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我想分配一个在点击通知时将触发的事件,而不显示活动(如此所述)。

我怎样才能做到这一点?

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1 回答 1

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您可以通过注册广播接收器来做到这一点。这样,您可以在单击通知时在接收器中运行代码,而无需显示任何 UI。

基本接收器代码:

public class Provider extends BroadcastReceiver {

    @Override
    public void onReceive(Context context, Intent intent) {
        super.onReceive(context, intent);
        //Your code here

    }
}

在清单中:

<receiver android:name="Provider" />

通知的示例代码:

public static void showNotification(Context context) {
    CharSequence title = "title";
    CharSequence message = "text";

    NotificationManager notificationManager = (NotificationManager) context
            .getSystemService(Context.NOTIFICATION_SERVICE);
    Notification notification = new Notification(<image>, title, System.currentTimeMillis());
    notification.flags = Notification.FLAG_ONGOING_EVENT;

    Intent notificationIntent = new Intent(context, Provider.class);
    PendingIntent pendingIntent = PendingIntent.getBroadcast(context, 0, notificationIntent, 0);

    notification.setLatestEventInfo(context, title, message, pendingIntent);
    notificationManager.notify(<notification id>, notification);
}
于 2013-01-19T16:40:47.713 回答