无法从列表视图中获取项目位置：尝试在空对象引用上调用虚拟方法...答案

【问题标题】：Can't get item position from listview: Attempt to invoke virtual method ... on a null object reference无法从列表视图中获取项目位置：尝试在空对象引用上调用虚拟方法...
【发布时间】：2021-07-24 06:48:01
【问题描述】：

当我单击 ListView 上的一项时，我试图让一个弹出菜单出现，但我无法获得该项的位置。

        listax.setOnItemClickListener(new AdapterView.OnItemClickListener() {
            public void onItemClick(AdapterView<?> adapterView, View view, int position, long id) {

                ItemList item= adapter.getItem(position);

                Toast.makeText(getContext(), item.getTxtNomeItem().toString(),   Toast.LENGTH_SHORT).show();
                //showPopup(adapterView);

            }
        });

该代码给了我“尝试在空对象引用上调用虚拟方法'java.lang.Object com.example.securityaplication.ItemArrayAdapter.getItem(int)'”错误。

我在另一个项目中做了同样的事情，效果很好：

        listax.setOnItemClickListener(new AdapterView.OnItemClickListener() {
            public void onItemClick(AdapterView<?> parent, View view,
                                    int position, long id) {

                modelo item = adapter.getItem(position);


                Toast.makeText(getApplicationContext(),
                        item.getNome().toString (), Toast.LENGTH_SHORT).show();

            }
        });

唯一的区别是我使用“getContext”而不是“getApplicationContext()”，因为它不接受“getApplicationContext()”，并且第一个在片段中，而后一个在活动中。

我的 Fragment 给了我错误：

public class HomeFragment extends Fragment implements PopupMenu.OnMenuItemClickListener  {

    ItemArrayAdapter adapter;
    private ListView listax;


    public HomeFragment() {

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


        View view = inflater.inflate(R.layout.fragment_home, container, false);
        listax = (ListView)view.findViewById(R.id.listViewx);

        ArrayList<ItemList> ItemList = new ArrayList<>();

        ItemArrayAdapter itemArrayAdapter = new ItemArrayAdapter(getActivity(), R.layout.adapter_view_layout,ItemList);
        listax.setAdapter(itemArrayAdapter);
        


        listax.setOnItemClickListener(new AdapterView.OnItemClickListener() {
            public void onItemClick(AdapterView<?> adapterView, View view, int position, long id) {

                ItemList item= adapter.getItem(position);

                Toast.makeText(getContext(), item.getTxtNomeItem().toString(),   Toast.LENGTH_SHORT).show();
                //showPopup(adapterView);

            }
        });

        return view;
    }

    public void showPopup(View v){
        PopupMenu popup = new PopupMenu(getActivity(), v);
        popup.setOnMenuItemClickListener(this);
        popup.inflate(R.menu.menupopup);
        popup.show();
    }

    @Override
    public boolean onMenuItemClick(MenuItem menuItem) {
        switch(menuItem.getItemId()){
            case R.id.itemAlterar:
                Toast.makeText(getActivity(), "Item 1 clicked", Toast.LENGTH_LONG).show();
                return true;
            case R.id.itemExcluir:
                Toast.makeText(getActivity(), "Item 2 clicked", Toast.LENGTH_LONG).show();
                return true;
            case R.id.itemEmprestar:
                Toast.makeText(getActivity(), "Item 3 clicked", Toast.LENGTH_LONG).show();
                return true;
            case R.id.itemRecuperar:
                Toast.makeText(getActivity(), "Item 3 clicked", Toast.LENGTH_LONG).show();
                return true;
            default:
                return false;
    }}
}

【问题讨论】：

标签： java android listview android-fragments

【解决方案1】：

在设置 ListAdapter 时不要忘记使用类成员变量“适配器”。

//ItemArrayAdapter itemArrayAdapter = new ItemArrayAdapter(getActivity(), R.layout.adapter_view_layout,ItemList);
//listax.setAdapter(itemArrayAdapter);

adapter = new ItemArrayAdapter(getActivity(), R.layout.adapter_view_layout,ItemList);
listax.setAdapter(adapter);

【讨论】：