Question

您好我现在制作了Fragment我试图将所有联系人姓名显示给我的ListView我做了所有事情，但我仍然得到Cursor null我也打电话给{{} 1}}。仍然是错误相同。

getActivity()

例外：

public class AllContacts extends Fragment {


public AllContacts() {
    // Required empty public constructor

}


@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
                         Bundle savedInstanceState) {

    View view = inflater.inflate(R.layout.fragment_all_contacts, container, false);
    ListView list = (ListView) view.findViewById(R.id.listview);

    ArrayList<String> arraylist= new ArrayList<>();


    String sorting = ContactsContract.Contacts.DISPLAY_NAME+ "DESC";
    Uri uri = Uri.parse("content://ContactsContract");
    Cursor c = getActivity().getContentResolver().query(uri,null,null,null,sorting);
    while (c.moveToNext()) {

        String contactName = c.getString(c.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME));

        arraylist.add(contactName);
    }//end loop
       c.close();
    ArrayAdapter adapter = new ArrayAdapter(getContext(),android.R.layout.simple_list_item_1,arraylist);

    list.setAdapter(adapter);
    return view;
}

}

Answer 1

您的查询存在两个问题，第一个问题是DESC关键字之前缺少空格，第二个问题是无效Uri。

更改以下内容：

String sorting = ContactsContract.Contacts.DISPLAY_NAME+ "DESC";
Uri uri = Uri.parse("content://ContactsContract");
Cursor c = getActivity().getContentResolver().query(uri,null,null,null,sorting);

对此：

String sorting = ContactsContract.Contacts.DISPLAY_NAME + " DESC";
Cursor c = getActivity().getContentResolver()
    .query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
        null, null, null, sorting);

片段getActivity（）。getContentResolver（）。query（）返回null

1 个答案: