【发布时间】:2018-06-18 21:17:56
【问题描述】:
我正在尝试创建一个将 UIViewController 作为函数的函数。这样做的原因是可以传递多个自定义视图控制器。这是我当前的函数,它可以工作,但使用 switch 语句和枚举:
enum controllerTypes {
case First, Second
}
extension UIViewController {
func presentViewController(storyBoardName: String, storyBoardIdentifier: String, controllerType: controllerTypes, completion:(() -> Void)?) {
switch controllerType {
case .First:
let firstVC = UIStoryboard(name: storyBoardName, bundle: nil).instantiateViewController(withIdentifier: storyBoardIdentifier) as? FirstViewController
if let firVC = firstVC {
self.present(firVC, animated: true, completion: nil)
}
case .Second:
let secondVC = UIStoryboard(name: storyBoardName, bundle: nil).instantiateViewController(withIdentifier: storyBoardIdentifier) as? SecondViewController
if let secVC = secondVC {
self.present(secVC, animated: true, completion: nil)
}
}
completion?()
}
}
我没有为参数传递“controllerTypes”枚举,而是将任何类型的 UIViewController 传递给它,当我尝试执行此操作时,出现以下错误:
func presentViewController(storyBoardName: String, storyBoardIdentifier: String, controllerType: UIViewController, completion:(() -> Void)?) {
let sampleVC = UIStoryboard(name: storyBoardName, bundle: nil).instantiateViewController(withIdentifier: storyBoardIdentifier) as? controllerType//error - use of undeclared type 'controllerType'
if let samVC = sampleVC {
self.present(samVC, animated: true, completion: nil)
}
}
知道是否可以这样做吗?
【问题讨论】:
-
你是根据VC的类型来赋值还是直接存在?
-
@Sh_Khan 到目前为止它刚刚呈现
-
我希望使用一个参数来接受它将呈现的 UIViewController
标签: ios swift uiviewcontroller presentviewcontroller