firebase - Firebase，我如何捕捉异常并告诉用户它？

Question

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

Answer 1

您可以使用 a 捕获异常并以try/catch多种方式将其显示在 UI 上。例子

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);
    }
  }