从脚手架内部显示小吃栏在浮动操作按钮上按下回调

Question

我正在尝试调用

Scaffold.of(context).showSnackBar(SnackBar(
content: Text("Snack text"),
));

脚手架floatingActionButton的onPressed内。

我收到此错误

I/flutter (18613): Scaffold.of() called with a context that does not contain a Scaffold.
I/flutter (18613): No Scaffold ancestor could be found starting from the context that was passed to 
....

当你在身体内部调用Scaffold.of(context)时，它指向了一个解决方案。

https://docs.flutter.io/flutter/material/Scaffold/of.html

但是，如果您在内部调用它，则相同的解决方案不起作用FloatingActionButton

Answer 1

更新：第二个解决方案比这个解决方案更好。

您应该将浮动的ActionButton小部件放在Builder Widget中。 以下代码应该有效：

@override
Widget build(BuildContext context) {
return new Scaffold(
floatingActionButton: new Builder(builder: (BuildContext context) {
return new FloatingActionButton(onPressed: () {
Scaffold
.of(context)
.showSnackBar(new SnackBar(content: new Text('Hello!')));
});
}),
body: new Container(
padding: new EdgeInsets.all(32.0),
child: new Column(
children: <Widget>[
new MySwitch(
value: _switchValue,
onChanged: (bool value) {
if (value != _switchValue) {
setState(() {
_switchValue = value;
});
}
},
)
],
),
),
);

Answer 2

添加一个基架状态的全局键并使用它 显示小吃店如下，

GlobalKey<ScaffoldState> scaffoldState;
Scaffold {
key: scaffoldState,
....

scaffoldState.currentState.showSnackBar(new SnackBar(content: new Text('Hello!')));

Answer 3

这更简单。试过了。将浮动操作按钮创建为单独的无状态控件。从基架调用此无状态小部件。

class Abc extends StatelessWidget
{
Widget build(BuildContext context)
{
return Scaffold(
appBar:AppBar(title:Text("Long List View")),
body:SomeOtherWidget(),
floatingActionButton:MyFAB()  
);
}
}
class MyFAB extends StatelessWidget
{
Widget build(BuildContext context)
{
return FloatingActionButton(
onPressed:(){
showSnackBarHandler(context);
},
child:Icon(Icons.add),
tooltip:"Press to Add More"
);
}
}
void showSnackBarHandler(BuildContext context){
var snackBar = SnackBar(
content:Text("Hello")
);
Scaffold.of(context).showSnackBar(snackBar);
}