I'm a beginner in iOS programming, so please don't kill me :]。 I have application with authorization and when I successful authorize I would like go to second view。 When authorization fails I show alert。 My code for the authorization:

- (IBAction)LoginClick:(id)sender {

    for(User *user in self.allUsers){
        if([user.userName isEqualToString:self.usernameTextField.text] && [user.password isEqualToString:self.passwordTextField.text]){
            // Logged in --> go to second view (TableView)
            [self performSegueWithIdentifier:@"LoginSegue" sender:self];
        }
        else{
            // Wrong password or username
            Alerts *alert = [Alerts new];
            [alert showAlert:LOGIN_ERROR];
        }
    }
}

澳门百家乐玩法大全I tried to look a simple way for custom transition (segue) to second view, when I successful log in, but I didn't find。 Now I have this storyboard:

And code for transition between View is:

[self performSegueWithIdentifier:@"LoginSegue" sender:self];

The result is error:

* Terminating app due to uncaught exception 'NSGenericException', reason: 'Could not find a navigation controller for segue 'LoginSegue'. Push segues can only be used when the source controller is managed by an instance of UINavigationController.'

What is the problem? Why I need something like "Navigation controller". I need to create a class for segue? What is the simplest way to solve this problem? Thank you.

澳门百家乐玩法大全EDIT ------- SOLUTION

Thanks to all for help. I added navigation controller to my application as initial view. And I found, that use of a custom segue and making class for this custom segue is unnecessary. The simplest way for solution my problem is connect the ViewControllers (NOT BUTTON and second ViewController, so I had it first time...) and name the push segue. The name I used here in code:

- (IBAction)LoginClick:(id)sender {

    for(User *user in self.allUsers){
        if([user.userName isEqualToString:self.usernameTextField.text] && [user.password isEqualToString:self.passwordTextField.text]){
            // Logged in --> go to second view (TableView)
            [self performSegueWithIdentifier:@"LoginSegue" sender:self];
        }
        else{
            // Wrong password or username
            Alerts *alert = [Alerts new];
            [alert showAlert:LOGIN_ERROR];
        }
    }
}

Concrete here: [self performSegueWithIdentifier:@"LoginSegue" sender:self];

And all works great。 When I login with correctly pass and username I'm redirected to second View。

The error is telling you you can't use that type of transition without using a navigation controller。

Simply drag and drop a navigation controller from the right menu and set it as the Initial Controller in the properties section on the right. like so:

Then connect your login controller as its root view controller。 Then this will work for you。

The end result should look something like:

本文地址：IT屋 » Custom segue after successful login

我在的iOS编程初学者，所以请不要杀我：]。我有授权的，当我成功的授权，我想转到第二个视图的应用程序。当授权失败我显示警告。我的code的授权：

   - （IBAction为）LoginClick：（ID）发送{    对（在self.allUsers用户*用户）{
        如果（[user.userName isEqualToString：self.usernameTextField.text]放大器;＆安培; user.password的isEqualToString：self.passwordTextField.text]）{
            //已登录 - ＆GT;到第二个视图（TableView中）
            [个体经营performSegueWithIdentifier：@LoginSegue”发件人：个体经营];
        }
        其他{
            // 用户名或密码错误
            快讯*警报= [快讯新]
            [警报showAlert：LOGIN_ERROR];
        }
    }
}
 

我想看看一个简单的方法进行自定义转换（SEGUE），以第二种观点，当我成功登录，但我没有找到。现在，我有这样的情节提要：

的

和code为浏览之间的过渡是：

  [自performSegueWithIdentifier：@LoginSegue”发件人：个体经营];
 

结果是错误：

的*终止应用程序由于未捕获的异常'NSGenericException，理由是：”找不到赛格瑞导航控制器LoginSegue”。推时，源控制器由UINavigationController的一个实例管理塞格斯只能用'

这是什么问题？为什么我需要的东西，如导航器”。我需要创建一个赛格瑞类？什么是解决这一问题最简单的方法？谢谢你。

编辑------- SOLUTION

感谢所有的帮助。我添加导航控制器到我作为初始视图的应用程序。而且我发现，这个定制赛格瑞使用自定义SEGUE与制作类是不必要的。最简单的方法解决了我的问题是连接ViewControllers（NOT BUTTON和第二的ViewController，所以我有它第一次...），并命名为推动SEGUE。这个名字我以前在这里code：

   - （IBAction为）LoginClick：（ID）发送{    对（在self.allUsers用户*用户）{
        如果（[user.userName isEqualToString：self.usernameTextField.text]放大器;＆安培; user.password的isEqualToString：self.passwordTextField.text]）{
            //已登录 - ＆GT;到第二个视图（TableView中）
            [个体经营performSegueWithIdentifier：@LoginSegue”发件人：个体经营];
        }
        其他{
            // 用户名或密码错误
            快讯*警报= [快讯新]
            [警报showAlert：LOGIN_ERROR];
        }
    }
}
 

混凝土进入：[自performSegueWithIdentifier：@LoginSegue”发件人：个体经营];

和所有伟大工程。当我登录正确传递和用户名，我重定向到第二个视图。