【发布时间】:2016-09-25 04:03:33
【问题描述】:
我正在使用 fire base,我正在关注在他们的网站上创建用户帐户代码 ->https://www.firebase.com/docs/android/guide/login/password.html.
public class RegisterFragment extends Fragment {
private final String FIREBASE_URL = "(MyFirebaseURL)";
private Firebase fRef;
private EditText etUserName;
private EditText etEmail;
private EditText etPassword;
private EditText etConfirmPassword;
private Button bSubmit;
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState){
View v = inflater.inflate(R.layout.fragment_register,container,false);
Firebase.setAndroidContext(getActivity());
fRef = new Firebase(FIREBASE_URL);
fRef.createUser("(hardcodedEmail)", "(hardcodedPassword)", new Firebase.ValueResultHandler<Map<String, Object>>() {
@Override
public void onSuccess(Map<String, Object> result) {
Toast toast = Toast.makeText(getActivity(),"Registration Successful! UID: " + result.get("uid"), Toast.LENGTH_LONG);
toast.show();
}
@Override
public void onError(FirebaseError firebaseError) {
Toast toast = Toast.makeText(getActivity(),"There was an error", Toast.LENGTH_LONG);
toast.show();
}
});
...由于某种原因 onError() 方法被调用。 这是为什么呢?
如果这有帮助,我首先使用“Firebase.setAndroidContext(getActivity());”在我的 LoginFragment 中。
【问题讨论】:
-
你为什么不检查
FirebaseError firebaseError变量,不知何故我认为它会回答你的问题。
标签: java android database firebase android-studio-2.1