处理新 Firebase 和 Swift 中的错误

Question

我正在尝试在使用 swift 和 firebase 在 iOS 项目中创建用户按钮时添加错误处理：

这是按钮的代码：

     @IBAction func Register(sender: AnyObject) {

    if NameTF.text == "" || EmailTF.text == "" || PasswordTF.text == "" || RePasswordTF == "" || PhoneTF.text == "" || CityTF.text == ""
    {
        let alert = UIAlertController(title: "عذرًا", message:"يجب عليك ملىء كل الحقول المطلوبة", preferredStyle: .Alert)
        alert.addAction(UIAlertAction(title: "نعم", style: .Default) { _ in })
        self.presentViewController(alert, animated: true){}

    } else {

        if PasswordTF.text != RePasswordTF.text {
            let alert = UIAlertController(title: "عذرًا", message:"كلمتي المرور غير متطابقتين", preferredStyle: .Alert)
            alert.addAction(UIAlertAction(title: "نعم", style: .Default) { _ in })
            self.presentViewController(alert, animated: true){}

        } else {


            FIRAuth.auth()?.createUserWithEmail(EmailTF.text!, password: PasswordTF.text!, completion: { user, error in
                print(error)

                if error != nil {

                    let errorCode = FIRAuthErrorNameKey

                    switch errorCode {
                    case "FIRAuthErrorCodeEmailAlreadyInUse":
                        let alert = UIAlertController(title: "عذرًا", message:"الإيميل مستخدم", preferredStyle: .Alert)
                        alert.addAction(UIAlertAction(title: "نعم", style: .Default) { _ in })
                        self.presentViewController(alert, animated: true){}

                    case "FIRAuthErrorCodeUserNotFound":
                        let alert = UIAlertController(title: "عذرًا", message:"المستخدم غير موجود", preferredStyle: .Alert)
                        alert.addAction(UIAlertAction(title: "نعم", style: .Default) { _ in })
                        self.presentViewController(alert, animated: true){}

                    case "FIRAuthErrorCodeInvalidEmail":
                        let alert = UIAlertController(title: "عذرًا", message:"الإيميل غير صحيح", preferredStyle: .Alert)
                        alert.addAction(UIAlertAction(title: "نعم", style: .Default) { _ in })
                        self.presentViewController(alert, animated: true){}

                    case "FIRAuthErrorCodeNetworkError":
                        let alert = UIAlertController(title: "عذرًا", message:"خطأ في الاتصال بالانترنت", preferredStyle: .Alert)
                        alert.addAction(UIAlertAction(title: "نعم", style: .Default) { _ in })
                        self.presentViewController(alert, animated: true){}

                    default:
                        let alert = UIAlertController(title: "عذرًا", message:"خطأ غير معروف", preferredStyle: .Alert)
                        alert.addAction(UIAlertAction(title: "نعم", style: .Default) { _ in })
                        self.presentViewController(alert, animated: true){}



                    }


                } else {

                    FIRAuth.auth()?.signInWithEmail(self.EmailTF.text!, password: self.PasswordTF.text!, completion: { (user: FIRUser?, error: NSError?) in
                        if let error = error {
                            print(error.localizedDescription)
                        } else {

                           self.ref.child("UserProfile").child(user!.uid).setValue([
                                "email": self.EmailTF.text!,
                                "name" : self.NameTF.text!,
                                "phone": self.PhoneTF.text!,
                                "city" : self.CityTF.text!,
                                ])
                            print("Sucess")
                          //  self.performSegueWithIdentifier("SignUp", sender: nil)

                        }
                    })

                } //else
            })

        } //Big else


    } //Big Big else
}


}//end of

我不确定switch语句中的错误语法是否正确！

因为当我在模拟器中测试它时，它总是给我默认情况，这是未知错误！ + 我在文档中找不到语法： https://firebase.google.com/docs/auth/ios/errors

