当底部导航被点击时颤振弹出到根目录

Question

我想要实现功能，当用户点击BottomNavigationBarItem 弹出到根目录时，如果当前页面索引等于像普通 iOS 应用程序那样点击的索引。 我尝试如下。我在MaterialApp 和HomeScreen 中设置所有根页面路由，如果currentIndex 等于index，popUntil 到根页面。 但是错误说

flutter：处理手势时引发以下 StateError：
颤振：坏状态：未来已经完成

我怎样才能做到这一点？

代码：

// MyApp class 
    class MyApp extends StatelessWidget {
      @override
      Widget build(BuildContext context) {
        return new MaterialApp(
          routes: <String, WidgetBuilder>{
            '/Page1': (BuildContext context) => new Page1(),
            '/Page2': (BuildContext context) => new Page2(),
            '/Page3': (BuildContext context) => new Page3(),
            '/Page4': (BuildContext context) => new Page4(),
          },
          title: 'Flutter Example',
          home: new HomeScreen(),
        );
      }
    }


  // MyHome class
    class HomeScreen extends StatefulWidget {
      @override
      _HomeScreenState createState() => new _HomeScreenState();
    }

    class _HomeScreenState extends State<HomeScreen> {
      final List<StatelessWidget> pages = [
        new Page1(),
        new Page2(),
        new Page3(),
        new Page3(),
      ];
      final List<String> routes = [
        '/Page1',
        '/Page2',
        '/Page3',
        '/Page4',
      ];

      int currentIndex = 0;

      @override
      Widget build(BuildContext context) {
        return new WillPopScope(
          onWillPop: () => new Future<bool>.value(true),
          child: new CupertinoTabScaffold(
            tabBar: new CupertinoTabBar(
              onTap: (index) {
                if (index == currentIndex) {
                  Navigator.popUntil(context, ModalRoute.withName(routes[index]));
                }
                currentIndex = index;
              },
              backgroundColor: Colors.white,
              activeColor: Colors.blue,
              inactiveColor: Colors.grey,
              items: const <BottomNavigationBarItem>[
                BottomNavigationBarItem(
                  icon: Icon(Icons.looks_one),
                  title: Text('Page1'),
                ),
                BottomNavigationBarItem(
                  icon: Icon(Icons.looks_two),
                  title: Text('Page2'),
                ),
                BottomNavigationBarItem(
                  icon: Icon(Icons.looks_3),
                  title: Text('Page3'),
                ),
                BottomNavigationBarItem(
                  icon: Icon(Icons.looks_4),
                  title: Text('Page4'),
                ),
              ],
            ),
            tabBuilder: (BuildContext context, int index) {
              return new DefaultTextStyle(
                style: const TextStyle(
                  fontFamily: '.SF UI Text',
                  fontSize: 17.0,
                  color: CupertinoColors.black,
                ),
                child: new CupertinoTabView(
                  builder: (BuildContext context) {
                    return pages[index];
                  },
                ),
              );
            },
          ),
        );
      }
    }

Answer 1

自从 Flutter v1.1.1 将 navigatorKey 添加到 CupertinoTabView 以来，现在可以实现此行为 (https://github.com/flutter/flutter/commit/65df90d8b5e1145e1022c365bb0465aa8c30dcdf)

首先，您必须为每个选项卡声明一个GlobalKey<NavigatorState>，然后将其传递给相应的CupertinoTabView 构造函数。 然后，在CupertinoTabBar 的onTap 方法中，如果用户在同一选项卡中点击，您可以使用firstTabNavKey.currentState.popUntil((r) => r.isFirst) 弹出到根目录。

下面的完整示例：

void main() {
  runApp(MyApp());
}
class MyApp extends StatefulWidget {
  @override
  _MyAppState createState() => _MyAppState();
}
class _MyAppState extends State<MyApp> {
  int currentIndex = 0;

  final GlobalKey<NavigatorState> firstTabNavKey = GlobalKey<NavigatorState>();
  final GlobalKey<NavigatorState> secondTabNavKey = GlobalKey<NavigatorState>();
  final GlobalKey<NavigatorState> thirdTabNavKey = GlobalKey<NavigatorState>();

  @override
  Widget build(BuildContext context) {
    return MaterialApp(
      title: 'My App',
      home: CupertinoTabScaffold(
        tabBuilder: (BuildContext context, int index) {
          switch (index) {
            case 0:
              return CupertinoTabView(
                navigatorKey: firstTabNavKey,
                builder: (BuildContext context) => FirstTab(),
              );
              break;
            case 1:
              return CupertinoTabView(
                navigatorKey: secondTabNavKey,
                builder: (BuildContext context) => SecondTab(),
              );
              break;
            case 2:
              return CupertinoTabView(
                navigatorKey: thirdTabNavKey,
                builder: (BuildContext context) => ThirdTab(),
              );
              break;
          }
          return null;
        },
        tabBar: CupertinoTabBar(
          items: <BottomNavigationBarItem> [
            BottomNavigationBarItem(
              title: Text('Tab 1'),
            ),
            BottomNavigationBarItem(
              title: Text('Tab 2'),
            ),
            BottomNavigationBarItem(
              title: Text('Tab 3'),
            ),
          ],
          onTap: (index){
            // back home only if not switching tab
            if(currentIndex == index) {
              switch (index) {
                case 0:
                  firstTabNavKey.currentState.popUntil((r) => r.isFirst);
                  break;
                case 1:
                  secondTabNavKey.currentState.popUntil((r) => r.isFirst);
                  break;
                case 2:
                  thirdTabNavKey.currentState.popUntil((r) => r.isFirst);
                  break;
              }
            }
            currentIndex = index;
          },
        ),
      ),
    );
  }
}

Answer 2

您可以使用pushAndRemoveUntil 达到相同的效果（如果我正确理解您的需求）。

final PageRouteBuilder _homeRoute = new PageRouteBuilder(
  pageBuilder: (BuildContext context, _, __) {
    return HomeScreen();
  }, 
);

void _goHome() {
  Navigator.pushAndRemoveUntil(context, _homeRoute, (Route<dynamic> r) => false);
}

这具有能够利用 PageRouteBuilder 的其他属性的额外好处。

Answer 3

这个简单的代码对我来说很有效

onPressed: () {
     Navigator.of(context).pushNamedAndRemoveUntil('/', ModalRoute.withName('/'));
            }

Answer 4

以下工作，在我的设置中，所有路线都被命名：

Navigator.of(context)
         .pushNamedAndRemoveUntil('homeScreen', (route) => false);

感谢@blaneyneil 和@Juanma Menendez 的回答