Flutter：为什么没有捕获异常？

Question

我想在发生错误时显示小吃栏。异常正在运行，但未在 main.dart 中捕获。你能看出问题是什么吗？我是 Flutter 新手，所以可能不太好。谢谢。

主程序.dart

Future<void> _signInWithPhoneNumber() async {
    try {
      postPhoneInfo(
          tokenValue,
          _packageInfo.appName,
          _packageInfo.version,
          Theme.of(context).platform.toString().substring(15),
          "phone number");
      print("debug : try");
    } catch (e) {
      print("debug : catch");
      // print(e);
      ScaffoldMessenger.of(context).showSnackBar(SnackBar(content: Text("${e.toString()}")));
    }
  }

这是 postPhoneInfo()

Future<phoneInfo> postPhoneInfo(String token, String appName, String appVersion,
    String platform, String phone) async {
  var queryParameters = {
    'param1': token,
    'param2': appName,
    'param3': appVersion,
    'param4': platform,
    'param5': phone
  };
  var uri =
      Uri.http('~~', '~~', queryParameters);
  final response = await http.post(uri, headers: <String, String>{
    'Content-Type': 'application/json; charset=UTF-8',
  });

  if (response.statusCode == 200) {
    final parsedJson = jsonDecode(response.body);

    if (parsedJson["Res"] == "1") {
      print("user is registered");
      return phoneInfo.fromJson(jsonDecode(response.body));
    } else {
      print("user is not registered");
      throw Exception("user is not registered");
    }
  } else {
    print(response.statusCode);
    throw Exception('not connected');
  }
}

Answer 1

您需要在 中调用await该函数。否则，将在上下文中引发异常。postPhoneInfo_signInWithPhoneNumberasync

你必须用runZoned它来捕捉那些。文档： https: //api.flutter.dev/flutter/dart-async/runZoned.html