StreamProvider<User>监听的_MapStream<FirebaseUser, User>抛出异常,但没有提供`catchError`

An exception was throw by _MapStream<FirebaseUser, User> listened by StreamProvider<User>, but no `catchError` was provided

<--- 更新 --->

找到解决方案。检查下面的答案

<--- 更新 --->

知道为什么会发生这种情况,是什么导致了错误,我该如何解决?

我在 YouTube 上关注 TheNetNin​​ja 的教程,我看到人们遇到了类似的问题,但又不完全相同。

当应用程序突然关闭(使用Android Studio 中的停止按钮)而不注销然后重新启动时抛出异常,仅当使用电子邮件并通过登录时。对于 Google 登录,它似乎工作正常。此外,如果我在 Firebase 控制台中手动创建新用户并首次登录,则不会发生这种情况。

例外情况:

════════ Exception caught by provider ══════════════════════════════════════════════════════════════
The following assertion was thrown:
An exception was throw by _MapStream<FirebaseUser, User> listened by

StreamProvider<User>, but no `catchError` was provided.

Exception:
RangeError (index): Invalid value: Only valid value is 0: 1

════════════════════════════════════════════════════════════════════════════════════════════════════

我在 pubspec.yaml 中使用 firebase_auth: ^0.16.0,我认为这可能与版本有关。

当我登录时,我得到这个输出

I/BiChannelGoogleApi(17821): [FirebaseAuth: ] getGoogleApiForMethod() returned Gms: com.google.firebase.auth.api.internal.zzaq@cd1fa83
D/FirebaseAuth(17821): Notifying id token listeners about user ( 1EMwk8483hQWnSQwJeQYm9AcpiD2 ).

看起来帐户已通过验证并登录,但随后屏幕没有变化,所以我猜它在验证过程中卡在了某个地方。

奇怪的是,如果我正常登录但出现异常并重新启动应用程序,它就可以正常工作。如果我一直重启它,它工作得很好,但是如果我突然停止它,我又会出现异常。

这是我的验证码

import 'package:firebase_auth/firebase_auth.dart';
import 'package:flutter/material.dart';
import 'package:flutter/services.dart';
import 'package:google_sign_in/google_sign_in.dart';
import 'package:common_ui_module/Utils.dart';
import 'package:user_module/Role.dart';
import 'package:user_module/User.dart';
import 'package:localization_module/AppLocalizations.dart';

class AuthService {
  final FirebaseAuth _auth = FirebaseAuth.instance;

    return user != null
        ? User(uid: user.uid)
        : null;
  }

  Stream<User> get user {
    // map FireBaseUsers to User objects
    return _auth.onAuthStateChanged.map(_userFromFireBaseUser);
  }


// sign in with email and pass
  Future<User> signInWithEmailAndPassword(String email, String pass) async {
    try {
      AuthResult result =
          await _auth.signInWithEmailAndPassword(email: email, password: pass);
      FirebaseUser user = result.user;
      assert(await user.getIdToken() != null);
      print(
          "Signed in: ${user.email} , pass: $pass, phone: ${user.phoneNumber}");

      /// return the mapped user
      return _userFromFireBaseUser(user);
    } catch (e) {
      print(e.toString());
      return null;
    }
  }


// sign out
  Future signOut() async {
    try {
      return await _auth.signOut();
    } catch (e) {
      print(e.toString());
      return null;
    }
  }
}

这是在登录和主屏幕之间切换的包装器

class Wrapper extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    final User user = Provider.of<User>(context);

    if (user != null) {
      return RestaurantOrdersScreen();
    } else {
      return LoginScreen();
    }
  }
}

我已经将主应用程序包装在 MultiProvider 中:

Widget build(BuildContext context) {
    return MultiProvider(
      providers: [
        /// PROVIDER FOR AUTHENTICATION
        StreamProvider<User>.value(
          value: AuthService().user,
        ),

       //...
       // some other providers
      ],
      child: MaterialApp(
        home: SplashScreen(),
      ),
    );
  }

终于找到了解决方法!

对我来说,当我试图操纵 displayName 时,就是这个分裂。这就是我得到索引超出范围错误的原因。

List<String> name = ["", ""];
    if (user != null && user.displayName != null) {
      name = user.displayName.split(" ");
    }

    return user != null
        ? User(
            id: 0,
            token: user.uid,
            firstName: name[0],
            lastName: name[1],
...

因此,如果您遇到此类错误,请在尝试将 FirebaseUser 转换为您自己的自定义用户对象时仔细检查您的用户属性操作。