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
检查Android源代码,你会发现ScrollView已经有一个成员函数scrollToChild(View),它做的正是所请求的。不幸的是,这个函数由于某种模糊的原因被标记为private。在这个函数的基础上,我写了下面的函数,它可以找到指定为参数的视图上方的第一个ScrollView,并滚动它,使它在ScrollView中可见:
private void make_visible(View view)
{
int vt = view.getTop();
int vb = view.getBottom();
View v = view;
for(;;)
{
ViewParent vp = v.getParent();
if(vp == null || !(vp instanceof ViewGroup))
break;
ViewGroup parent = (ViewGroup)vp;
if(parent instanceof ScrollView)
{
ScrollView sv = (ScrollView)parent;
// Code based on ScrollView.computeScrollDeltaToGetChildRectOnScreen(Rect rect) (Android v5.1.1):
int height = sv.getHeight();
int screenTop = sv.getScrollY();
int screenBottom = screenTop + height;
int fadingEdge = sv.getVerticalFadingEdgeLength();
// leave room for top fading edge as long as rect isn't at very top
if(vt > 0)
screenTop += fadingEdge;
// leave room for bottom fading edge as long as rect isn't at very bottom
if(vb < sv.getChildAt(0).getHeight())
screenBottom -= fadingEdge;
int scrollYDelta = 0;
if(vb > screenBottom && vt > screenTop)
{
// need to move down to get it in view: move down just enough so
// that the entire rectangle is in view (or at least the first
// screen size chunk).
if(vb-vt > height) // just enough to get screen size chunk on
scrollYDelta += (vt - screenTop);
else // get entire rect at bottom of screen
scrollYDelta += (vb - screenBottom);
// make sure we aren't scrolling beyond the end of our content
int bottom = sv.getChildAt(0).getBottom();
int distanceToBottom = bottom - screenBottom;
scrollYDelta = Math.min(scrollYDelta, distanceToBottom);
}
else if(vt < screenTop && vb < screenBottom)
{
// need to move up to get it in view: move up just enough so that
// entire rectangle is in view (or at least the first screen
// size chunk of it).
if(vb-vt > height) // screen size chunk
scrollYDelta -= (screenBottom - vb);
else // entire rect at top
scrollYDelta -= (screenTop - vt);
// make sure we aren't scrolling any further than the top our content
scrollYDelta = Math.max(scrollYDelta, -sv.getScrollY());
}
sv.smoothScrollBy(0, scrollYDelta);
break;
}
// Transform coordinates to parent:
int dy = parent.getTop()-parent.getScrollY();
vt += dy;
vb += dy;
v = parent;
}
}
其他回答
scrollView.post(new Runnable() {
@Override
public void run() {
scrollView.smoothScrollTo(0, myTextView.getTop());
}
});
回答来自我的实际项目。
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));
}
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]);
}
});
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>
另一种说法是:
scrollView.postDelayed(new Runnable()
{
@Override
public void run()
{
scrollView.smoothScrollTo(0, img_transparent.getTop());
}
}, 200);
或者可以使用post()方法。