小编Oli*_* U.的帖子

如何在弹出菜单中自定义menuItem？我需要第一个menuitem的开关.这是我到目前为止所得到的:

menu.xml文件

<menu xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:app="http://schemas.android.com/apk/res-auto"
    xmlns:tools="http://schemas.android.com/tools"
    tools:context="android.oli.com.fitnessapp.activities.LiveSelectActivity">

    <item
        android:icon="@drawable/ic_access_alarm_white_24dp"
        android:orderInCategory="100"
        app:showAsAction="always"
        android:visible="true"
        android:title="@string/action_settings"
        android:onClick="showPopup"/>

</menu>

menu_popup.xml

<?xml version="1.0" encoding="utf-8"?>
<menu xmlns:android="http://schemas.android.com/apk/res/android">

    <item
        android:id="@+id/one"
        android:title="One"/>

    <item
        android:id="@+id/setTime"
        android:title="Two"
        android:onClick="showTimePickerDialog"/>


</menu>

活动代码段

public void showPopup(MenuItem menuItem){
        View view = findViewById(R.id.action_alarm);
        PopupMenu popup = new PopupMenu(this, view);
        MenuInflater inflater = popup.getMenuInflater();
        inflater.inflate(R.menu.menu_popup, popup.getMenu());
        popup.show();
    }