Android ArrayAdapter NullPointerException（但不爲空）

Question

我試圖創建一個與firebase聊天，試圖讓添加的用戶朋友ListView，但是當我把.setAdapter（），它說我有一個Java。 .setAdapter行上的lang.NullPointerException。

package studio.brunocasamassa.superchat.fragments; 


import android.content.Context; 
import android.os.Bundle; 
import android.support.v4.app.Fragment; 
import android.view.LayoutInflater; 
import android.view.View; 
import android.view.ViewGroup; 
import android.widget.ArrayAdapter; 
import android.widget.ListView; 

import com.google.firebase.database.DatabaseReference; 

import java.util.ArrayList; 

import studio.brunocasamassa.superchat.R; 

import static android.R.layout.activity_list_item; 
import static android.R.layout.simple_list_item_2; 

/** 
* A simple {@link Fragment} subclass. 
*/ 


public class ContatosFragment extends Fragment { 

    private ListView listview_nomes; 
    private ArrayList<String> arraylist_nomes; 
    private ArrayAdapter<String> adapter_nomes = null; 
    private DatabaseReference firebaseDatabase; 


    public ContatosFragment() { 
     // Required empty public constructor 
    } 


    public String insertContact(String nomeContato) { 
     arraylist_nomes = new ArrayList<String>(); 
     arraylist_nomes.add(nomeContato); 
     adapter_nomes = new ArrayAdapter<String>(getContext(), 
       android.R.layout.simple_list_item_2, 
       android.R.id.text2, 
       arraylist_nomes); 
     return nomeContato; 
    } 


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


     if (adapter_nomes == null) { 
      insertContact("teste"); 
     } 

     listview_nomes = (ListView) getActivity().findViewById(R.id.ListContatos); 
     System.out.println(adapter_nomes); 
> the problem: 
     listview_nomes.setAdapter(adapter_nomes); 


     // Inflate the layout for this fragment 
     return inflater.inflate(R.layout.fragment_contatos, container, false); 

    } 


} 

我打電話insertContact methos在另一個類，如下所示：

 Contato contato = new Contato(); 
             contato.setidUser(idContact); 
             contato.setEmail(emailContact); 
             contato.setNome(usuarioContato.getNome()); 
      ContatosFragment contact = new ContatosFragment(); 
              contact.insertContact(usuarioContato.getNome()); 
System.out.println("nome usuario: "+usuarioContato.getNome()); 

我不知道如何解決它，如果有人知道， 感謝

已更新

我只是改變了我的insertContact方法（因爲這個問題看起來就像是它裏面一個ArrayAdapter變量）：

public String insertContact(String nomeContato) { 
    arraylist_nomes = new ArrayList<String>(); 
    arraylist_nomes.add(nomeContato); 
    System.out.println("nomes_array: " + arraylist_nomes); 

    adapter_nomes = new ArrayAdapter<String>(
      getContext().getApplicationContext(), 
      android.R.layout.simple_list_item_2, 
      android.R.id.text2, 
      arraylist_nomes); 

    return nomeContato; 
} 

Answer 1

你onCreateView()應該是這樣的：

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

     View v = inflater.inflate(R.layout.fragment_contatos, container, false); 
     if (adapter_nomes == null) { 
      insertContact("teste"); 
     } 

     listview_nomes = (ListView) v.findViewById(R.id.ListContatos); 

     System.out.println(adapter_nomes); 
     listview_nomes.setAdapter(adapter_nomes); 
     return v; 
} 

你必須得到ListView從當前的佈局視圖中未從getActivity()其給出一個空引用

listview_nomes = (ListView) v.findViewById(R.id.ListContatos); 

Answer 2

中onCreateVIew做到這一點（）

View root = inflater.inflate(R.layout.fragment_contatos, container, false); 
listview_nomes = (ListView) root.findViewById(R.id.ListContatos); 
listview_nomes.setAdapter(adapter_nomes); 


return root; 

問題是

listview_nomes = (ListView) getActivity().findViewById(R.id.ListContatos); 

試圖找到在胡亞蓉佈局，而不是片段佈局列表視圖。因此你得到一個空指針。您的列表視圖參考​​爲空。