我有一个很长的带有滚动视图的活动。它是一个包含用户必须填写的各种字段的表单。我在表单的中间有一个复选框,当用户选中它时,我想滚动到视图的特定部分。是否有办法以编程方式滚动到EditText对象(或任何其他视图对象)?
此外,我知道这是可能的使用X和Y坐标,但我想避免这样做,因为形式可能会从用户到用户的变化。
我有一个很长的带有滚动视图的活动。它是一个包含用户必须填写的各种字段的表单。我在表单的中间有一个复选框,当用户选中它时,我想滚动到视图的特定部分。是否有办法以编程方式滚动到EditText对象(或任何其他视图对象)?
此外,我知道这是可能的使用X和Y坐标,但我想避免这样做,因为形式可能会从用户到用户的变化。
当前回答
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>
其他回答
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));
}