Return 类型无效但不知道为什么？ Flutter/Dart

Question

嘿，我是 flutter 的新手，最近我一直在开发一个移动应用程序，它通过 BLE 从 ESP32 接收数据，但我遇到了一个问题，如果我想让用户像这样断开与设备的连接:

Future<bool> _onWillPop() {
    return showDialog(
        context: context,
        builder: (context) =>
            new AlertDialog(
              title: Text('Are you sure?'),
              content: Text('Do you want to disconnect device 
              and go back?'),

             actions: <Widget>[
                new ElevatedButton(
                    onPressed: () => 
                      Navigator.of(context).pop(false),
                    child: new Text('No')),
                new ElevatedButton(
                    onPressed: () {
                      disconnectFromDevice();
                      Navigator.of(context).pop(true);
                    },
                    child: new Text('Yes')),
              ],
            ) ??
            false);
  }

它给我错误警告：

A value of type 'Future<dynamic>' can't be returned from the method '_onWillPop' because it has a return type of 'Future<bool>'.

The return type 'Object' isn't a 'Widget', as required by the closure's context.

但以我目前的知识，我不知道如何解决我的问题。如果有人能帮助我，我将非常感激 :) 对于任何语法错误，我深表歉意

Answer 1

Future<T?> showDialog

Future<T?> showDialog<T>(

    {required BuildContext context,
    required WidgetBuilder builder,
    bool barrierDismissible = true,

不需要 return

Answer 2

错误说 showDialog 的 return 类型不是 bool。相反，您可以只替换为 await 然后 return bool.

下面是您可以提供的代码。

Future<bool> _onWillPop(BuildContext context) async {
    bool shouldPop = false;
    await showDialog(
        context: context,
        builder: (context) =>
            AlertDialog(
              title: const Text('Are you sure?'),
              content: const Text('Do you want to disconnect device and go back?'),
             actions: <Widget>[
                ElevatedButton(
                    onPressed: () {
                      // shouldPop is already false
                    },
                    child: const Text('No')),
                ElevatedButton(
                    onPressed: () async {
                      await disconnectFromDevice();
                      Navigator.of(context).pop();
                      shouldPop = true;
                    },
                    child: const Text('Yes')),
              ],
            ));
    return shouldPop;
  }

我稍微更改了代码，如果您不想弹出，则 return 为假，如果您想弹出，则 return 为真。您可以根据需要更改代码。

Return 类型无效但不知道为什么？ Flutter/Dart

Return of invalid type but don't know why? Flutter/Dart

dart

bluetooth-lowenergy

flutter