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

我的答案是Firebase和位置0是一个变通办法

Parcelable state;

DatabaseReference everybody = db.getReference("Everybody Room List");
    everybody.addValueEventListener(new ValueEventListener() {
        @Override
        public void onDataChange(@NonNull DataSnapshot dataSnapshot) {
            state = listView.onSaveInstanceState(); // Save
            progressBar.setVisibility(View.GONE);
            arrayList.clear();
            for (DataSnapshot messageSnapshot : dataSnapshot.getChildren()) {
                Messages messagesSpacecraft = messageSnapshot.getValue(Messages.class);
                arrayList.add(messagesSpacecraft);
            }
            listView.setAdapter(convertView);
            listView.onRestoreInstanceState(state); // Restore
        }

        @Override
        public void onCancelled(@NonNull DatabaseError databaseError) {
        }
    });

和convertView

位置0 a添加一个您不使用的空白项

public class Chat_ConvertView_List_Room extends BaseAdapter {

private ArrayList<Messages> spacecrafts;
private Context context;

@SuppressLint("CommitPrefEdits")
Chat_ConvertView_List_Room(Context context, ArrayList<Messages> spacecrafts) {
    this.context = context;
    this.spacecrafts = spacecrafts;
}

@Override
public int getCount() {
    return spacecrafts.size();
}

@Override
public Object getItem(int position) {
    return spacecrafts.get(position);
}

@Override
public long getItemId(int position) {
    return position;
}

@SuppressLint({"SetTextI18n", "SimpleDateFormat"})
@Override
public View getView(final int position, View convertView, ViewGroup parent) {
    if (convertView == null) {
        convertView = LayoutInflater.from(context).inflate(R.layout.message_model_list_room, parent, false);
    }

    final Messages s = (Messages) this.getItem(position);

    if (position == 0) {
        convertView.getLayoutParams().height = 1; // 0 does not work
    } else {
        convertView.getLayoutParams().height = RelativeLayout.LayoutParams.WRAP_CONTENT;
    }

    return convertView;
}
}

我已经看到这个工作暂时不打扰用户,我希望它为您工作

其他回答

难道不是简单的android:saveEnabled="true"在ListView xml声明足够吗?

我发现了一些有趣的事情。

我尝试了setSelection和scrolltoXY,但它根本不起作用,列表仍然在相同的位置,经过一些尝试和错误,我得到了以下代码,确实工作

final ListView list = (ListView) findViewById(R.id.list);
list.post(new Runnable() {            
    @Override
    public void run() {
        list.setSelection(0);
    }
});

如果不是发布Runnable,你尝试runOnUiThread,它也不工作(至少在一些设备上)

这是一个非常奇怪的变通方法,应该是直截了当的。

我的答案是Firebase和位置0是一个变通办法

Parcelable state;

DatabaseReference everybody = db.getReference("Everybody Room List");
    everybody.addValueEventListener(new ValueEventListener() {
        @Override
        public void onDataChange(@NonNull DataSnapshot dataSnapshot) {
            state = listView.onSaveInstanceState(); // Save
            progressBar.setVisibility(View.GONE);
            arrayList.clear();
            for (DataSnapshot messageSnapshot : dataSnapshot.getChildren()) {
                Messages messagesSpacecraft = messageSnapshot.getValue(Messages.class);
                arrayList.add(messagesSpacecraft);
            }
            listView.setAdapter(convertView);
            listView.onRestoreInstanceState(state); // Restore
        }

        @Override
        public void onCancelled(@NonNull DatabaseError databaseError) {
        }
    });

和convertView

位置0 a添加一个您不使用的空白项

public class Chat_ConvertView_List_Room extends BaseAdapter {

private ArrayList<Messages> spacecrafts;
private Context context;

@SuppressLint("CommitPrefEdits")
Chat_ConvertView_List_Room(Context context, ArrayList<Messages> spacecrafts) {
    this.context = context;
    this.spacecrafts = spacecrafts;
}

@Override
public int getCount() {
    return spacecrafts.size();
}

@Override
public Object getItem(int position) {
    return spacecrafts.get(position);
}

@Override
public long getItemId(int position) {
    return position;
}

@SuppressLint({"SetTextI18n", "SimpleDateFormat"})
@Override
public View getView(final int position, View convertView, ViewGroup parent) {
    if (convertView == null) {
        convertView = LayoutInflater.from(context).inflate(R.layout.message_model_list_room, parent, false);
    }

    final Messages s = (Messages) this.getItem(position);

    if (position == 0) {
        convertView.getLayoutParams().height = 1; // 0 does not work
    } else {
        convertView.getLayoutParams().height = RelativeLayout.LayoutParams.WRAP_CONTENT;
    }

    return convertView;
}
}

我已经看到这个工作暂时不打扰用户,我希望它为您工作

这里提供的解决方案似乎都不适合我。在我的情况下,我有一个ListView在一个片段,我替换在一个FragmentTransaction,所以一个新的片段实例创建每次片段显示,这意味着ListView状态不能存储为片段的成员。

相反,我最终将状态存储在我的自定义Application类中。下面的代码应该会让你了解它是如何工作的:

public class MyApplication extends Application {
    public static HashMap<String, Parcelable> parcelableCache = new HashMap<>();


    /* ... code omitted for brevity ... */
}

 

public class MyFragment extends Fragment{
    private ListView mListView = null;
    private MyAdapter mAdapter = null;


    @Override
    public void onViewCreated(View view, @Nullable Bundle savedInstanceState) {
        super.onViewCreated(view, savedInstanceState);

        mAdapter = new MyAdapter(getActivity(), null, 0);
        mListView = ((ListView) view.findViewById(R.id.myListView));

        Parcelable listViewState = MyApplication.parcelableCache.get("my_listview_state");
        if( listViewState != null )
            mListView.onRestoreInstanceState(listViewState);
    }


    @Override
    public void onPause() {
        MyApplication.parcelableCache.put("my_listview_state", mListView.onSaveInstanceState());
        super.onPause();
    }

    /* ... code omitted for brevity ... */

}

基本思想是将状态存储在片段实例之外。如果您不喜欢在应用程序类中拥有静态字段的想法,我猜您可以通过实现一个片段接口并将状态存储在您的活动中来实现它。

另一种解决方案是将其存储在SharedPreferences中,但这有点复杂,您需要确保在应用程序启动时清除它,除非您希望在应用程序启动时保持状态。

 

另外,为了避免“当第一项可见时滚动位置不保存”,你可以显示一个0px高度的虚拟第一项。这可以通过重写适配器中的getView()来实现,如下所示:

@Override
public View getView(int position, View convertView, ViewGroup parent) {
    if( position == 0 ) {
        View zeroHeightView = new View(parent.getContext());
        zeroHeightView.setLayoutParams(new ViewGroup.LayoutParams(0, 0));
        return zeroHeightView;
    }
    else
        return super.getView(position, convertView, parent);
}

为了澄清Ryan Newsom的精彩回答并针对片段进行调整通常情况下,我们想要从主ListView片段导航到细节片段然后再返回主ListView片段

    private View root;
    public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState)
        {
           if(root == null){
             root = inflater.inflate(R.layout.myfragmentid,container,false);
             InitializeView(); 
           } 
           return root; 
        }

    public void InitializeView()
    {
        ListView listView = (ListView)root.findViewById(R.id.listviewid);
        BaseAdapter adapter = CreateAdapter();//Create your adapter here
        listView.setAdpater(adapter);
        //other initialization code
    }

这里的“神奇”是,当我们从细节片段导航回ListView片段时,视图不会被重新创建,我们不设置ListView的适配器,所以一切都保持不变!