我正在尝试这个 -
final Future<FirebaseUser> user = auth.currentUser();
但问题是,不是通过“userid”制作文档,而是制作一个名为的文档 -
Instance of 'Future<FirebaseUser>'
这就是我现在的文档名称,但我想特别指出它是用户标识。
我该怎么办?
uid是FirebaseUser对象的属性。由于auth.currentUser()返回一个future,你必须等待才能得到这样的用户对象:
void inputData() async {
final FirebaseUser user = await auth.currentUser();
final uid = user.uid
// here you write the codes to input the data into firestore
}
您需要等待异步操作完成。
final FirebaseUser user = awit auth.currentUser();
final userid = user.uid
或者您可以使用then样式语法:
final FirebaseUser user = auth.currentUser().then((FirebaseUser user) {
final userid = user.uid;
// rest of the code
});
如果您使用Google登录,则会获得此用户信息。
final FirebaseAuth firebaseAuth = FirebaseAuth.instance;
final GoogleSignIn _googleSignIn = new GoogleSignIn();
void initState(){
super.initState();
firebaseAuth.onAuthStateChanged
.firstWhere((user) => user != null)
.then((user) {
String user_Name = user.displayName;
String image_Url = user.photoUrl;
String email_Id = user.email;
String user_Uuid = user.uid; // etc
}
// Give the navigation animations, etc, some time to finish
new Future.delayed(new Duration(seconds: 2))
.then((_) => signInWithGoogle());
}
Future<FirebaseUser> signInWithGoogle() async {
// Attempt to get the currently authenticated user
GoogleSignInAccount currentUser = _googleSignIn.currentUser;
if (currentUser == null) {
// Attempt to sign in without user interaction
currentUser = await _googleSignIn.signInSilently();
}
if (currentUser == null) {
// Force the user to interactively sign in
currentUser = await _googleSignIn.signIn();
}
final GoogleSignInAuthentication googleAuth =
await currentUser.authentication;
// Authenticate with firebase
final FirebaseUser user = await firebaseAuth.signInWithGoogle(
idToken: googleAuth.idToken,
accessToken: googleAuth.accessToken,
);
assert(user != null);
assert(!user.isAnonymous);
return user;
}
这是另一种解决方法:
Future<String> inputData() async {
final FirebaseUser user = await FirebaseAuth.instance.currentUser();
final String uid = user.uid.toString();
return uid;
}
它将uid
作为String返回