如何在 flutter 中从 class（小部件外部）打开屏幕

Question

所以，我的 lab_widget 有一个普通按钮，它从另一个名为 not_plugin 的文件中调用一个函数。我希望每次单击按钮时，not_pugin 中的函数都会打开一个确定的屏幕。这就是我在 not_plugin:

中的内容


Future<dynamic> onClick(String payload) async {
     //Do other things

     Navigator.push(context, MaterialPageRoute(builder: (context) => LabMdScreen()));
}

但它抛出以下错误：

46:24: Error: The getter 'context' isn't defined for the class 'NotPlugin'.

'NotPlugin' is from 'not_plugin.dart'. Try correcting the name to the name of an existing getter, or defining a getter or field named 'context'. Navigator.push(context, MaterialPageRoute(builder: (context) => MyScreen()));

那我该如何解决呢？

更新

传递 BuildContext 后，在另一个函数中抛出此错误：

Future<void> initLocalNotification() async {
    androidInitializationSettings =
        AndroidInitializationSettings('@mipmap/ic_launcher');//here
    iosInitializationSettings = IOSInitializationSettings(
        onDidReceiveLocalNotification: onDidReceiveLocalNotification);
    initializationSettings = InitializationSettings(
        android: androidInitializationSettings, iOS: iosInitializationSettings);

    await flutterLocalNotificationsPlugin.initialize(initializationSettings,
        onSelectNotification: onClick());
  }

最后一行(onSelectNotification: onClick());)

error: The argument type 'Future' can't be assigned to the parameter type 'Future Function(String)'

Answer 1

您需要将 BuildContext 传递给您的函数：

void onClick(String payload, BuildContext context) {
     //Do other things

     Navigator.push(context, MaterialPageRoute(builder: (context) => LabMdScreen()));
}

如何在 flutter 中从 class（小部件外部）打开屏幕

How to open a screen from a class (outside a widget) in flutter

widget

dart

flutter