我有PopUpViewControllerSwift
我希望一次又一次地弹出,直到alreadyMatched
索引达到零。这就是我执行弹出窗口的方式,代码:
var alreadyMatched = [0,1,2]
class QuestionsGame: UIViewController {
var popUpViewController = PopUpViewControllerSwift()
override func viewDidLoad() {
super.viewDidLoad()
matched()
}
func matched() {
var a = alreadyMatched.count
if a > 0 {
self.view.addSubview(self.popUpViewController.view)
self.addChildViewController(self.popUpViewController)
self.popUpViewController.setValues(UIImage(named: "hot.png"), messageText: "You have matched!!", congratsText: "Snap!")
self.popUpViewController.didMoveToParentViewController(self)
alreadyMatched.removeLast()
}
}
}
并且PopUpViewControllerSwift
代码为:
@objc class PopUpViewControllerSwift : UIViewController {
var popUpUserImage: UIImageView!
var messageLabel: UILabel!
var popUpView: UIView!
var congratsLabel: UILabel!
var matchedOrNot = 2
var matchedUser : PFUser!
override func viewDidLoad() {
super.viewDidLoad()
}
func setValues(image : UIImage!, messageText : String, congratsText : String) {
self.popUpUserImage!.image = image
self.messageLabel!.text = messageText
self.congratsLabel.text = congratsText
}
func showAnimate()
{
self.view.transform = CGAffineTransformMakeScale(1.3, 1.3)
self.view.alpha = 0.0;
UIView.animateWithDuration(0.25, animations: {
self.view.alpha = 1.0
self.view.transform = CGAffineTransformMakeScale(1.0, 1.0)
});
}
func removeAnimate()
{
UIView.animateWithDuration(0.25, animations: {
self.view.transform = CGAffineTransformMakeScale(1.3, 1.3)
self.view.alpha = 0.0;
}, completion:{(finished : Bool) in
if (finished)
{
self.view.removeFromSuperview()
let sb = UIStoryboard(name: "Main", bundle: nil)
let questionsVC = sb.instantiateViewControllerWithIdentifier("Questions") as! QuestionsGame
questionsVC.timer()
}
})
}
}
出于某种原因,这只会弹出一次,不会反复出现?我确定它与ParentViewController有关吗?
答案 0 :(得分:0)
您的第一个QuestionsGame
实例正在屏幕上显示。在viewDidLoad中,执行matched()
并将弹出窗口的视图放在此实例中,该实例显示弹出窗口。这可以正常工作。
现在您要再次展示它的部分:您有一个按钮,使用以下代码创建QuestionsGame
的新实例:
let questionsVC = sb.instantiateViewControllerWithIdentifier("Questions") as! QuestionsGame
您再次手动访问matched()
(这导致它被调用两次,一次在viewDidLoad中,一次在手动中)此方法会将您的popover视图放在 new 实例中,但是你不会看到这一切,因为新实例不在视野中。
编辑:
如果要创建新的弹出窗口,使用委托将是一种简单的方法。我已经解释了在另一个answer中使用代理人的问题。我不确定你是如何解雇你的弹出窗口,但这是另一个问题。如果要在弹出窗口中创建新的弹出窗口,可以使用委托来访问方法matched()
,以便新的弹出窗口显示在当前视图的顶部。