Question

我正在制作BackBtn类，该类在此应用程序中的许多地方都使用。

设计是相同的，只是按下时的行为是不同的。

因此，我想在构造函数中传递该函数，该函数在按下按钮时使用。

但是它显示下面的错误

flutter: ══╡ EXCEPTION CAUGHT BY WIDGETS LIBRARY ╞═══════════════════════════════════════════════════════════
flutter: The following assertion was thrown building BackBtn(dirty):
flutter: setState() or markNeedsBuild() called during build.
flutter: This Overlay widget cannot be marked as needing to build because the framework is already in the
flutter: process of building widgets.  A widget can be marked as needing to be built during the build phase
flutter: only if one of its ancestors is currently building. This exception is allowed because the framework
flutter: builds parent widgets before children, which means a dirty descendant will always be built.
flutter: Otherwise, the framework might not visit this widget during this build phase.
flutter: The widget on which setState() or markNeedsBuild() was called was:
flutter:   Overlay-[LabeledGlobalKey<OverlayState>#320d3]
flutter: The widget which was currently being built when the offending call was made was:
flutter:   BackBtn

这些是我编写的代码。

class BackBtn extends StatelessWidget{

  final Function onPressed;  
  const BackBtn({ Key key ,this.onPressed}) : super(key: key);

  @override
  Widget build(BuildContext context) {
    return Container(
      alignment: Alignment(1.0,-1.0), 
      child: FlatButton(
        onPressed: onPressed(),      
        padding: EdgeInsets.all(50),
        child:  Image.asset('images/BackIcon.png')
      )
    );
  }
}

BackBtn(
    onPressed: () => Navigator.push(context,MaterialPageRoute(
    builder: (context) => MyHomePage())),
),

Answer 1

将onPressed()回调分配给FlatButton()后，您已经在执行该回调。

尝试像这样去除括号：

class BackBtn extends StatelessWidget{

  final Function onPressed;  
  const BackBtn({ Key key ,this.onPressed}) : super(key: key);

  @override
  Widget build(BuildContext context) {
    return Container(
      alignment: Alignment(1.0,-1.0), 
      child: FlatButton(
        onPressed: onPressed, // Removed the Braces ()      
        padding: EdgeInsets.all(50),
        child:  Image.asset('images/BackIcon.png')
      )
    );
  }
}

Tldr;

当离开括号时，它将在构建窗口小部件时立即执行onPressed处理程序。这意味着您将在构建过程中点击导航器和所有其他内容，而不仅仅是按下实际按钮时。那会导致您的错误。

在构建错误期间调用setState（）或markNeedsBuild（）

1 个答案: