我正在尝试使用测试值实现 ListView。每个列表元素应该显示 2 个字符串。但我得到一个 NullPointerException。
我有一个 ListActivity，调用一个适配器。如果我评论这个活动的最后一行，我没有任何错误，所以我猜我的适配器有问题。但是我在网上搜索了解决方案，不幸的是，由于 Java（和 OOP）技能非常低，我无法解决我的问题。你能告诉我我在哪里犯了错误吗？
public class ListActivity extends Activity {
//private final String TAG = ListActivity.class.getSimpleName();
private ListView listMessView;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_list);
List<Message> listMessages = new ArrayList<Message>();
listMessages.add(new Message("London", "aze"));
listMessages.add(new Message("Rome", "azeeaze"));
listMessages.add(new Message("Paris", "qsdqsdqsd"));
listMessView = (ListView) findViewById(R.id.message_list);
listMessView.setAdapter(new ListAdaptater(this, R.layout.list_row, listMessages));
}
}
public class ListAdaptater extends ArrayAdapter<Message>{
private int resource;
private LayoutInflater inflater;
private Context context;
public ListAdaptater ( Context ctx, int resourceId, List<Message> objects) {
super( ctx, resourceId, objects );
resource = resourceId;
inflater = LayoutInflater.from( context );
context=ctx;
}
@Override
public View getView ( int position, View convertView, ViewGroup parent ) {
convertView = ( RelativeLayout ) inflater.inflate( resource, null );
Message message = (Message) getItem( position );
TextView txtName = (TextView) convertView.findViewById(R.id.user_name);
txtName.setText(message.getUserName());
TextView txtMsg = (TextView) convertView.findViewById(R.id.message);
txtMsg.setText(message.getMessage());
return convertView;
}
}