在Flutter中，如果你的页面由 A->B->C->D->E ，然后在E页面需要返回到B页面，其余的页面依次返回，你可以在D页面跳转到E页面的时候这样写：
Get.offUntil(
GetPageRoute<EPage>(
settings: RouteSettings(
name: '/EPage',
arguments: arguments,
),
page: () => EPage()),
(route) =>
(route as GetPageRoute).routeName ==
'/BPage');
查看Get.offUntil的源码，它是这样写的：
/// **Navigation.pushAndRemoveUntil()** shortcut.<br><br>
///
/// Push the given `page`, and then pop several pages in the stack until
/// [predicate] returns true
///
/// [id] is for when you are using nested navigation,
/// as explained in documentation
///
/// Obs: unlike other get methods, this one you need to send a function
/// that returns the widget to the page argument, like this:
/// Get.offUntil(GetPageRoute(page: () => HomePage()), predicate)
///
/// [predicate] can be used like this:
/// `Get.offUntil(page, (route) => (route as GetPageRoute).routeName == '/home')`
/// to pop routes in stack until home,
/// or also like this:
/// `Get.until((route) => !Get.isDialogOpen())`, to make sure the dialog
/// is closed
Future<T?>? offUntil<T>(Route<T> page, RoutePredicate predicate, {int? id}) {
// if (key.currentState.mounted) // add this if appear problems on future with route navigate
// when widget don't mounted
return global(id).currentState?.pushAndRemoveUntil<T>(page, predicate);
}
由于我本地没有写路由，所以我用GetPageRoute生成了它需要的route。
(route) => (route as GetPageRoute).routeName == '/BPage'); 这里其实是在判断，如果routeName == /BPage ,那么返回到此为止 ，否则会一直往前面的页面返回。
