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ios - 使用 Firebase 创建用户时，用户变量为零

问题描述

我正在学习教程，似乎无法将我的用户注册为 Firebase 中的用户变量。createUser 方法似乎为零。因此，当我打开它时，我得到一个错误。

我已经阅读了很多文档，并检查了许多与我类似的其他问题，但似乎没有任何效果

import UIKit
import Firebase
import SwiftKeychainWrapper

class ViewController: UIViewController {

    @IBOutlet weak var userImgView: UIImageView!
    @IBOutlet weak var usernameField: UITextField!
    @IBOutlet weak var emailField: UITextField!
    @IBOutlet weak var passwordField: UITextField!

    var imagePicker: UIImagePickerController!
    var selectedImage: UIImage!

    override func viewDidLoad() {
        super.viewDidLoad()
        imagePicker = UIImagePickerController()
        imagePicker.allowsEditing = true
        imagePicker.delegate = self
    }

    override func viewDidAppear(_ animated: Bool) {
        if let _ = KeychainWrapper.standard.string(forKey: "uid") {
            self.performSegue(withIdentifier: "toFeed", sender: nil)
        }
    }

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

    func setupUser(userUid: String) {
        if let imageData = self.userImgView.image!.jpegData(compressionQuality: 0.2) {
            let imgUid = NSUUID().uuidString
            let metaData = StorageMetadata()
            Storage.storage().reference().child(imgUid).putData(imageData, metadata: metaData) { (metadata, error) in
                let downloadURL = metadata
                let userData = [
                    "username": self.usernameField.text!,
                    "userImg": downloadURL!
                    ] as [String : Any]

                Database.database().reference().child("users").child(userUid).setValue(userData)
                self.performSegue(withIdentifier: "toFeed", sender: nil)
            }
        }
    }

    @IBAction func signInPressed(_ sender: Any) {
        if let email = emailField.text, let password = passwordField.text {
            Auth.auth().signIn(withEmail: email, password: password) { user, error in
                if error != nil && !(self.usernameField.text?.isEmpty)! {
                    Auth.auth().createUser(withEmail: email, password: password) { (user, error) in
                        self.performSegue(withIdentifier: "toFeed", sender: nil)
                        let userID = (user?.user.uid)!
                        self.setupUser(userUid: userID)
                        KeychainWrapper.standard.set(userID, forKey: "uid")
                    }
                } else {
                    if let userID = (user?.user.uid) {
                        KeychainWrapper.standard.set((userID), forKey: "uid")
                        self.performSegue(withIdentifier: "toFeed", sender: nil)
                    }
                }
            }
        }
    }

    @IBAction func getPhoto (_ sender: AnyObject) {
        present(imagePicker, animated: true, completion: nil)
    }
}

extension ViewController: UIImagePickerControllerDelegate, UINavigationControllerDelegate {
    internal func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {
        if let image = info[.originalImage] as? UIImage {
            userImgView.image = image
        } else {
            print("image wasnt selected")
        }
        imagePicker.dismiss(animated: true, completion: nil)
    }
}

我得到的错误是“let userID = (user?.user.uid)！”之一。这是

线程 1：EXC_BAD_INSTRUCTION（代码=EXC_I386_INVOP，子代码=0x0）

标签： iosswiftfirebaseauthenticationfirebase-authentication

解决方案

用a或acreateUser(withEmail:,password:)调用完成块。这就是为什么，正如 Joshua 评论的那样，您应该在访问任何用户属性之前检查是否是。AuthResult.user errorerrornil

来自Swift 的身份验证快速入门：

Auth.auth().createUser(withEmail: email, password: password) { authResult, error in
  strongSelf.hideSpinner {
    guard let user = authResult?.user, error == nil else {
      strongSelf.showMessagePrompt(error!.localizedDescription)
      return
    }
    print("\(user.email!) created")
    strongSelf.navigationController?.popViewController(animated: true)
  }
}

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