那么，使用新的 firebase 和 swift 添加错误处理的正确语法是什么！

Answer 1

实际上，我已经为此苦苦挣扎了很长时间，并发现了问题所在。我已经尝试了上面答案中发布的代码，error.code 行给了我一个错误。它确实适用于error._code。换句话说，对保罗的原始答案稍作修改。这是我的最终代码（我会针对所有错误进行编辑）：

if let errCode = AuthErrorCode(rawValue: error!._code) {

    switch errCode {
        case .errorCodeInvalidEmail:
            print("invalid email")
        case .errorCodeEmailAlreadyInUse:
            print("in use")
        default:
            print("Create User Error: \(error)")
    }    
}

Answer 2

为 Swift 4 + Firebase 4 + UIAlertController 更新

extension AuthErrorCode {
    var errorMessage: String {
        switch self {
        case .emailAlreadyInUse:
            return "The email is already in use with another account"
        case .userNotFound:
            return "Account not found for the specified user. Please check and try again"
        case .userDisabled:
            return "Your account has been disabled. Please contact support."
        case .invalidEmail, .invalidSender, .invalidRecipientEmail:
            return "Please enter a valid email"
        case .networkError:
            return "Network error. Please try again."
        case .weakPassword:
            return "Your password is too weak. The password must be 6 characters long or more."
        case .wrongPassword:
            return "Your password is incorrect. Please try again or use 'Forgot password' to reset your password"
        default:
            return "Unknown error occurred"
        }
    }
}


extension UIViewController{
    func handleError(_ error: Error) {
        if let errorCode = AuthErrorCode(rawValue: error._code) {
            print(errorCode.errorMessage)
            let alert = UIAlertController(title: "Error", message: errorCode.errorMessage, preferredStyle: .alert)

            let okAction = UIAlertAction(title: "Ok", style: .default, handler: nil)

            alert.addAction(okAction)

            self.present(alert, animated: true, completion: nil)

        }
    }

}

使用示例：

    Auth.auth().signIn(withEmail: email, password: password, completion: { (user, error) in

        if error != nil {
            print(error!._code)
            self.handleError(error!)      // use the handleError method
            return
        }
        //successfully logged in the user

    })

Answer 3

尽管这已被正确回答，但还是想分享一个我们添加到项目中的不错的实现。

这也可以用于其他错误类型，但我们只需要它用于FIRAuthErrorCodes。

如果您将 FIRAuthErrorCode 扩展为具有字符串类型的变量 errorMessage，您可以为用户提供自己的错误消息：

extension FIRAuthErrorCode {
    var errorMessage: String {
        switch self {
        case .errorCodeEmailAlreadyInUse:
            return "The email is already in use with another account"
        case .errorCodeUserDisabled:
            return "Your account has been disabled. Please contact support."
        case .errorCodeInvalidEmail, .errorCodeInvalidSender, .errorCodeInvalidRecipientEmail:
            return "Please enter a valid email"
        case .errorCodeNetworkError:
            return "Network error. Please try again."
        case .errorCodeWeakPassword:
            return "Your password is too weak"
        default:
            return "Unknown error occurred"
        }
    }
}

您可以像我们上面那样只自定义一些，并将其余的分组在“未知错误”下。

使用此扩展程序，您可以处理 Vladimir Romanov 的回答中所示的错误：

func handleError(_ error: Error) {
    if let errorCode = FIRAuthErrorCode(rawValue: error._code) {
        // now you can use the .errorMessage var to get your custom error message
        print(errorCode.errorMessage)
    }
}

Answer 4

FIRAuthErrorCode 是一个 int 枚举而不是一个字符串枚举。执行以下操作：

if let error = error {
        switch FIRAuthErrorCode(rawValue: error.code) !{
                case .ErrorCodeInvalidEmail:

更多信息请参阅answer。

Answer 5

我正在使用 Swift 5 和 Firebase 6.4.0，对我来说，以上都没有真正奏效。经过一番尝试，我想出了这个：

Auth.auth().createUser(withEmail: emailTextfield.text!, password: passwordTextfield.text!) { (user, error) in
        if error!= nil{

                let alert = UIAlertController(title: "Error", message: error!.localizedDescription, preferredStyle: .alert)
                let okAction = UIAlertAction(title: "Ok", style: .default, handler: nil)
                alert.addAction(okAction)
                self.present(alert,animated: true)

        }