如何使用swift将UIVIewController名称作为参数传递给特定函数?

问题描述 投票:0回答:1

在我的情况下,我需要将UIVIewController名称和更多字符串值传递给特定的function。我尝试了下面的代码,但没有得到结果。

将参数传递给特定功能

self.accountoptionscall(vcName: UIViewController(), vcIdentifier: "profileviewcontroller", popUpVC: ProfileViewController.self)

func accountoptionscall<T: UIViewController>(vcName: UIViewController,vcIdentifier: String, popUpVC: T.self) {
            let viewcontrollers = self.storyboard!.instantiateViewController(withIdentifier: vcIdentifier) as! vcName
        let navController = UINavigationController(rootViewController: viewcontrollers)
        self.present(navController, animated:true, completion: nil)
    }
ios swift
1个回答
0
投票

我在应用程序中使用它,我认为它可以为您提供帮助。

extension UIViewController {

  /// Load UIViewController type from UIStoryboard
  class func loadFromStoryboard<T: UIViewController>() -> T {
    let name = String(describing: T.self)
    let storybord = UIStoryboard(name: name, bundle: nil)
    if let viewController = storybord.instantiateInitialViewController() as? T {
      return viewController
    } else {
      fatalError("Error:  No initial view controller in \(name) storyboard!")
    }
  }
}

这里如何使用它:

  func loadVC<T: UIViewController>(controller: T) {
    let vc: T = T.loadFromStoryboard()
    let navigationVC = UINavigationController(rootViewController: vc)
    self.present(navController, animated:true, completion: nil)
    // or if use in appDelegate you can do it: window?.rootViewController = navigationVC
  }
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