ios - 关闭弹出 View

Question

我以编程方式添加了一个弹出 View ，当我单击弹出窗口中的按钮时，它将显示另一个警报 View Controller 。但问题是我无法忽略最初添加的弹出 View 。它仍然显示在警报 View 后面。谁能帮我解决这个问题吗？

使用下面的代码，我创建了弹出 View 。

   let storyboard : UIStoryboard = UIStoryboard(name: "Main",bundle: nil)
        let menuViewController: PopupViewController = storyboard.instantiateViewControllerWithIdentifier("Popup") as! PopupViewController
        menuViewController.modalPresentationStyle = UIModalPresentationStyle.Popover
        menuViewController.preferredContentSize = CGSizeMake(300, 275)

        let popoverMenuViewController = menuViewController.popoverPresentationController
        popoverMenuViewController?.delegate = self
        popoverMenuViewController?.sourceView = self.view
        // If wants the arrow head remove the below line code
        popoverMenuViewController?.permittedArrowDirections = UIPopoverArrowDirection()
        popoverMenuViewController?.sourceRect = CGRectMake(CGRectGetMidX(self.view.bounds), CGRectGetMidY(self.view.bounds),0,0)
        presentViewController(menuViewController,animated: true,completion: nil)

Answer 1

这可能对你有帮助

let alertController = UIAlertController(title: "My alert", message: "some text...", preferredStyle: .Alert)
// .Default will dismiss the alert automatically
let myAlertAction = UIAlertAction(title: "Alert", style: .Default, handler: { (UIAlertAction) in
    // Some stuff, present your other alert
    // self.myMainViewController.presentViewController...
 })
alertController.addAction(myAlertAction)

我就是这么做的，效果很好