android - Flutter中如何将onTap函数传递给ListTile？

Question

我的抽屉内容：

drawer: Drawer(
    elevation: 0.0,
    child: Container(
      color: kBackgroundColor,
      child: ListView(
        padding: EdgeInsets.zero,
        children: <Widget>[
          DrawerHeader(
            decoration: BoxDecoration(
              color: Colors.grey,
              image: DecorationImage(
                alignment: Alignment.topRight,
                image: AssetImage('images/banner.png'),
                fit: BoxFit.cover,
              ),
            ),
            child: Text('App Test'),
          ),
          ReusableOptionTile(
            textData: 'Home',
            icon: Icons.home,
            onTouch: () {
              print('Home Clicked');
              Navigator.pop(context);
            },
          ),
        ]
      ),
    ),
),

提取的列表磁贴内容：

    class ReusableOptionTile extends StatelessWidget {
 const ReusableOptionTile(
      {required this.icon, required this.textData, required this.onTouch});
  final IconData icon;
  final String textData;
  final Function onTouch;

  @override
  Widget build(BuildContext context) {
    return ListTile(
      title: Row(
        children: [
          Icon(icon),
          SizedBox(
            width: 5.0,
          ),
          Text('$textData'),
        ],
      ),
    onTap: onTouch,   // ERROR POINT
    );
  }
}

我从 Drawer 中提取了 List Tile 以使代码更具可重用性，但现在无法为不同的 tile 传递 on tap 函数。在错误点显示错误“无法将参数类型'Function'分配给参数类型'void Function()？'。”

如何解决这个问题？

标签： androidflutterdartvisual-studio-codeflutter-layout

Answer 1

onTouch应提取为 void 函数

void onTouch(){
 print('Home Clicked');
 Navigator.pop(context);
}

然后

ReusableOptionTile(
            textData: 'Home',
            icon: Icons.home,
            onTouch: onTouch,
          ),

并打电话给

ListTile(
      title: Row(
        children: [
          Icon(icon),
          SizedBox(
            width: 5.0,
          ),
          Text('$textData'),
        ],
      ),
    onTap: onTouch(),
    );