnable() {
            public void run() { 
                mObservableScrollView.fullScroll([View_FOCUS][1]); 
            }
        });

你可以像这样使用ObjectAnimator:

ObjectAnimator.ofInt(yourScrollView, "scrollY", yourView.getTop()).setDuration(1500).start();

我的解决方案是:

            int[] spinnerLocation = {0,0};
            spinner.getLocationOnScreen(spinnerLocation);

            int[] scrollLocation = {0, 0};
            scrollView.getLocationInWindow(scrollLocation);

            int y = scrollView.getScrollY();

            scrollView.smoothScrollTo(0, y + spinnerLocation[1] - scrollLocation[1]);

我认为我已经找到了更优雅、更不容易出错的解决方案

ScrollView.requestChildRectangleOnScreen

它不涉及数学运算，与其他提出的解决方案相反，它将正确地处理上下滚动。

/**
 * Will scroll the {@code scrollView} to make {@code viewToScroll} visible
 * 
 * @param scrollView parent of {@code scrollableContent}
 * @param scrollableContent a child of {@code scrollView} whitch holds the scrollable content (fills the viewport).
 * @param viewToScroll a child of {@code scrollableContent} to whitch will scroll the the {@code scrollView}
 */
void scrollToView(ScrollView scrollView, ViewGroup scrollableContent, View viewToScroll) {
    Rect viewToScrollRect = new Rect(); //coordinates to scroll to
    viewToScroll.getHitRect(viewToScrollRect); //fills viewToScrollRect with coordinates of viewToScroll relative to its parent (LinearLayout) 
    scrollView.requestChildRectangleOnScreen(scrollableContent, viewToScrollRect, false); //ScrollView will make sure, the given viewToScrollRect is visible
}

如果ScrollView正在被更改，那么将它封装到postDelayed中以使其更加可靠是个好主意

/**
 * Will scroll the {@code scrollView} to make {@code viewToScroll} visible
 * 
 * @param scrollView parent of {@code scrollableContent}
 * @param scrollableContent a child of {@code scrollView} whitch holds the scrollable content (fills the viewport).
 * @param viewToScroll a child of {@code scrollableContent} to whitch will scroll the the {@code scrollView}
 */
private void scrollToView(final ScrollView scrollView, final ViewGroup scrollableContent, final View viewToScroll) {
    long delay = 100; //delay to let finish with possible modifications to ScrollView
    scrollView.postDelayed(new Runnable() {
        public void run() {
            Rect viewToScrollRect = new Rect(); //coordinates to scroll to
            viewToScroll.getHitRect(viewToScrollRect); //fills viewToScrollRect with coordinates of viewToScroll relative to its parent (LinearLayout) 
            scrollView.requestChildRectangleOnScreen(scrollableContent, viewToScrollRect, false); //ScrollView will make sure, the given viewToScrollRect is visible
        }
    }, delay);
}

如果ScrollView是ChildView的直接父类，上面的答案就可以很好地工作。如果你的ChildView被包装在ScrollView中的另一个ViewGroup中，它将导致意外的行为，因为View.getTop()获得相对于其父的位置。在这种情况下，你需要实现这个:

public static void scrollToInvalidInputView(ScrollView scrollView, View view) {
    int vTop = view.getTop();

    while (!(view.getParent() instanceof ScrollView)) {
        view = (View) view.getParent();
        vTop += view.getTop();
    }

    final int scrollPosition = vTop;

    new Handler().post(() -> scrollView.smoothScrollTo(0, scrollPosition));
}