Firebase，我如何捕获异常并告诉用户它？ [复制]

Posted 2023-02-23

【中文标题】Firebase，我如何捕获异常并告诉用户它？ [复制]

【英文标题】：Firebase, how do I catch a exception and tell the user about it? [duplicate]

【发布时间】：2021-10-23 19:23:04

【问题描述】：

Future<Users?>createUserWithEmailAndPassword(String email, String password) async 
    final credential = await _firebaseAuth.createUserWithEmailAndPassword(
      email: email, 
      password: password 
    ); 
    return _userFromFirebase(credential.user); 
 

【参考方案1】：

您可以使用 try/catch 捕获异常，并以多种方式将其显示在 UI 上。示例

Future<Users?> createUserWithEmailAndPassword(
    String email,
    String password,
  ) async 
    try 
      final credential = await _firebaseAuth.createUserWithEmailAndPassword(
        email: email,
        password: password,
      );

      return _userFromFirebase(credential.user);
     on FirebaseException catch (e) 
      // FirebaseException
      print(e.message);
     catch (e) 
      // all other exceptions
      print(e);
    
  

【讨论】：

swedischeef thnks 在 loginPage onPressed 按钮上工作 onPressed: () async if(_formKey.currentState!.validate()) try await authService.signInWithEmailPassword( emailController.text, passwordController.text ); Navigator.pop(上下文); catch(e) await showDialog(context: context, builder: (BuildContext) => AlertDialog( title: Text('ERROR'), content: Text(e.toString()), actions: [ TextButton(孩子：Text（'Ok'），onPressed：Navigator.of（context）.pop，）]，））;

很高兴听到它有效。介意接受我的回答吗？