在不包含Scaffold的上下文中调用Scaffold.of() [英] Scaffold.of() called with a context that does not contain a Scaffold

问题描述

如您所见，我的按钮在Scaffold的体内.但是我得到了这个例外:

As you can see my button is inside Scaffold's body. But I get this exception:

Scaffold.of().

import 'package:flutter/material.dart';

void main() => runApp(MyApp());

class MyApp extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return MaterialApp(
      title: 'Flutter Demo',
      theme: ThemeData(
        primarySwatch: Colors.blue,
      ),
      home: HomePage(),
    );
  }
}

class HomePage extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text('SnackBar Playground'),
      ),
      body: Center(
        child: RaisedButton(
          color: Colors.pink,
          textColor: Colors.white,
          onPressed: _displaySnackBar(context),
          child: Text('Display SnackBar'),
        ),
      ),
    );
  }
}

_displaySnackBar(BuildContext context) {
  final snackBar = SnackBar(content: Text('Are you talkin\' to me?'));
  Scaffold.of(context).showSnackBar(snackBar);
}

我找到了解决该问题的另一种方法.如果我们为Scaffold提供一个名为GlobalKey的键，则可以如下所示显示SnackBar，而无需使用Builder小部件来包装我们的主体.返回Scaffold的小部件应该是有状态的小部件.:

I found an another solution to this problem. If we give Scaffold a key which is GlobalKey, we can display the SnackBar as following without need to wrap our body with Builder widget. The widget which return Scaffold should be Stateful widget though.:

 _scaffoldKey.currentState.showSnackBar(snackbar); 

推荐答案

发生此异常是因为您正在使用实例化Scaffold的窗口小部件的context.不是Scaffold子级的context.

This exception happens because you are using the context of the widget that instantiated Scaffold. Not the context of a child of Scaffold.

您可以通过使用其他上下文来解决此问题:

You can solve this by just using a different context :

Scaffold(
    appBar: AppBar(
        title: Text('SnackBar Playground'),
    ),
    body: Builder(
        builder: (context) => 
            Center(
            child: RaisedButton(
            color: Colors.pink,
            textColor: Colors.white,
            onPressed: () => _displaySnackBar(context),
            child: Text('Display SnackBar'),
            ),
        ),
    ),
);

请注意，虽然我们在这里使用Builder，但这并不是获取其他BuildContext的唯一方法.

Note that while we're using Builder here, this is not the only way to obtain a different BuildContext.

还可以将子树提取到其他Widget中(通常使用extract widget重构)

It is also possible to extract the subtree into a different Widget (usually using extract widget refactor)

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