如何在扑扑中捕捉异常？

Question

这是我的异常类。异常类已经由flutter的抽象异常类实现。我错过了什么吗？

class FetchDataException implements Exception {
 final _message;
 FetchDataException([this._message]);

String toString() {
if (_message == null) return "Exception";
  return "Exception: $_message";
 }
}


void loginUser(String email, String password) {
  _data
    .userLogin(email, password)
    .then((user) => _view.onLoginComplete(user))
    .catchError((onError) => {
       print('error caught');
       _view.onLoginError();
    });
}

Future < User > userLogin(email, password) async {
  Map body = {
    'username': email,
    'password': password
  };
  http.Response response = await http.post(apiUrl, body: body);
  final responseBody = json.decode(response.body);
  final statusCode = response.statusCode;
  if (statusCode != HTTP_200_OK || responseBody == null) {
    throw new FetchDataException(
      "An error occured : [Status Code : $statusCode]");
   }
  return new User.fromMap(responseBody);
}

当状态不是200时，CatchError不会捕获错误。捕获的Inshort错误不会被打印。

Answer 1

假设这是您的函数抛出异常：

Future<void> foo() async {
  throw Exception('FooException');
}

您可以使用try-catchblock 或catchErroron the，Future因为两者都执行相同的操作。

• 使用try-catch

  try {
    await foo();
  } on Exception catch (e) {
    print(e); // Only catches an exception of type `Exception`.
  } catch (e) {
    print(e); // Catches all types of `Exception` and `Error`.
  }
  

• 使用catchError

  await foo().catchError(print);
  

Answer 2

尝试

void loginUser(String email, String password) async {
  try {
    var user = await _data
      .userLogin(email, password);
    _view.onLoginComplete(user);
      });
  } on FetchDataException catch(e) {
    print('error caught: $e');
    _view.onLoginError();
  }
}

catchError有时候做起来有点棘手。使用async/，await您可以将try/ catch喜欢与同步代码配合使用，通常更容易正确。