android - 如何使用异步返回字符串

Question

我正在努力从异步方法返回字符串值。

这是我的实用程序类:

class Utilities {

  Future<PackageInfo> getPackageInfo() async {
    PackageInfo packageInfo = await PackageInfo.fromPlatform();
    return packageInfo;
  }

  String getVersion() {
    String version;
    getPackageInfo().then((PackageInfo pi) {
      version = pi.version;
    });
    return version;
  }

}

这是我的UI，我在其中调用getVersion方法:

class abc extends StatelessWidget {

  Utilities _utilities = new Utilities();

  @override
  Widget build(BuildContext context) {
  ...
  Text(_utilities.getVersion()),
  ...

}

我总是在Text()小部件上收到空值错误。

我究竟做错了什么？

Answer 1

您将返回正常的String，而应将其包装在Future中:

Future<String> getVersion() async => getPackageInfo().then((PackageInfo pi) => pi.version);

然后 Text小部件变为:

FutureBuilder<String>(
                    future: getVersion(),
                    builder: (BuildContext context, AsyncSnapshot<String> snapshot) {
                     if(snapshot.connectionState==ConnectionState.waiting)
                       return CircularProgressIndicator();
                     else if(snapshot.connectionState==ConnectionState.done && snapshot.hasData)
                       return Text(snapshot.data);
                     //handle other connection statuses
                    }
                ),