Changing Widget Layout Programmatically(以编程方式更改小部件布局)
问题描述
假设我有两个小部件布局:Layout1 和 Layout2.小部件的默认值为 Layout1,但我允许用户选择他们希望小部件成为哪种布局.因此,如果用户更改为Layout2,如何以编程方式将布局更改为 Layout2?
Let's say that I have two layouts for a widget: Layout1 and Layout2. The default for the widget is Layout1, but I allow the user to choose which layout they want the widget to be. So if the user changes to Layout2, how do I programmatically change the layout to Layout2?
小部件没有像活动那样的 setContentView 方法.
There isn't a setContentView method for widgets like there is for Activities.
谢谢
推荐答案
在构建 remoteView 时必须选择布局.在我的小部件代码中:
You have to choose the layout when you're building your remoteView. In my widget code:
public static RemoteViews buildUpdate(Context context, String action) {
RemoteViews updateViews;
SharedPreferences prefs = context.getSharedPreferences(PREFS_NAME, 0);
String typeface = prefs.getString("typeface", "sans");
int layoutId = R.layout.widget_sans;
if ("monospace".equals(typeface)){
layoutId = R.layout.widget_mono;
} else if ("serif".equals(typeface)){
layoutId = R.layout.widget_serif;
}
updateViews = new RemoteViews(context.getPackageName(),
layoutId);
//actually do things here
//then finally, return our remoteView
AppWidgetManager.getInstance(context).updateAppWidget(
new ComponentName(context, FuzzyWidget.class), updateViews);
}
这篇关于以编程方式更改小部件布局的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持编程学习网!
本文标题为:以编程方式更改小部件布局
基础教程推荐
- iPhone - 获取给定地点/时区的当前日期和时间并将其与同一地点的另一个日期/时间进行比较的正确方法 2022-01-01
- AdMob 广告未在模拟器中显示 2022-01-01
- libGDX 从精灵或纹理中获取像素颜色 2022-01-01
- 通过重定向链接在 Google Play 中打开应用 2022-01-01
- 如何从 logcat 中删除旧数据? 2022-01-01
- iOS4 创建后台定时器 2022-01-01
- NSString intValue 不能用于检索电话号码 2022-01-01
- Cocos2d iPhone 非矩形精灵触摸检测 2022-01-01
- Android:getLastKnownLocation(LocationManager.NETWORK_PROVIDER 2022-01-01
- navigator.geolocation.getCurrentPosition 在 Android 浏览器上 2022-01-01
