Flutter：为什么没有捕获到异常？

Question

我想在发生错误时显示一个 snackbar。异常有效但未在 main.dart 中捕获。你能看出问题是什么吗？我是 Flutter 的新手，所以我可能不太擅长。谢谢。

main.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

ScaffoldMessenger.of(context) 必须在脚手架小部件中调用。检查它来自的上下文。

Answer 2

您需要 await 在 _signInWithPhoneNumber 中调用 postPhoneInfo 函数。否则，将在 async 上下文中抛出异常。

您必须使用 runZoned 才能捕捉到它们。文档：https://api.flutter.dev/flutter/dart-async/runZoned.html

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