在 Swift 中实例化并呈现视图控制器

Question

问题

我开始查看

Swift Programming Language

，不知怎的，我无法从特定的

UIViewController

正确输入

UIStoryboard

的初始化。

在

Objective-C

中我简单地写：

UIStoryboard *storyboard = [UIStoryboard storyboardWithName:@"StoryboardName" bundle:nil];
UIViewController *viewController = [storyboard instantiateViewControllerWithIdentifier:@"ViewControllerID"];
[self presentViewController:viewController animated:YES completion:nil];

任何人都可以帮助我如何在 Swift 上实现这一目标吗？

Answer 1

此答案最后针对 Swift 5.4 和 iOS 14.5 SDK 进行了修订。

---

这都是新语法和稍微修改的 API 的问题。 UIKit 的底层功能没有改变。对于绝大多数 iOS SDK 框架都是如此。

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

确保在故事板内的“故事板 ID”下设置

myVCID

。

Answer 2

对于使用

@akashivskyy 的答案 来实例化 UIViewController

 并有例外的人：

致命错误：对类使用未实现的初始化程序“init(coder:)”

快速提示：

在您尝试实例化的目的地

required init?(coder aDecoder: NSCoder)

 手动实现 

UIViewController

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

如果您需要更多描述，请参阅我的回答

这里

Answer 3

此链接有两种实现：

斯威夫特：

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故事板中启动视图控制器的代码

/*
 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)

}

Answer 4

akashivskyy 的答案效果很好！但是，如果您在从呈现的视图控制器返回时遇到一些问题，则此替代方案可能会有所帮助。这对我有用！

斯威夫特：

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-C：

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

Answer 5

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)

Answer 6

// "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)

当我将它放入 AppDelegate 时，这对我来说效果很好

Answer 7

如果你想以模态方式呈现它，你应该有如下所示的内容：

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

Answer 8

我想建议一种更干净的方法。当我们有多个故事板时这将很有用

1.用所有故事板创建一个结构

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)
  }
}

将故事板标识符作为类名，并使用以下代码实例化

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

Answer 9

无论我尝试什么，它对我来说都不起作用 - 没有错误，但我的屏幕上也没有新的视图控制器。不知道为什么，但将其包装在超时函数中最终使其工作：

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)
}

Answer 10

斯威夫特3

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

})

Answer 11

斯威夫特 4：

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

Answer 12

如果你有一个不使用任何storyboard/Xib的Viewcontroller，你可以像下面的调用一样推送到这个特定的VC：

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

Answer 13

斯威夫特5

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

Answer 14

我知道这是一个旧线程，但我认为当前的解决方案（对给定视图控制器使用硬编码字符串标识符）很容易出错。

我创建了一个构建时脚本（您

可以在此处访问），它将创建一个编译器安全的方法，用于从给定项目中的所有故事板访问和实例化视图控制器。

例如，

Main.storyboard中名为vc1的视图控制器将像这样实例化：

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

Answer 15

我创建了一个库，可以通过更好的语法更轻松地处理这个问题：

https://github.com/Jasperav/Storyboardable

只需更改 Storyboard.swift 并让

ViewControllers

 符合 

Storyboardable

。

Answer 16

guard let vc = storyboard?.instantiateViewController(withIdentifier: "add") else { return }
        vc.modalPresentationStyle = .fullScreen
        present(vc, animated: true, completion: nil)

Answer 17

我使用这个助手：

struct Storyboard<T: UIViewController> {
    
    static var storyboardName: String {
        return String(describing: T.self)
    }
    
    static var viewController: T {
        let storyboard = UIStoryboard(name: "Main", bundle: nil)
        
        guard let vc = storyboard.instantiateViewController(withIdentifier: Self.storyboardName) as? T else {
            fatalError("Could not get controller from Storyboard: \(Self.storyboardName)")
        }
        
        return vc
    }
}

使用（Storyboard ID 必须与 UIViewController 类名匹配）

let myVC = Storyboard.viewController as MyViewController

Answer 18

if let destinationVC = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "DestinationVC") as? DestinationVC {
    let nav = self.navigationController
    //presenting
    nav?.present(destinationVC, animated: true, completion: { })
    //push
    nav?.pushViewController(destinationVC, animated: true)
}