我有一个很长的ListView,用户可以在返回前一个屏幕之前滚动它。当用户再次打开这个ListView时,我希望列表被滚动到与之前相同的位置。关于如何实现这一点,你有什么想法吗?
当前回答
Parcelable state;
@Override
public void onPause() {
// Save ListView state @ onPause
Log.d(TAG, "saving listview state");
state = listView.onSaveInstanceState();
super.onPause();
}
...
@Override
public void onViewCreated(final View view, Bundle savedInstanceState) {
super.onViewCreated(view, savedInstanceState);
// Set new items
listView.setAdapter(adapter);
...
// Restore previous state (including selected item index and scroll position)
if(state != null) {
Log.d(TAG, "trying to restore listview state");
listView.onRestoreInstanceState(state);
}
}
其他回答
一个非常简单的方法:
/** Save the position **/
int currentPosition = listView.getFirstVisiblePosition();
//Here u should save the currentPosition anywhere
/** Restore the previus saved position **/
listView.setSelection(savedPosition);
方法setSelection将把列表重置为所提供的项。如果不是在触摸模式,项目将实际被选中,如果在触摸模式,项目将只定位在屏幕上。
一个更复杂的方法:
listView.setOnScrollListener(this);
//Implements the interface:
@Override
public void onScroll(AbsListView view, int firstVisibleItem,
int visibleItemCount, int totalItemCount) {
mCurrentX = view.getScrollX();
mCurrentY = view.getScrollY();
}
@Override
public void onScrollStateChanged(AbsListView view, int scrollState) {
}
//Save anywere the x and the y
/** Restore: **/
listView.scrollTo(savedX, savedY);
试试这个:
// save index and top position
int index = mList.getFirstVisiblePosition();
View v = mList.getChildAt(0);
int top = (v == null) ? 0 : (v.getTop() - mList.getPaddingTop());
// ...
// restore index and position
mList.setSelectionFromTop(index, top);
Explanation: ListView.getFirstVisiblePosition() returns the top visible list item. But this item may be partially scrolled out of view, and if you want to restore the exact scroll position of the list you need to get this offset. So ListView.getChildAt(0) returns the View for the top list item, and then View.getTop() - mList.getPaddingTop() returns its relative offset from the top of the ListView. Then, to restore the ListView's scroll position, we call ListView.setSelectionFromTop() with the index of the item we want and an offset to position its top edge from the top of the ListView.
我使用的是FirebaseListAdapter,不能让任何解决方案工作。我最后做了这个。我猜有更优雅的方式,但这是一个完整的和有效的解决方案。
在onCreate之前:
private int reset;
private int top;
private int index;
FirebaseListAdapter内部:
@Override
public void onDataChanged() {
super.onDataChanged();
// Only do this on first change, when starting
// activity or coming back to it.
if(reset == 0) {
mListView.setSelectionFromTop(index, top);
reset++;
}
}
启动时间:
@Override
protected void onStart() {
super.onStart();
if(adapter != null) {
adapter.startListening();
index = 0;
top = 0;
// Get position from SharedPrefs
SharedPreferences sharedPref = PreferenceManager.getDefaultSharedPreferences(this);
top = sharedPref.getInt("TOP_POSITION", 0);
index = sharedPref.getInt("INDEX_POSITION", 0);
// Set reset to 0 to allow change to last position
reset = 0;
}
}
停止:
@Override
protected void onStop() {
super.onStop();
if(adapter != null) {
adapter.stopListening();
// Set position
index = mListView.getFirstVisiblePosition();
View v = mListView.getChildAt(0);
top = (v == null) ? 0 : (v.getTop() - mListView.getPaddingTop());
// Save position to SharedPrefs
SharedPreferences sharedPref = PreferenceManager.getDefaultSharedPreferences(this);
sharedPref.edit().putInt("TOP_POSITION" + "", top).apply();
sharedPref.edit().putInt("INDEX_POSITION" + "", index).apply();
}
}
因为我还必须解决这个FirebaseRecyclerAdapter,我在这里发布的解决方案:
在onCreate之前:
private int reset;
private int top;
private int index;
FirebaseRecyclerAdapter内部:
@Override
public void onDataChanged() {
// Only do this on first change, when starting
// activity or coming back to it.
if(reset == 0) {
linearLayoutManager.scrollToPositionWithOffset(index, top);
reset++;
}
}
启动时间:
@Override
protected void onStart() {
super.onStart();
if(adapter != null) {
adapter.startListening();
index = 0;
top = 0;
// Get position from SharedPrefs
SharedPreferences sharedPref = PreferenceManager.getDefaultSharedPreferences(this);
top = sharedPref.getInt("TOP_POSITION", 0);
index = sharedPref.getInt("INDEX_POSITION", 0);
// Set reset to 0 to allow change to last position
reset = 0;
}
}
停止:
@Override
protected void onStop() {
super.onStop();
if(adapter != null) {
adapter.stopListening();
// Set position
index = linearLayoutManager.findFirstVisibleItemPosition();
View v = linearLayoutManager.getChildAt(0);
top = (v == null) ? 0 : (v.getTop() - linearLayoutManager.getPaddingTop());
// Save position to SharedPrefs
SharedPreferences sharedPref = PreferenceManager.getDefaultSharedPreferences(this);
sharedPref.edit().putInt("TOP_POSITION" + "", top).apply();
sharedPref.edit().putInt("INDEX_POSITION" + "", index).apply();
}
}
Parcelable state;
@Override
public void onPause() {
// Save ListView state @ onPause
Log.d(TAG, "saving listview state");
state = listView.onSaveInstanceState();
super.onPause();
}
...
@Override
public void onViewCreated(final View view, Bundle savedInstanceState) {
super.onViewCreated(view, savedInstanceState);
// Set new items
listView.setAdapter(adapter);
...
// Restore previous state (including selected item index and scroll position)
if(state != null) {
Log.d(TAG, "trying to restore listview state");
listView.onRestoreInstanceState(state);
}
}
我采用了@(Kirk Woll)建议的解决方案,它对我很有效。我还在“联系人”应用程序的Android源代码中看到,他们使用了类似的技术。我还想补充一些具体情况: 在我的listactivity派生类的顶部:
private static final String LIST_STATE = "listState";
private Parcelable mListState = null;
然后,一些方法重写:
@Override
protected void onRestoreInstanceState(Bundle state) {
super.onRestoreInstanceState(state);
mListState = state.getParcelable(LIST_STATE);
}
@Override
protected void onResume() {
super.onResume();
loadData();
if (mListState != null)
getListView().onRestoreInstanceState(mListState);
mListState = null;
}
@Override
protected void onSaveInstanceState(Bundle state) {
super.onSaveInstanceState(state);
mListState = getListView().onSaveInstanceState();
state.putParcelable(LIST_STATE, mListState);
}
当然,“loadData”是我从DB中检索数据并将其放入列表的函数。
在我的Froyo设备上,当你改变手机方向时,当你编辑一个项目并返回列表时,这都是有效的。
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