Swift

Question

我正在尝试从扩展程序的 class 方法中获取 class 名称。这可能吗？我最终以 "DRHT" 作为 class

的名称

扩展

extension UIViewController {

    public class func instanceFromStoryboard(storyboardIdentifier: String = "Main") -> UIViewController {
        let storyboard = UIStoryboard(name: storyboardIdentifier, bundle: nil)
        let controllerIdentifier = NSStringFromClass(self)
        return storyboard.instantiateViewControllerWithIdentifier(controllerIdentifier) as UIViewController
    }

}

用法

let homeViewController = HomeViewController.instanceFromStoryboard()

Answer 1

我最终采用了不同的方式来使其更加类型安全并消除转换。

extension UIStoryboard {

    public class func instantiateViewController <T: UIViewController>(type: T.Type, storyboardIdentifier: String = "Main") -> T {
        let storyboard = UIStoryboard(name: storyboardIdentifier, bundle: nil)
        return storyboard.instantiateViewController(type)
    }

    public func instantiateViewController <T: UIViewController>(type: T.Type) -> T {
        return instantiateViewControllerWithIdentifier(String(type)) as! T
    }

}

extension UIViewController {

    public class func instantiateFromStoryboard(storyboardName: String = "Main") -> Self {
        return UIStoryboard.instantiateViewController(self, storyboardIdentifier: storyboardName)
    }
}

用法

let vc = UIStoryboard.instantiateViewController(HomeViewController.self)

或

let vc = UIStoryboard.instantiateViewController(HomeViewController.self, storyboardIdentifier: "Storybaord_Name")

或

let vc = storyboard.instantiateViewController(HomeViewController.self)

或

let vc = HomeViewController.instantiateViewController()

Answer 2

感谢MartinR and his ，我知道答案了：

extension UIViewController
{
    class func instantiateFromStoryboard() -> Self
    {
        return instantiateFromStoryboardHelper(self, storyboardName: "Main")
    }

    class func instantiateFromStoryboard(storyboardName: String) -> Self
    {
        return instantiateFromStoryboardHelper(self, storyboardName: storyboardName)
    }

    private class func instantiateFromStoryboardHelper<T>(type: T.Type, storyboardName: String) -> T
    {
        var storyboardId = ""
        let components = "\(type.dynamicType)".componentsSeparatedByString(".")

        if components.count > 1
        {
            storyboardId = components[1]
        }
        let storyboad = UIStoryboard(name: storyboardName, bundle: nil)
        let controller = storyboad.instantiateViewControllerWithIdentifier(storyboardId) as! T

        return controller
    }
}

用法：

let vc = HomeViewController.instantiateFromStoryboard()

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