ios - 在 Swift 2.2 中呈现弹出窗口时在闭包错误中隐式使用 'self'

Question

如果用户成功重置密码，我会尝试将 View Controller 呈现为弹出窗口。基本上，如果通过电子邮件发送密码重置说明电子邮件一切正常，那么将运行以下代码。但是，我收到了错误

闭包中“self”的隐式用户；使用“.self”使捕获语义明确

在下面的第一行:

    let VC = storyboard?.instantiateViewControllerWithIdentifier("ResetPasswordSuccessPopOver") as! ResetPasswordSuccessPopOverViewController
    VC.preferredContentSize = CGSize(width: UIScreen.mainScreen().bounds.width, height: 100)
    let navController = UINavigationController(rootViewController: VC)
    navController.modalPresentationStyle = UIModalPresentationStyle.Popover
    let popover = navController.popoverPresentationController
    popover?.delegate = self

   self.presentViewController(navController, animated: true, completion: nil)

Answer 1

试试这个

let viewController = UIStoryboard(name: "Main", bundle: nil).instantiateViewControllerWithIdentifier("ResetPasswordSuccessPopOver") as ResetPasswordSuccessPopOverViewController

或者试试这个

  let viewController = self.storyboard!.instantiateViewControllerWithIdentifier("ResetPasswordSuccessPopOver") as ResetPasswordSuccessPopOverViewController