[英]Return of invalid type but don't know why? Flutter/Dart
嘿,我对 flutter 完全陌生,最近我一直在开发一款通过 BLE 从 ESP32 接收数据的移动应用程序,但我遇到的问题是,如果我想要求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);
}
它给了我错误警告:
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.
但是以我目前的知识,我不知道如何解决我的问题。 如果有人可以帮助我,我将非常感激:) 并对任何语法错误感到抱歉
未来<T?> showDialog
Future<T?> showDialog<T>(
{required BuildContext context,
required WidgetBuilder builder,
bool barrierDismissible = true,
无需return
该错误表明showDialog
的return
类型不是 bool。 相反,您可以只用 await 替换,然后返回布尔值。
以下是您可以发布的代码。
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;
}
我对代码进行了一些更改,以便在您不想弹出时返回 false,如果您想弹出则返回 true。 您可以根据您的要求更改代码。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.