交互式过渡：Segue 声明

Question

理论问题：在下面的代码中，为什么在“performSegue ....”行没有完全执行segue？我的意思是，它怎么知道这是一个交互式过渡，为什么方法“performSegue...”的效果与在其他地方声明的效果不同？

FirstViewController.swift

let transitionManager = TransitionManager()


override func viewDidLoad() {
    super.viewDidLoad()

    transitionManager.sourceViewController = self

    // Do any additional setup after loading the view.
}

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject!) {

    let dest = segue.destinationViewController as UIViewController
    dest.transitioningDelegate = transitionManager
    transitionManager.destViewController = dest

}

故事板

segue 是从 FirstViewController 到 SecondViewController。

• 标识符：“segueIdentifier”

• 转场：模态

• 目的地：当前

TransitionManager.swift

class TransitionManager: UIPercentDrivenInteractiveTransition,UIViewControllerAnimatedTransitioning, UIViewControllerTransitioningDelegate,UIViewControllerInteractiveTransitioning {

    var interactive = false
    var presenting = false

    var panGesture : UIPanGestureRecognizer!

    var sourceViewController : UIViewController! {
        didSet {

            panGesture = UIPanGestureRecognizer(target: self, action: "gestureHandler:")
            sourceViewController.view.addGestureRecognizer(panGesture)

        }
    }


    func gestureHandler(pan : UIPanGestureRecognizer) {

        let translation = pan.translationInView(pan.view!)

        let d =  translation.x / pan.view!.bounds.width * 0.5        

        switch pan.state {

        case UIGestureRecognizerState.Began :

            interactive = true

            sourceViewController.performSegueWithIdentifier("segueIdentifier", sender: self)


        case UIGestureRecognizerState.Changed :

            self.updateInteractiveTransition(d)

        default :

            interactive = false

            if d > 0.2 || velocity.x > 0 {
                self.finishInteractiveTransition()
            }
            else {
                self.cancelInteractiveTransition()
            }
        }

    }

    func animateTransition(transitionContext: UIViewControllerContextTransitioning) {

        //Something....


        UIView.animateWithDuration(1.0, animations: {

            //Something....

            }, completion: {}

        })

    }

    func transitionDuration(transitionContext: UIViewControllerContextTransitioning) -> NSTimeInterval {
        return 1
    }

    // MARK: UIViewControllerTransitioningDelegate protocol methods

    func animationControllerForPresentedController(presented: UIViewController, presentingController presenting: UIViewController, sourceController source: UIViewController) -> UIViewControllerAnimatedTransitioning? {
        self.presenting = true
        return self
    }

    func animationControllerForDismissedController(dismissed: UIViewController) -> UIViewControllerAnimatedTransitioning? {
        self.presenting = false
        return self
    }

    func interactionControllerForPresentation(animator: UIViewControllerAnimatedTransitioning) -> UIViewControllerInteractiveTransitioning? {
        return self.interactive ? self : nil
    }

    func interactionControllerForDismissal(animator: UIViewControllerAnimatedTransitioning) -> UIViewControllerInteractiveTransitioning? {
        return self.interactive ? self : nil
    }





}

事实上，我有很多类似的代码，这个正在工作，但其他非常相似的代码正是我所解释的：segue 只是在没有交互转换的情况下执行。谁能解释一下？

感谢您的帮助

Answer 1

如果问题是您的交互式过渡动画不是交互式的，我建议在interactionControllerForPresentation 中放置一个断点：

• 确保interactionControllerForPresentation 的正确实例以您认为应该的方式调用。

  例如，确保目标的viewDidLoad 没有替换您在源控制器中仔细设置的prepareForSegue 中的transitioningDelegate ...如果从A 进行交互式转换，这很容易做到到 B，然后再从 B 到 C，您在 B 到 C 的预期中设置的内容可能会干扰您打算从 A 到 B 执行的操作。

• 确保 interactionControllerForPresentation 返回您认为应该返回的值（即确保 interactive 绝对是 true）。

每次我看到手势的这种行为没有以交互方式启动转换，就像我认为应该的那样，这是因为当在目标控制器的 transitioningDelegate 中调用 interactionControllerForPresentation 时，它没有返回有效的UIViewControllerInteractiveTransitioning（如UIPercentDrivenInteractiveTransition），如我所愿。

无论如何，您可以通过在interactionControllerForPresentation 中添加一条日志语句来诊断此问题，并且您会看到（a）由于某种原因它没有被调用；或 (b) self.interactive 不是 true。