任何人都可以解释为什么这说我无法使用 getRef 来获取被单击的位置吗?从我查到的所有内容来看,这应该有效,但由于某种原因却无效。
public class DeleteChoiceListFragment extends Fragment {
DatabaseReference mRootRef = FirebaseDatabase.getInstance().getReference();
DatabaseReference mRestReference = mRootRef.child("restaurants");
List<String> listofrest = new ArrayList<String>();
ListView restaurantListView;
ListAdapter restaurantListAdapter;
public DeleteChoiceListFragment() {
}
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
View view = inflater.inflate(R.layout.restaurant_selection_list_frag, container, false);
restaurantListView = (ListView) view.findViewById(R.id.restaurantListView);
restaurantListAdapter = new FirebaseListAdapter<Restaurants>(getActivity(), Restaurants.class, R.layout.individual_restaurant_name_nocheckbox, mRestReference) {
@Override
protected void populateView(View v, Restaurants model, final int position) {
TextView restName = (TextView) v.findViewById(R.id.restname);
restName.setText(model.getName());
listofrest.add(position, model.getName());
}
};
restaurantListView.setAdapter(restaurantListAdapter);
restaurantListView.setOnItemClickListener(new AdapterView.OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> adapterView, View view, int i, long l) {
Firebase itemtoremove = restaurantListAdapter.getRef(i);
//I get an error here saying I can't use .getRef()
itemtoremove.removeValue();
}
});
return view;
}
}
最佳答案
您已声明restaurantListAdapter 为ListAdapter
。即使您将 FirebaseListAdapter
对象放入该字段,您也只能访问 ListAdapter
上定义的方法。
解决方案是将字段声明为 FirebaseListAdapter
:
FirebaseListAdapter restaurantListAdapter;
关于java - 无法使用 getRef() 吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41077476/