背景：
showCupertinoModalPopup中使用Navigator.of(context).pushNamedAndRemoveUntil(RouteName.login, (route) => false);
报错：Looking up a deactivated widget’s ancestor is unsafe

解决方案：先将CupertinoModal关闭，再使用不同的的context去跳转；
注意： _context和context是不同的

Navigator.pop(_context); Future.delayed(Duration(milliseconds: 300), () { Navigator.of(context) .pushNamedAndRemoveUntil(RouteName.login, (Route<dynamic> route) => false); });

void _showAlertDialog(BuildContext context) {showCupertinoModalPopup<void>(context: context,builder: (BuildContext _context) => CupertinoAlertDialog(title: const Text('退出后不删除任何历史数据， 请确认是否退出'),// content: const Text('Proceed with destructive action?'),actions: <CupertinoDialogAction>[CupertinoDialogAction(/// This parameter indicates this action is the default,/// and turns the action's text to bold text.isDefaultAction: true,onPressed: () {Navigator.pop(_context);},child: const Text('取消'),),CupertinoDialogAction(/// This parameter indicates the action would perform/// a destructive action such as deletion, and turns/// the action's text color to red.isDestructiveAction: true,onPressed: () async {// final isLogout = await loginOut();// await UserUtil.cleanToKen();Navigator.pop(_context);Future.delayed(Duration(milliseconds: 300), () {Navigator.of(context).pushNamedAndRemoveUntil(RouteName.login, (Route<dynamic> route) => false);});},child: const Text('退出登录'),),],),);}