我正在实施数据列表采用者uisng" simple_list_item_2"内置但我得到错误。它说"来自片段的getview无法应用"它也无法解决" get"得到的符号。(位置)
public class AboutFragment extends Fragment {
ListView listView2;
String[] items = {"Friendly Map", "Inc"};
public AboutFragment() {
// Required empty public constructor
}
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
Bundle savedInstanceState) {
// Inflate the layout for this fragment
View v = inflater.inflate(R.layout.fragment_about, container, false);
listView2 = (ListView)v.findViewById(R.id.listView2);
ArrayAdapter arrayAdapter = new ArrayAdapter(getContext(), android.R.layout.simple_list_item_2, items);
listView2.setAdapter(arrayAdapter);
return v;
}
@Override
public View getView(int position, View convertView, ViewGroup parent){
View view = super.getView(position, convertView, parent);
String[] entry = listView2.get(position);
return view;
}
}
答案 0 :(得分:0)
试试这个。
public class AboutFragment extends Fragment {
ListView listView2;
String[] items = {"Friendly Map", "Inc"};
public AboutFragment() {
// Required empty public constructor
}
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
Bundle savedInstanceState) {
// Inflate the layout for this fragment
View v = inflater.inflate(R.layout.fragment_about, container, false);
listView2 = (ListView) v.findViewById(R.id.listView2);
ArrayAdapter arrayAdapter = new ArrayAdapter(getContext(), android.R.layout.simple_list_item_2, items);
listView2.setAdapter(arrayAdapter);
return v;
}
@Override
public View getView(int position, View convertView, ViewGroup parent){
View view = convertView;
String entry = listView2.get(position);
return view;
}
}