Android 单击弹出菜单的侦听器

如何在弹出菜单项“单击”上启动新的意图。以下是我尝试过的代码：

public class Profile extends Fragment {
    public Profile() { }

    View vi;
    public static final String ARG_SECTION_NUMBER = "section_number";

    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container,
            Bundle savedInstanceState) {
        vi = inflater.inflate(R.layout.profile, container, false);
        TextView iv = (TextView) vi.findViewById(R.id.btnprofile_settings);

        iv.setOnClickListener(new OnClickListener() {
            @Override
            public void onClick(View v) {
                PopupMenu homepopup = new PopupMenu(getActivity(), v);
                MenuInflater inflater = homepopup.getMenuInflater();
                inflater.inflate(R.menu.account_settings, homepopup.getMenu());

                homepopup.show();
            }
        });
        return vi;
    }
}

有什么想法吗

---

谢谢，

您必须使用

弹出菜单上的
设置菜单项ClickListener
，例如：

homepopup.setOnMenuItemClickListener(new OnMenuItemClickListener() {

        @Override
        public boolean onMenuItemClick(android.view.MenuItem item) {
            return true;
        }
    })