Question

我想知道是否有一种方法可以使用枚举并为其分配不同类的不同值，例如，我有3个实例化View Controller的类，我总是像这样启动它们

let placeAreaC = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "PlaceAreaC") as! PlaceAreaC

let notificationC = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "NotificationC") as! NotificationC

let reportC = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "ReportC") as! ReportC

每次我想要一个新的控制器，我必须这样做，我想知道我是否可以创建一个可以保存这些值的枚举，现在我有这个代码

class ControllerEnum: NSObject {

    enum InstantiateEnum: String {
        case Report = "Report" // UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "ReportC") as! ReportC
        case Notification = "Notification" // UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "NotificationC") as! NotificationC
        case PlaceArea = "PlaceArea" // UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "ReportC") as! ReportC

    }
  }

我知道类型是String但我想知道如何才能使类型接受上面的3个不同的类？那么我就可以像

那样引用它们了

InstantiateEnum.Notification

这将非常有用，因为现在我可以在一个地方更改这些值。任何建议都会很棒。

Answer 1

您可以拥有在枚举

中返回ViewController的属性

enum InstantiateEnum: String {
    case Report = "Report" // UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "ReportC") as! ReportC
    case Notification = "Notification" // UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "NotificationC") as! NotificationC
    case PlaceArea = "PlaceArea" // UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "ReportC") as! ReportC

   var viewController:UIViewController {
       switch self  {
           case  Report :
             return UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: self.rawValue)
           ...

       }

   }

}

OR

您可以在每个View Controller返回自我对象中使用一个类方法

喜欢

//--------------------------------------------------------------------------------

// MARK: - Abstract Methods

//--------------------------------------------------------------------------------

class func viewController() -> ReportC {
    return StoryBoard.main.instantiateViewController(withIdentifier: StoryBoard.controller.ReportC) as! ReportC
}


//--------------------------------------------------------------------------------

创建标识符的常量

struct StoryBoard {

    static let main = UIStoryboard.init(name: "Main", bundle: nil)
    static let otherStoryboard = UIStoryboard.init(name: "otherStoryboard", bundle: nil)

    struct controller {
        static let ReportC                                  =   "Report"
     }
}

Answer 2

我使用的方法更好：

第1步：只需使用扩展程序复制此枚举。并使用项目中的那些更改故事板名称

enum AppStoryboard : String {

case Main
case Notification
case Profile
case Login
}

extension AppStoryboard {

var instance : UIStoryboard {

    return UIStoryboard(name: self.rawValue, bundle: Bundle.main)
}

func viewController<T : UIViewController>(viewControllerClass : T.Type, function : String = #function, line : Int = #line, file : String = #file) -> T {

    let storyboardID = (viewControllerClass as UIViewController.Type).storyboardID

    guard let scene = instance.instantiateViewController(withIdentifier: storyboardID) as? T else {

        fatalError("ViewController with identifier \(storyboardID), not found in \(self.rawValue) Storyboard.\nFile : \(file) \nLine Number : \(line) \nFunction : \(function)")
    }

    return scene
}

func initialViewController() -> UIViewController? {

    return instance.instantiateInitialViewController()
}
}

extension UIViewController {

static func instantiate(fromAppStoryboard appStoryboard: AppStoryboard) -> Self {

    return appStoryboard.viewController(viewControllerClass: self)
}

}

第2步： 在故事板中分配与视图控制器的类名相同的故事板ID。

第3步： 现在创建viewcontroller的对象，如：

let notificationVC = NotificationVC.instantiate(fromAppStoryboard: .Notification)

/* Syntax:
let <yourObject> = <YourViewController's class name>.instantiate(fromAppStoryboard: <respected storyboard>) */

notificationVC将成为NotificationVC

类型的对象

现在玩你的对象。：）

IOS Swift如何让枚举返回不同的类

2 个答案: