在Swift中实例化并呈现一个viewController

问题

我开始浏览上的新Swift功能Xcode 6，并尝试了一些演示项目和教程。现在我被困在：

实例化，然后呈现viewController特定情节提要中的

Objective-C解决方案

UIStoryboard *storyboard = [UIStoryboard storyboardWithName:@"myStoryboardName" bundle:nil];
UIViewController *vc = [storyboard instantiateViewControllerWithIdentifier:@"myVCID"];
[self presentViewController:vc animated:YES completion:nil];

如何在Swift上实现这一目标？

此答案最近针对Swift 5.2和iOS 13.4 SDK进行了修订。

这都是新语法和经过稍微修改的API的问题。UIKit的基本功能没有改变。对于绝大多数iOS SDK框架来说都是如此。

let storyboard = UIStoryboard(name: "myStoryboardName", bundle: nil)
let vc = storyboard.instantiateViewController(withIdentifier: "myVCID")
self.present(vc, animated: true)

如果您遇到问题init(coder:)，请参阅EridB的答案。

对于使用@akashivskyy的答案进行实例化的人UIViewController，但有例外：

致命错误：对类使用未实现的初始化程序“ init（coder :)”

小建议：

required init?(coder aDecoder: NSCoder)在您UIViewController要实例化的目的地手动实现

required init?(coder aDecoder: NSCoder) {
    super.init(coder: aDecoder)
}

如果您需要更多说明，请在这里参考我的回答

此链接具有两种实现：

迅速：

let viewController:UIViewController = UIStoryboard(name: "Main", bundle: nil).instantiateViewControllerWithIdentifier("ViewController") as UIViewController
self.presentViewController(viewController, animated: false, completion: nil)

目标C

UIViewController *viewController = [[UIStoryboard storyboardWithName:@"MainStoryboard" bundle:nil] instantiateViewControllerWithIdentifier:@"ViewController"];

此链接包含用于在同一情节提要中启动viewcontroller的代码

/*
 Helper to Switch the View based on StoryBoard
 @param StoryBoard ID  as String
*/
func switchToViewController(identifier: String) {
    let viewController = self.storyboard?.instantiateViewControllerWithIdentifier(identifier) as! UIViewController
    self.navigationController?.setViewControllers([viewController], animated: false)

}

akashivskyy的答案很好！但是，如果您从显示的视图控制器返回时遇到麻烦，此替代方法可能会有所帮助。它为我工作！

迅速：

let storyboard = UIStoryboard(name: "MyStoryboardName", bundle: nil)
let vc = storyboard.instantiateViewControllerWithIdentifier("someViewController") as! UIViewController
// Alternative way to present the new view controller
self.navigationController?.showViewController(vc, sender: nil)

对象：

UIStoryboard *storyboard = [UIStoryboard storyboardWithName:@"MyStoryboardName" bundle:nil];
UIViewController *vc = [storyboard instantiateViewControllerWithIdentifier:@"someViewController"];
[self.navigationController showViewController:vc sender:nil];

Swift 4.2更新的代码是

let storyboard = UIStoryboard(name: "StoryboardNameHere", bundle: nil)
let controller = storyboard.instantiateViewController(withIdentifier: "ViewControllerNameHere")
self.present(controller, animated: true, completion: nil)

// "Main" is name of .storybord file "
let mainStoryboard: UIStoryboard = UIStoryboard(name: "Main", bundle: nil)
// "MiniGameView" is the ID given to the ViewController in the interfacebuilder
// MiniGameViewController is the CLASS name of the ViewController.swift file acosiated to the ViewController
var setViewController = mainStoryboard.instantiateViewControllerWithIdentifier("MiniGameView") as MiniGameViewController
var rootViewController = self.window!.rootViewController
rootViewController?.presentViewController(setViewController, animated: false, completion: nil)

当我将其放在AppDelegate中时，这对我来说效果很好

如果要模态呈现，则应具有以下波纹管：

let vc = self.storyboard!.instantiateViewControllerWithIdentifier("YourViewControllerID")
self.showDetailViewController(vc as! YourViewControllerClassName, sender: self)

我想提出一种更清洁的方法。当我们有多个故事板时，这将非常有用

1.使用所有故事板创建结构

struct Storyboard {
      static let main = "Main"
      static let login = "login"
      static let profile = "profile" 
      static let home = "home"
    }

2.创建一个像这样的UIStoryboard扩展

extension UIStoryboard {
  @nonobjc class var main: UIStoryboard {
    return UIStoryboard(name: Storyboard.main, bundle: nil)
  }
  @nonobjc class var journey: UIStoryboard {
    return UIStoryboard(name: Storyboard.login, bundle: nil)
  }
  @nonobjc class var quiz: UIStoryboard {
    return UIStoryboard(name: Storyboard.profile, bundle: nil)
  }
  @nonobjc class var home: UIStoryboard {
    return UIStoryboard(name: Storyboard.home, bundle: nil)
  }
}

提供情节提要标识符作为类名，并使用以下代码实例化

let loginVc = UIStoryboard.login.instantiateViewController(withIdentifier: "\(LoginViewController.self)") as! LoginViewController

斯威夫特4：

    let storyboard = UIStoryboard(name: "Main", bundle: nil)
    let yourVC: YourVC = storyboard.instantiateViewController(withIdentifier: "YourVC") as! YourVC

如果您的Viewcontroller不使用任何Storyboard / Xib，则可以像下面的调用一样推送到该特定的VC：

 let vcInstance : UIViewController   = yourViewController()
 self.present(vcInstance, animated: true, completion: nil)

迅捷3

let settingStoryboard : UIStoryboard = UIStoryboard(name: "SettingViewController", bundle: nil)
let settingVC = settingStoryboard.instantiateViewController(withIdentifier: "SettingViewController") as! SettingViewController
self.present(settingVC, animated: true, completion: {

})

我知道这是一个旧线程，但是我认为当前的解决方案（对给定的视图控制器使用硬编码的字符串标识符）非常容易出错。

我已经创建了一个构建时间脚本（您可以在此处访问），它将创建一种编译器安全的方法，用于从给定项目中的所有故事板上访问和实例化视图控制器。

例如，名为视图控制器VC1在Main.storyboard将被实例化，如下所示：

let vc: UIViewController = R.storyboard.Main.vc1^  // where the '^' character initialize the controller

无论我尝试了什么，它都对我不起作用-没有错误，但屏幕上也没有新的视图控制器。不知道为什么，但是将其包装在超时函数中终于使它起作用：

DispatchQueue.main.asyncAfter(deadline: .now() + 0.0) {
    let storyboard = UIStoryboard(name: "Main", bundle: nil)
    let controller = storyboard.instantiateViewController(withIdentifier: "TabletViewController")
    self.present(controller, animated: true, completion: nil)
}

斯威夫特5

let vc = self.storyboard!.instantiateViewController(withIdentifier: "CVIdentifier")
self.present(vc, animated: true, completion: nil)

我创建了一个库，它将以更好的语法更轻松地处理此问题：

https://github.com/Jasperav/Storyboardable

只需更改Storyboard.swift并使其ViewControllers符合即可Storyboardable。