使用Flutter应用+ Deno + MongoDB后端出现奇怪错误

Question

我想写一个简单的注册函数，这是我的代码:

auth_screen.dart:

Future<void> _submit() async {
if (!_formKey.currentState!.validate()) {
// Invalid!
return;
}
_formKey.currentState!.save();
setState(() {
_isLoading = true;
});
try {
if (_authMode == AuthMode.Login) {
// Log user in
await Provider.of<Auth>(context, listen: false).login(
_authData['email'] as String,
_authData['password'] as String,
);
} else {
// Sign user up
await Provider.of<Auth>(context, listen: false).signup(
_authData['email'] as String,
_authData['password'] as String,

);
}
} on HttpException catch (error) {
var errorMessage = 'Authentication failed';
print("this is the auth data");
print(_authData);
if (error.toString().contains('EMAIL_EXISTS')) {
errorMessage = 'This email address is already in use.';
} else if (error.toString().contains('INVALID_EMAIL')) {
errorMessage = 'This is not a valid email address';
} else if (error.toString().contains('WEAK_PASSWORD')) {
errorMessage = 'This password is too weak.';
} else if (error.toString().contains('EMAIL_NOT_FOUND')) {
errorMessage = 'Could not find a user with that email.';
} else if (error.toString().contains('INVALID_PASSWORD')) {
errorMessage = 'Invalid password.';
}
_showErrorDialog(errorMessage);
} catch (error) {
var errorMessage = 'Could not authenticate you. Please try again later.' +
error.toString();
_showErrorDialog(errorMessage);
}
setState(() {
_isLoading = false;
});
}

auth.dart:

Future<void> signup(String email, String password) async {
return _authenticate(email, password, 'register');
}
Future<void> _authenticate(
String email, String password, String urlSegment) async {
final url = Uri.parse('http://10.0.2.2:8000/api/$urlSegment');
// final url = Uri.http('http://localhost:8000/api/', 'urlSegment');
try {
final response = await http.post(
url,
body: json.encode(
{
'email': email,
'password': password,
//'returnSecureToken': true,
},
),
);
final responseData = json.decode(response.body);
if (responseData['error'] != null) {
throw HttpException(responseData['error']['message']);
}
_token = responseData['idToken'];
_userId = responseData['localId'];
_expiryDate = DateTime.now().add(
Duration(
seconds: int.parse(
responseData['expiresIn'],
),
),
);
_autoLogout();
notifyListeners();
final prefs = await SharedPreferences.getInstance();
final userData = json.encode(
{
'token': _token,
'userId': _userId,
'expiryDate': _expiryDate!.toIso8601String(),
},
);
prefs.setString('userData', userData);
} catch (error) {
throw error;
}
}

项目的后端(Deno)部分:

auth_controller.ts:

async register(ctx: RouterContext) {
const { email, password } = await ctx.request.body().value;
let user = await User.findOne({ email });
if (user) {
ctx.response.status = 422;
ctx.response.body = { message: "Email is already exist" };
return;
}
const hashedPassword = hashSync(password);
user = new User({ email, password: hashedPassword });
await user.save();
ctx.response.status = 201;
ctx.response.body = {
id: user.id,
name: user.name,
email: user.email
};
}

这是user.ts类:

export default class User extends BaseModel {
public id: string = "";
public name: string = "";
public email: string = "";
public password: string = "";
constructor({ id = "", name = "", email = "", password = "" }) {
super();
this.id = id;
this.name = name;
this.email = email;
this.password = password;
}
static async findOne(params: object): Promise<User | null> {
const user = await userCollection.findOne(params);
if (!user) {
return null;
}
return new User(User.prepare(user));
}
async save() {
const _id = await userCollection.insertOne(this);
this.id = _id;
return this;
}
}

当我想在Android模拟器上测试应用程序时，我得到这个错误消息:

类型'Null'不是类型'String'的子类型

当我尝试后端服务器使用邮差和发送post请求到http://0.0.0.0:8000/api/register地址。我得到正确的响应，它的工作原理，但我不知道为什么我得到Null响应使用扑动应用程序?

我试图在应用程序的前端和后端打印变量，似乎它们都是好的和正确的，但我不明白为什么我得到这个错误信息?!

Answer 1

可能你在你的map中得到了一个空值，因为map在找不到键时返回一个空值，检查map值是否为空

try {
if (_authMode == AuthMode.Login) {
// Log user in
await Provider.of<Auth>(context, listen: false).login(
_authData['email'] as String,  // can return a null value
_authData['password'] as String,   // can return a null value
);
} else {
// Sign user up
await Provider.of<Auth>(context, listen: false).signup(
_authData['email'] as String,   // can return a null value
_authData['password'] as String,   // can return a null value

);
}
}

Answer 2

似乎您正在获得null作为响应值而不是String。

编辑:

Future<void> signup(String email, String password) async {
return _authenticate(email, password, 'register');
}

检查email和password不是null。并得到正确的值

Answer 3

实际上问题来自这两行代码:

_token = responseData['idToken'];_userId = responseData['localId'];

因为它没有在响应中从后端传递这些数据