如何判断子视图何时被移除 UIView
基本上我想实现一个弹出 UIView 所以我遵循了这里发布的内容 POP-UP UIView“IMDB App”样式
这非常有效。不过我有一个疑问。我的主要视图是 tableView。因此，当弹出视图时，我禁用表格中的滚动。现在，当弹出子视图被删除时，我需要重新启用滚动。我怎样才能做到这一点？我无法使用 willRemoveFromSuperview 因为弹出视图完全加载不同的 NIB。
我应该使用通知吗？
希望我能清楚地解释这个场景。
提前致谢！
Basically i wanted to implement a popup UIView so i followed what was posted here
POP-UP UIView "IMDB App" style
This works very well. However i have one query. My main view is a tableView. so when a view is popped up i disable scrolling in the table. Now when the popup subView is removed, i need to re-enable scrolling. How do i achieve that? i can't use willRemoveFromSuperview because the popup view is loading a different NIB altogether.
Should i use Notifications?
hope i was clear with explaining the scenario.
Thanks in advance!
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重罪猫有正确答案。这是在弹出视图中使用
@protocol以及注册委托的完美方式。有东西触发该弹出视图关闭。无论触发器是什么，调用协议，委托就可以根据需要处理情况。此外，当正确使用协议时，您的代码在一个项目以及其他项目中变得非常可重用。
Feloneous Cat has the correct answer. This is the perfect use a
@protocolin your popup view along with a registered delegate. Something is triggering that popup view to close. Whatever that trigger is, call the protocol and the delegate can handle the situation as needed.Furthermore, when protocols are used correctly, your code becomes very reusable within a project as well as in other projects.
您可以做的是子类
UIView并重写removeFromSuperview来发送通知。我不认为存在不使用removeFromSuperview方法就删除视图的情况。What you could do is subclass
UIViewand overrideremoveFromSuperviewto send a notification. I don't think there's ever a case where a view gets removed without using theremoveFromSuperviewmethod.