FirebaseAuthException捕捉到错误，但应用程序仍停止工作

Question

我正在登录/注册，当我尝试使用应用程序停止的同一封电子邮件注册时，我添加了这一行

case "email-already-exists":
            errorMessage = "The email is already being used.";
            break;

问题是，它仍然不工作，消息显示出来，但应用程序仍然停止工作。（但信息不同。）

这是我用于身份验证的代码

void signUp(String email, String password) async {
    if (_formKey.currentState!.validate()) {
      try {
        await _auth
            .createUserWithEmailAndPassword(email: email, password: password)
            .then((value) => {postDetailsToFirestore()})
            .catchError((e) {
          _toastInfo(e!.message);
        });
      } on FirebaseAuthException catch (error) {
        switch (error.code) {
          case "invalid-email":
            errorMessage = "Your email address appears to be malformed.";
            break;
          case "wrong-password":
            errorMessage = "Your password is wrong.";
            break;
          case "user-not-found":
            errorMessage = "User with this email doesn't exist.";
            break;
          case "user-disabled":
            errorMessage = "User with this email has been disabled.";
            break;
          case "too-many-requests":
            errorMessage = "Too many requests";
            break;
          case "operation-not-allowed":
            errorMessage = "Signing in with Email and Password is not enabled.";
            break;
          case "email-already-exists":
            errorMessage = "The email is already being used.";
            break;
          default:
            errorMessage = "An undefined Error happened.";
        }
        _toastInfo(errorMessage!);
        print(error.code);
      }
    }
  }

Answer 1

错误不在“FirebaseAuthException”上，但正如错误消息告诉您的：“Future.catchError的错误处理程序…”

 await _auth
            .createUserWithEmailAndPassword(email: email, password: password)
            .then((value) => {postDetailsToFirestore()})
            .catchError((e) {
          _toastInfo(e!.message);
        });

2事情没有做好。

1. 您同时使用wait和then，这不是处理异步函数的正确方式。在这种情况下，它们都做同样的事情，这不好。阅读此处了解有关异步函数的更多信息-这里是官方Flutter文档的链接

这是我处理UserAuth的方式

  /// [Creating account using Email & Password]
  Future<User?> createAccount(String email, String password) async {
    try {
      UserCredential userCredential = await FirebaseAuth.instance
          .createUserWithEmailAndPassword(email: email, password: password);
      return userCredential.user;
    } on FirebaseAuthException catch (e) {
      // Here you manage the errors
      return null;
    }
  }

Answer 2

错误可能是由以下代码引起的：

.catchError((e) {
          _toastInfo(e!.message);
        });

。catchError（）返回类型为future的future