android - 调用 notifyDataSetChanged 后 Android setOnClickListener 无法正常工作

Question

我确实有一个从数据库更新的 ArrayAdapter。当数据更新notifyDataSetChanged();时被调用并更新数据列表。我的问题是 onListItemClick 没有更新。当用户单击第一个列表项时，会检查每个 1-9-18.. 并更改背景颜色。

应该这样做是因为“ListView 会回收其行，因此您需要确保在适配器的 getView() 方法中重置视图的状态。” 关联

这是怎么做到的？

@Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container,Bundle savedInstanceState) {
        contacts.clear();
        activity = getActivity();
        final View rootView = inflater.inflate(R.layout.activity_tab_contacts, container, false);


        context=inflater.getContext();
        ListView listView=(ListView) rootView.findViewById(R.id.list);

        RequestParams params = new RequestParams();
        params.put("type", "get_users");

        clientGetUsers.post(address, params, new AsyncHttpResponseHandler() {
            @Override
            public void onSuccess(String response) {
                try {
                    JSONArray jsonArray = new JSONArray(response);
                    for (int i = 0; i < jsonArray.length(); i++) {
                        phoneNumberArray.add(jsonArray.getJSONObject(i).getString("mobile"));
                    }
                } catch (JSONException e) {
                    e.printStackTrace();
                }

                matchPhoneNumbersWithAppUsers();
                arrayAdapter.notifyDataSetChanged();

                Log.w("response", response);
            }
        });

        arrayAdapter = new InteractiveArrayAdapter(activity, R.layout.rowbuttonlayout,contacts);
        setListAdapter(arrayAdapter);

        return rootView;
    }

onListItemClick

@Override
    public void onListItemClick(ListView l, View v, int position, long id) {
       super.onListItemClick(l, v, position, id);
       Log.w("click2", "virker " + position);

        CheckBox checkbox = (CheckBox) v.findViewById(R.id.checkbox_users); 
        if (checkbox.isChecked() == false) {
            checkbox.setChecked(true); 
        } else {
            checkbox.setChecked(false);
            v.setBackgroundColor(Color.TRANSPARENT);
        }
    }