Android 错误:颤振实例';未来字符串';?
我怎么修理它?这是一个颤振认证错误 GPT此错误:颤振“未来字符串”实例 代码如下:Android 错误:颤振实例';未来字符串';?,android,flutter,Android,Flutter,我怎么修理它?这是一个颤振认证错误 GPT此错误:颤振“未来字符串”实例 代码如下: import 'package:firebase_auth/firebase_auth.dart'; import 'dart:async'; class Auth { FirebaseAuth _auth = FirebaseAuth.instance; Future<String> signIn(String email, String password) async { F
import 'package:firebase_auth/firebase_auth.dart';
import 'dart:async';
class Auth {
FirebaseAuth _auth = FirebaseAuth.instance;
Future<String> signIn(String email, String password) async {
FirebaseUser _user = await _auth.signInWithEmailAndPassword(email: email, password: password);
return _user != null ? _user.uid : null;
}
Future<String> currentUser() async {
FirebaseUser _user = await _auth.currentUser();
return _user != null ? _user.uid : null;
}
Future<void> signOut() async {
return _auth.signOut();
}
Future<void> resetAccount(String email) async {
await _auth.sendPasswordResetEmail(email: email);
}
}
import'package:firebase_auth/firebase_auth.dart';
导入“dart:async”;
类身份验证{
FirebaseAuth _auth=FirebaseAuth.instance;
未来登录(字符串电子邮件、字符串密码)异步{
FirebaseUser _user=wait _auth.sign with email and password(电子邮件:电子邮件,密码:密码);
返回_user!=null?_user.uid:null;
}
Future currentUser()异步{
FirebaseUser_user=wait_auth.currentUser();
返回_user!=null?_user.uid:null;
}
Future signOut()异步{
返回_auth.signOut();
}
未来重置帐户(字符串电子邮件)异步{
等待授权发送密码重置电子邮件(电子邮件:email);
}
}
这里是另一种避免“未来字符串”颤振实例的方法?
我建议您像这样使用代码
Future signIn(String email, String password) async {
FirebaseUser _user = await _auth.signInWithEmailAndPassword(email: email, password: password);
String data = _user != null ? _user.uid : null;
}
_performSignIn(_email, _password){
signIn(_email, _password).then((data), {
//data will contains either null or uid
print(data);
})
或者像这样调用sign-in方法
Future signIn(String email, String password) async {
FirebaseUser _user = await _auth.signInWithEmailAndPassword(email: email, password: password);
String data = _user != null ? _user.uid : null;
}
_performSignIn(_email, _password){
signIn(_email, _password).then((data), {
//data will contains either null or uid
print(data);
})
}这里有另一种避免“未来弦”颤振的方法? 我建议您像这样使用代码
Future signIn(String email, String password) async {
FirebaseUser _user = await _auth.signInWithEmailAndPassword(email: email, password: password);
String data = _user != null ? _user.uid : null;
}
_performSignIn(_email, _password){
signIn(_email, _password).then((data), {
//data will contains either null or uid
print(data);
})
或者像这样调用sign-in方法
Future signIn(String email, String password) async {
FirebaseUser _user = await _auth.signInWithEmailAndPassword(email: email, password: password);
String data = _user != null ? _user.uid : null;
}
_performSignIn(_email, _password){
signIn(_email, _password).then((data), {
//data will contains either null or uid
print(data);
})
}请添加错误代码行。请添加错误代码行。