iOS swift firebase auth不正确登录

Question

我使用Firebase身份验证设置了登录场景。在输入正确的细节时，场景应该通过执行“loggedIn”segue进行更改。在输入不正确的细节时，错误应该打印到控制台并且什么都不应该发生。

我发现，尽管发现错误（并将其打印到控制台），控制器仍然执行segue并移至下一个屏幕（但不会将“LOGGING IN”打印到控制台）。发现这种行为很奇怪，有什么想法？

我抄我的日志在视图控制器下面：

import Foundation 
import Firebase 
import FBSDKCoreKit 
import FBSDKLoginKit 

class LogInViewController: UIViewController{ 
@IBOutlet weak var Email: UITextField! 
@IBOutlet weak var Password: UITextField! 

@IBAction func Login(_ sender: AnyObject) { 
    let email = self.Email.text! 
    let password = self.Password.text! 

    if (email != "" && password != ""){ 
     FIRAuth.auth()?.signIn(withEmail: email, password: password, completion: { (user, error) in 
      if(error==nil){ 
       print("LOGGING IN") 
       self.performSegue(withIdentifier: "loggedIn", sender: nil) 
      } 
      else{ 
       print("COULDNT LOG IN") 
       print(error) 
      } 
     }) 
    } 
    else{ 
     let alert = UIAlertController(title: "Error", message: "Enter email and password", preferredStyle: .alert) 
     let action = UIAlertAction(title: "OK", style: .default, handler: nil) 
     alert.addAction(action) 
     self.present(alert, animated: true, completion: nil) 
    } 
} 
} 

Storyboard

编辑

我在加入故事板ID loggedInViewControllerVC_ID到mealViewController身份检查员。

这是我的新代码：

import Firebase 
import FBSDKCoreKit 
import FBSDKLoginKit 

class LogInViewController: UIViewController{ 

@IBOutlet weak var Email: UITextField! 
@IBOutlet weak var Password: UITextField! 
@IBAction func Login(_ sender: AnyObject) { 
    let email = self.Email.text! 
    let password = self.Password.text! 

    if (email != "" && password != ""){ 
     FIRAuth.auth()?.signIn(withEmail: email, password: password, completion: { (user, error) in 
      if(error==nil){ 
       print("LOGGING IN") 
       let nextViewController = self.navigationController?.storyboard?.instantiateViewController(withIdentifier: "loggedInViewControllerVC_ID") as! MealViewController 
       self.navigationController?.pushViewController(nextViewController, animated: true) 
       //self.performSegue(withIdentifier: "loggedIn", sender: nil) 
      } 
      else{ 
       print("COULDNT LOG IN") 
       print(error) 
      } 
     }) 
    } 
    else{ 
     let alert = UIAlertController(title: "Error", message: "Enter email and password", preferredStyle: .alert) 
     let action = UIAlertAction(title: "OK", style: .default, handler: nil) 
     alert.addAction(action) 
     self.present(alert, animated: true, completion: nil) 
    } 
} 
} 

现在我对成功输入登录信息

Answer 1

如果你已经在你的故事板通过赛格瑞连接您的按钮LoogedIn的viewController得到一个BAD_INSTRUCTION错误。它会延续UI。你可以做的是实例化你的viewController，然后来回延伸。

从故事板中删除该segue。

let loggedInScene = self.navigationController?.storyboard?.instantiateViewController(withIdentifier: "loggedInViewVontrollerVC_ID") as! loggedInViewVontroller 

self.navigationController?.pushViewController(loggedInScene, animated: true) 

PS： - loggedInViewVontrollerVC_ID是你对的loggedIn的viewController

storyboardID选择的viewController要Segue公司在故事板 - >身份监察长>故事板ID - >loggedInViewVontrollerVC_ID 。

并确保您已在第一个viewController中的导航控制器中嵌入。