为什么在构造函数参数更改时不会重置StatefulWidget的子状态?

问题描述 投票:0回答:1

我有一个自定义窗口小部件类:

class FancyButton extends StatefulWidget
{
  final Function() onTap;
  final String text;
  final IconData icon;
  bool enabled;

  FancyButton({this.onTap, this.text, this.icon, this.enabled = true})

  @override
  _FancyButtonState createState()
  {
    return _FancyButtonState(onTap, text, icon, enabled);
  }
}

class _FancyButtonState extends State<FancyButton> with SingleTickerProviderStateMixin
{
  Function() onTap;
  String text;
  IconData icon;
  bool enabled;
  _FancyButtonState(this.onTap, this.text, this.icon, this.enabled);

  @override
  Widget build(BuildContext context) {
    if(!enabled)
      {
        print("build disabled");
      }
    return GestureDetector(
      onTap: onTap,
      child: Container(
          color: enabled ? Colors.red : Colors.green,
          height: 20
      ),
    );
  }
}

我正在这样使用它:

Widget continueButton = new StreamBuilder(
          stream: viewModel.canFinishProfile.asBroadcastStream(),
          initialData: viewModel.canFinishProfile.value,
          builder: (builder, snapshot)
          {
            print("can finish profile " + viewModel.canFinishProfile.value.toString());
            return FancyButton(
              text: multilang.get("auth_continue"),
              enabled: viewModel.canFinishProfile.value,
              onTap: ()
              {
                viewModel.onClickedContinue.add(null);
              },
            );
          },
        )

因此,我希望小部件在canFinishProfile值更改后即可更新。问题在于,一旦canFinishProfile值更改,FancyButton的状态就不会更新,因此始终会绘制旧的窗口小部件视图。请注意,如果我将FancyButton更改为StatelessWidget,并将其构建代码放入StatelessWidget内,则它将重绘一次,然后可以更改canFinishProfile值。但是,我需要它成为StatefulWidget才能提供更复杂的功能。

flutter flutter-layout statefulwidget statelesswidget flutter-state
1个回答
0
投票

如果您使用State属性访问状态,则StatefulWidget类将能够侦听widget类中的状态更改。

您需要更改:

enabled ? Colors.red : Colors.green

widget.enabled ? Colors.red : Colors.green

希望有帮助。

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