Firebase邮箱注册登录功能遇类型错误：String无法赋值给int

问题根源 #

你的代码存在核心类型不匹配问题：

• Firebase Auth返回的用户uid是字符串类型，但UserModel中uid被定义为int类型
• Firestore文档ID要求传入String类型，却使用了int类型的userModel.uid
• 错误返回场景中，传入空字符串给int类型的uid参数

修复方案 #

核心修改是将UserModel中的uid类型从int改为String，同时调整相关代码适配该类型：

1. 修改UserModel类

import 'package:cloud_firestore/cloud_firestore.dart';

class UserModel {
  late final String uid; // 从int改为String
  late final String email;
  late final String displayName;
  late final String photoUrl;
  late final bool isAdmin;
  late final bool isGoogleUser;

  UserModel(
      {required this.uid,
      required this.email,
      required this.displayName,
      required this.photoUrl,
      this.isAdmin = false,
      this.isGoogleUser = false});

  factory UserModel.fromFirestore(DocumentSnapshot snapshot) {
    return UserModel(
      uid: snapshot['uid'] as String, // 明确类型为String
      email: snapshot['email'] as String,
      displayName: snapshot['displayName'] as String,
      photoUrl: snapshot['photoUrl'] as String,
      isAdmin: snapshot['isAdmin'] ?? false,
      isGoogleUser: snapshot['isGoogleUser'] ?? false,
    );
  }
}

2. 调整注册功能代码中的错误点

修改后，注册时传入Firebase的String类型uid不再报错，Firestore文档ID也能正常使用String类型的uid，错误返回场景的空字符串参数也符合类型要求：

//Registering a new user unto firebase....................................................
Future<AuthStatus> registerWithEmailAndPassword(
    String username, String email, String password) async {
  try {
    final UserCredential userCredential = await _auth
        .createUserWithEmailAndPassword(email: email, password: password);

    if (userCredential.user != null) {
      UserModel userModel = UserModel(
          uid: userCredential.user!.uid, // 现在uid是String类型，匹配参数
          email: email,
          displayName: username,
          photoUrl: '');

      // Add the UserModel data to a Firestore collection
      await _usersRef.doc(userModel.uid).set({ // uid为String，符合doc()参数要求
        'uid': userModel.uid,
        'email': userModel.email,
        'username': userModel.displayName,
        'photoUrl': userModel.photoUrl,
      });

      // Return a successful AuthStatus
      return AuthStatus(
        isSuccessful: true,
        errorMessage: '',
        userModel: userModel,
      );
    } else {
      // If userCredential.user is null, registration failed
      return AuthStatus(
        isSuccessful: false,
        errorMessage: 'User registration failed',
        userModel:
            UserModel(uid: '', email: '', displayName: '', photoUrl: ''),// 空字符串符合String类型
      );
    }
  } catch (e) {
    // Handle any errors during registration
    return AuthStatus(
      isSuccessful: false,
      errorMessage: e.toString(),
      userModel: UserModel(uid: '', email: '', displayName: '', photoUrl: ''),// 空字符串符合String类型
    );
  }
}

3. 调整登录功能代码中的错误点

//SignIn a registered member or user to firebase.........................................
Future<UserModel?> signInWithEmailAndPassword(
    String username, String password) async {
  try {
    final UserCredential userCredential =
        await _auth.signInWithEmailAndPassword(
      email: username,
      password: password,
    );

    if (userCredential.user != null) {
      UserModel userModel = UserModel(
        uid: userCredential.user!.uid, // 现在uid是String类型，匹配参数
        email: userCredential.user!.email ?? '',
        displayName: userCredential.user!.displayName ?? '',
        photoUrl: '',
      );

      return userModel;
    } else {
      return null;
    }
  } catch (e) {
    print('Sign-in error: $e');
    return null;
  }
}

关键修改总结 #

• 统一uid类型为String，匹配Firebase生态的类型规范
• 确保Firestore文档ID使用String类型的用户uid
• 错误场景中使用空字符串作为默认uid，符合类型要求

内容的提问来源于stack exchange，提问作者Fred_Wolfe