Flutter onPressed link instagram profile url with http

Question

我如何使用我的api instagram链接字段，当我点击它，它会打开instagram页面或任何url?

现在我的小部件变量给出这个错误
参数类型'Null Function(dynamic)'不能分配给参数类型'void Function()'。

IconButton(
icon: FaIcon(FontAwesomeIcons.facebook),
onPressed: (widget.company.social_instagram.toString()) { print("Pressed"); }
),

Answer 1

import 'package:url_launcher/url_launcher.dart';
IconButton(
icon: FaIcon(FontAwesomeIcons.facebook),
onPressed: () { 
print(Uri.parse(widget.company.social_instagram.toString()));
launchUrl(Uri.parse(widget.company.social_instagram.toString()));
print("Pressed"); 
}
),