如何创建简单的通知,点击后会关闭应用程序(多任务处理)?谢谢。 (Java,Android)
答案 0 :(得分:0)
首先我们应该创建一个调用活动的通知:
Intent intent = new Intent(this, YourClass.class);
intent.putExtra("NotiClick",true);
PendingIntent pIntent = PendingIntent.getActivity(this, 0, intent, PendingIntent.FLAG_UPDATE_CURRENT);
if (android.os.Build.VERSION.SDK_INT >= android.os.Build.VERSION_CODES.JELLY_BEAN) {
Notification Noti;
Noti = new Notification.Builder(this)
.setContentTitle("YourTitle")
.setContentText("YourDescription")
.setSmallIcon(R.mipmap.ic_launcher)
.setContentIntent(pIntent)
.setAutoCancel(true).build();
NotificationManager notificationManager =
(NotificationManager) getSystemService(NOTIFICATION_SERVICE);
notificationManager.notify(0, Noti);
}
然后在你班级的 onCreate()中执行以下操作:
Bundle extras = getIntent().getExtras();
if(extras == null)
{
//not being clicked on
}
else if (extras.getBoolean("NotiClick"))
{
//being clicked
}
对于完全关闭的应用程序,您可以终止进程:
android.os.Process.killProcess(android.os.Process.myPid());