Question

我在ListView上有一个上下文菜单。

我想为登录用户添加特殊情况上下文列表项。

假设有一个评论列表。但是，仅对于登录用户的评论，有一个特殊的上下文列表项称为“编辑”（显然，您不希望其他用户能够编辑自己的评论。

这是我的类扩展应用程序，我通常会为登录用户办理登机手续：

public class MyApp extends Application {

    public static boolean isUserLoggedIn = false;
    public static String username = null;
    public static SharedPreferences logInState;
    public static int ratescreen = 0;

    public static boolean userLogin() {

        return MyApp.isUserLoggedIn = true;
    }

    public static boolean userLogout() {

        return MyApp.isUserLoggedIn = false;
    }

    public static void setUser(String s) {

        MyApp.username = s;
    }

    @Override
    public void onCreate() {
        super.onCreate();
        String PREFS_NAME = "LoginState";
        logInState = getSharedPreferences(PREFS_NAME,
                MODE_PRIVATE);
    }

}

这是我的上下文菜单：

@Override
        public void onCreateContextMenu(ContextMenu menu, View v,
                ContextMenuInfo menuInfo) {

            super.onCreateContextMenu(menu, v, menuInfo);
            MenuInflater inflater = getActivity().getMenuInflater();
            inflater.inflate(R.menu.reviews_context, menu);
            menu.setHeaderTitle("Mark Comment as ...");
        }

        @Override
        public boolean onContextItemSelected(MenuItem item) {

            switch (item.getItemId()) {
            case R.id.helpful:

                new HelpfulTask().execute();

                Toast.makeText(getActivity(), "You have voted this up!",
                        Toast.LENGTH_SHORT).show();

                return true;
            case R.id.unhelpful:

                new UnHelpfulTask().execute();

                Toast.makeText(getActivity(), "You have voted this down!",
                        Toast.LENGTH_SHORT).show();

                return true;

            case R.id.spam:

                new SpamTask().execute();

                Toast.makeText(getActivity(),
                        "You have reported this as Spam or Offensive.",
                        Toast.LENGTH_SHORT).show();

                return true;

                    // Would like to add fourth option here but conditional if it is a comment from the currently logged in user.


            }
            return false;

        }

Answer 1

简单地添加Edit项取决于onCreateContextMenu

中的用户登录状态

    @Override
        public void onCreateContextMenu(ContextMenu menu, View v,
                ContextMenuInfo menuInfo) {
            // TODO Auto-generated method stub

            super.onCreateContextMenu(menu, v, menuInfo);
    // check if user logged-in so add Edit item to context menu.
            if (userLogin()) {
                menu.add(0, MENU_ITEM_EDIT, 0, R.string.menu_edit);
            }
//Add normall others menu items
            menu.add(0, MENU_ITEM_CALL, 0, R.string.menu_callContact);



        }

Answer 2

您可以将ContextMenuInfo转换为适用于ListView的AdapterView.AdapterContextMenuInfo，而不是更改其内容访问适配器（您自己的实现）。

AdapterView.AdapterContextMenuInfo contextMenuInfo = (AdapterView.AdapterContextMenuInfo)menuInfo;
        MyAdapter adapter = (MyAdapter)getListView().getAdapter();
adapter.remove(contextMenuInfo.position);

如何在Android ContextMenu中创建条件列表项

2 个答案: