flutter - 使用带有 RaisedButton onPressed() 的异步函数将对象传递给另一个路由

Question

我希望能够在将对象作为参数传递时按下按钮导航到另一个屏幕。该对象是使用不同 dart 文件中的 getPlayer() 函数创建的；因此，异步功能。每当我运行代码时，我都会收到错误消息：

Error: This expression has type 'void' and can't be used. onPressed: loadPlayerCard('nebula'),

这是代码：

    class _HomeState extends State<Home> {

  List<Player> players;

   void loadPlayerCard (String playerName) async {
    Player player = await getPlayer(playerName);
    players.add(player);

    Navigator.pushNamed(context, '/playerCard', arguments: player);
  }

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      backgroundColor: Colors.grey[900],
      appBar: AppBar(
        title: Text('Smash Tracker'),
        centerTitle: true,
        backgroundColor: Colors.grey[850],
        elevation: 0,
      ),
      body: Center(
        child: RaisedButton.icon(
            onPressed: loadPlayerCard('nebula'), //This is where the error message points to 
            icon: Icon(Icons.touch_app),
            label: Text('Nebula'),
        ),
      ),
    );
  }
}

任何帮助表示赞赏！

Answer 1

用括号括起来如下

RaisedButton.icon(
  onPressed:() async { loadPlayerCard('nebula'); }, // Fix for the issue
  icon: Icon(Icons.touch_app),
  label: Text('Nebula'),
),

Answer 2

您可以像这样简单地使用内联箭头函数 词法闭包。

RaisedButton.icon(
    onPressed: () => loadPlayerCard('nebula'), // Solved 
    icon: Icon(Icons.touch_app),
    label: Text('Nebula'),
),