减慢推推转场的过渡 [英] Slowing Down Transition of Push Segue
问题描述
我想知道是否有一种方法可以减慢'push segue&像Tinder& Snapchat确实如此.就像正常的segue，但它以某种方式缓慢地过渡.
I was wondering if there is a way to slow down 'push segue & back' like Tinder & Snapchat does. It's like normal segue but it's somehow transiting slowly.
我通常通过在Storyboard上设置Push通知并以编程方式调用segue(单击NavBarItem并将其推送到新的UIViewController)来处理push segues :
I normally handle push segues with setting up a Push notification on Storyboard and programatically calling the segue (clicked on NavBarItem and push segued to new UIViewController):
func settingsTapped() {
performSegueWithIdentifier("NewsToSettings", sender: nil)
}
有一种方法
It there a way to
-
为整个应用启用全局减慢速度吗?
enable slowing down segue globally for the whole app?
还是我应该每次都按部就班?如果是这样，适当的方法是什么?
or should I do it maually for each segue? If so, what is the appropriate way?
推荐答案
我想知道是否有一种方法可以减慢'push segue&像Tinder& Snapchat确实如此.就像正常的搜寻一样，但是它以某种方式缓慢地过渡.
I was wondering if there is a way to slow down 'push segue & back' like Tinder & Snapchat does. It's like normal segue but it's somehow transiting slowly.
如前所述，您只需编写自定义过渡动画.给导航控制器一个委托并实现navigationController:animationControllerForOperation:fromViewController:toViewController:.基本上，您只需要提供一个实现UIViewControllerAnimatedTransitioning协议的对象即可.作为其中的一部分，您可以指定动画的持续时间.
As you've been told, you simply write a custom transition animation. Give the navigation controller a delegate and implement navigationController:animationControllerForOperation:fromViewController:toViewController:. Basically then you just supply an object that implements the UIViewControllerAnimatedTransitioning protocol. As part of that, you get to specify the animation duration.
为整个应用启用全局减慢速度吗?
enable slowing down segue globally for the whole app?
如果所有导航控制器都具有相同的委托，或者如果所有委托都提供了相同的UIViewControllerAnimatedTransitioning对象，则所有push/pop转换将使用相同的代码，并且将以相同的方式工作.
If all your navigation controllers have this same delegate, or if all the delegates supply this same UIViewControllerAnimatedTransitioning object, all push / pop transitions will use the same code and will work the same way.
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