iOS Firebase 登录

Posted 2023-02-23

【中文标题】iOS Firebase 登录

【英文标题】：iOS firebase login

【发布时间】：2018-10-12 15:36:09

【问题描述】：

我的 firebase 登录屏幕在尝试登录时崩溃

这是错误：

2018-05-02 09:39:25.937258-0400 noteCollab[2418:625945] * 由于未捕获的异常“NSInvalidArgumentException”而终止应用程序，原因：“提供的链接对电子邮件/链接符号无效-在。在尝试将其用于电子邮件/链接登录之前，请通过调用 isSignInWithEmailLink:link: on Auth 检查链接。 * 首先抛出调用栈： （0x1837d6d8c 0x1829905ec 0x1837d6c6c 0x1008aa3e0 0x10089c0b8 0x10089cd08 0x10089b808 0x101a29260 0x101a29220 0x101a37e80 0x101a2c730 0x101a38dd8 0x101a3febc 0x1833fbe70 0x1833fbb08） libc++abi.dylib：以 NSException 类型的未捕获异常终止 (lldb)

这是我的代码：

    //
//  signinViewController.swift
//  noteCollab
//
//  Created by James Hall on 5/2/18.
//  Copyright © 2018 James Hall. All rights reserved.
//

import UIKit
import Firebase
import FirebaseAuth

class signinViewController: UIViewController 

    @IBOutlet weak var signInSelector: UISegmentedControl!
    @IBOutlet weak var signInLabel: UILabel!
    @IBOutlet weak var emailTextField: UITextField!
    @IBOutlet weak var passwordTextField: UITextField!
    @IBOutlet weak var signInButton: UIButton!


    var isSignIn:Bool = true

    override func viewDidLoad() 
        super.viewDidLoad()


        // Do any additional setup after loading the view.
    

    override func didReceiveMemoryWarning() 
        super.didReceiveMemoryWarning()
        // Dispose of any resources that can be recreated.
    

    @IBAction func signInSelectorChanged(_ sender: UISegmentedControl) 

        isSignIn = !isSignIn
        if isSignIn 
            signInLabel.text = "Sign In"
            signInButton.setTitle("Sign In", for: .normal)
        else
            signInLabel.text = "Register"
            signInButton.setTitle("Register", for: .normal)
        

    



    @IBAction func signInButtonTapped(_ sender: UIButton) 

        if let email = emailTextField.text, let pass = passwordTextField.text
            //check if its sign in or register
            if isSignIn
                //sign in the user with Firebase
                Auth.auth().signIn(withEmail: email, link: pass)  (user, error) in
                    // check that user isnt nil
                    if error != nil
                        print("cant sign in user")
                    else

                        self.performSegue(withIdentifier: "goToHome", sender: self)

                    
                
            else
                //register the user with Firebase

                Auth.auth().createUser(withEmail: email, password: pass)  (user, error) in
                    // check that user isnt nil
                    if let u = user 

                        //user is found, go to home
                        self.performSegue(withIdentifier: "goToHome", sender: self)

                    else
                        //error: check error and show message
                    
                

            
        



    

【问题讨论】：

您是否在 firebase 控制台的身份验证方法中启用了电子邮件登录？

是的，我使用的登录名和密码与注册时使用的登录名和密码相同

你为什么用signIn(withEmail, link)而不是密码？

我试图只使用电子邮件和密码感谢 Adeel 你真的帮助了我

乐于助人????

【参考方案1】：

您的登录代码正在寻找电子邮件链接而不是密码。只需将“链接：”一词更改为“密码：”，您就应该一切就绪。

Auth.auth().signIn(withEmail: email, password: 密码)

【讨论】：

很高兴我能帮上忙！