Firebase Auth Login must allow single device login

问题:

I am developing app with the help of Firebase backend and I am using Firebase Auth for login into my application. I have done all integration and every thing and my app is working fine.

But I want only single session with single user as right now with single userId I am able to login through multiple devices.

So I want to restrict user that at a time user can login in in single device.

I am using Custom auth with username password login :

Auth.auth().signIn(withCustomToken: customToken ?? "") { (user, error) in
  // ...
}

If user login with same id in another device I want to show alert that "You are already logged in another device".

Is there any possibility in Firebase Auth lib for single user single session?

Edit : Suggested duplicate question will not solve my query fully though it help me to understand scenireo and help to solve my problem.

Thanks @Frenk for pointing this out.

回答1:

I search a lot on above issue which I was facing through firebase authentication and after lots of research I ended up with below solution which was working as per my requirements.

First of all firebase not providing this in their library so we need to apply our custom logic here to achieve this 1 session user login in our app.

Step 1: You need to add new child "SignIn" at your root of Database.

Step 2: While Auth.auth().signIn() return success in that block we need to check below Flag that is User already signIn in any other device ? for that I have create one method as mention below.

func alreadySignedIn() {
        // [START single_value_read]
        let userID = Auth.auth().currentUser?.uid
        ref.child("SignIn").child(userID!).observeSingleEvent(of: .value, with: { (snapshot) in
            // Get user value

            if let dict = snapshot.value as? [String: Any] {
                if let signedIn = dict["signIn"] as? Bool {
                    if signedIn {
                        self.signOut()
                    }
                    else {
                        // change the screen like normal
                        print("First Session of user")
                        self.writeNewUserSigin(withUserID: userID!)
                    }
                }else{
                    self.writeNewUserSigin(withUserID: userID!)
                }
            }else{
                print(snapshot)
                self.writeNewUserSigin(withUserID: userID!)
            }
        }) { (error) in
            print(error.localizedDescription)
        }
        // [END single_value_read]
    }

By this method we are checking that current user uId have in our SignIn Child with True value if data is there in our database with Boll value True we need to handle that and show some alert and signOut from firebase.

Note : As we allowed user to sign-in and than we are checking that user already signin in any other device so if its returning True we need to SignOut() from firebase.

Now last step while user manually signOut from the app

Step 3: While user click on SignOut button in app we need to update our Child with False value in it so after onwards user can able to SignIn in any other device. For that we can use below method.

func updateUserSigIn(withUserID userID: String) {
        //Update SignIn Child with flase value on current UID
        // [START write_fan_out]

        let post = ["signIn": false]
        let childUpdates = ["/SignIn/\(userID)": post]
        let ref = Database.database().reference()
        ref.updateChildValues(childUpdates) { (error, refDatabase) in
            if (error != nil) {
                print("error \(String(describing: error))")
            }else {
                print("New user Saved successfully")
                self.signOut()
            }
        }
        // [END write_fan_out]
    }

Thats it now only one app user session will allow.

Hope this will helps others.

Thanks for this thread as I got some hints from this answer.