blocks|key|1461798|text|是的，superview是包含您的视图的视图。你的视图不应该确切知道哪个是它的视图控制器，因为这会违反MVC原则。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1461799|另一方面，控制器知道它负责哪个视图(self.view+=+myView)，通常，此视图将方法/事件委托给控制器处理。|1461800|通常，您应该有一个指向控制器的指针，而不是指向视图的指针，控制器可以执行某些控制逻辑，也可以将某些内容传递给视图。|1461801|entityMap^0|3|9|0|I|I|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|R|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|K|$]]

Yes, the <code>superview</code> is the view that contains your view. Your view shouldn't know which exactly is its view controller, because that would break MVC principles.

The controller, on the other hand, knows which view it's responsible for (<code>self.view = myView</code>), and usually, this view delegates methods/events for handling to the controller.

Typically, instead of a pointer to your view, you should have a pointer to your controller, which in turn can either execute some controlling logic, or pass something to its view.

blocks|key|1461839|text|我认为你可以将点击传递给视图控制器，让它来处理它。这是一种更容易接受的方法。至于从视图访问视图控制器，您应该维护对视图控制器的引用，因为没有其他方法。看看这个帖子，它可能会有帮助：Accessing+view+controller+from+a+view|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1461840|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/1405942/iphone-sdk-calling-a-method-of-views-superviews-viewcontroller/1406502#1406502^0|2I|11|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

I think you can propagate the tap to the view controller and let it handle it. This is more acceptable approach. As for accessing a view controller from its view, you should maintain a reference to a view controller, since there is no another way. See this thread, it might help:
<a href="https://stackoverflow.com/questions/1405942/iphone-sdk-calling-a-method-of-views-superviews-viewcontroller/1406502#1406502">Accessing view controller from a view</a>

blocks|key|1461878|text|如果设置了断点，则可以将其粘贴到调试器中以打印视图层次结构：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1461879|po+[[UIWindow+keyWindow]+recursiveDescription]|code-block|syntax|javascript|1461880|您应该能够在混乱中的某个地方找到视图的父视图:)|1461881|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you set a breakpoint, you can paste this into the debugger to print the view hierarchy:

<pre><code>po [[UIWindow keyWindow] recursiveDescription]
</code></pre>

You should be able to find your view's parent somewhere in that mess :)

blocks|key|1247262|text|要获得具有UIView的UIViewController的引用，您可以扩展UIResponder+(这是UIView和UIViewController的超类)，它允许通过响应器链向上移动，从而到达UIViewController(否则返回nil)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1247263|extension+UIResponder+{
++++func+getParentViewController()+->+UIViewController?+{
++++++++if+self.nextResponder()+is+UIViewController+{
++++++++++++return+self.nextResponder()+as?+UIViewController
++++++++}+else+{
++++++++++++if+self.nextResponder()+!=+nil+{
++++++++++++++++return+(self.nextResponder()!).getParentViewController()
++++++++++++}
++++++++++++else+{return+nil}
++++++++}
++++}
}

//Swift+3
extension+UIResponder+{
++++func+getParentViewController()+->+UIViewController?+{
++++++++if+self.next+is+UIViewController+{
++++++++++++return+self.next+as?+UIViewController
++++++++}+else+{
++++++++++++if+self.next+!=+nil+{
++++++++++++++++return+(self.next!).getParentViewController()
++++++++++++}
++++++++++++else+{return+nil}
++++++++}
++++}
}

let+vc+=+UIViewController()
let+view+=+UIView()
vc.view.addSubview(view)
view.getParentViewController()+//provide+reference+to+vc|code-block|syntax|javascript|1247264|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

To get reference to UIViewController having UIView, you could make extension of UIResponder (which is super class for UIView and UIViewController), which allows to go up through the responder chain and thus reaching UIViewController (otherwise returning nil).

<pre><code>extension UIResponder {
 func getParentViewController() -&gt; UIViewController? {
 if self.nextResponder() is UIViewController {
 return self.nextResponder() as? UIViewController
 } else {
 if self.nextResponder() != nil {
 return (self.nextResponder()!).getParentViewController()
 }
 else {return nil}
 }
 }
}

//Swift 3
extension UIResponder {
 func getParentViewController() -&gt; UIViewController? {
 if self.next is UIViewController {
 return self.next as? UIViewController
 } else {
 if self.next != nil {
 return (self.next!).getParentViewController()
 }
 else {return nil}
 }
 }
}

let vc = UIViewController()
let view = UIView()
vc.view.addSubview(view)
view.getParentViewController() //provide reference to vc
</code></pre>

blocks|key|1461965|text|Swift+3.0的更多类型安全代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1461966|extension+UIResponder+{
++++func+owningViewController()+->+UIViewController?+{
++++++++var+nextResponser+=+self
++++++++while+let+next+=+nextResponser.next+{
++++++++++++nextResponser+=+next
++++++++++++if+let+vc+=+nextResponser+as?+UIViewController+{
++++++++++++++++return+vc
++++++++++++}
++++++++}
++++++++return+nil
++++}
}|code-block|syntax|javascript|1461967|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

More type safe code for Swift 3.0

<pre><code>extension UIResponder {
 func owningViewController() -&gt; UIViewController? {
 var nextResponser = self
 while let next = nextResponser.next {
 nextResponser = next
 if let vc = nextResponser as? UIViewController {
 return vc
 }
 }
 return nil
 }
}
</code></pre>

blocks|key|1461976|text|Swift+3中快速而通用的方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1461977|extension+UIResponder+{
++++func+parentController<T:+UIViewController>(of+type:+T.Type)+->+T?+{
++++++++guard+let+next+=+self.next+else+{
++++++++++++return+nil
++++++++}
++++++++return+(next+as?+T)+??+next.parentController(of:+T.self)
++++}
}

//Use:
class+MyView:+UIView+{
++++...
++++let+parentController+=+self.parentController(of:+MyViewController.self)
}|code-block|syntax|javascript|1461978|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The fast and generic way in Swift 3:

<pre><code>extension UIResponder {
 func parentController&lt;T: UIViewController&gt;(of type: T.Type) -&gt; T? {
 guard let next = self.next else {
 return nil
 }
 return (next as? T) ?? next.parentController(of: T.self)
 }
}

//Use:
class MyView: UIView {
 ...
 let parentController = self.parentController(of: MyViewController.self)
}
</code></pre>

blocks|key|1462037|text|遗憾的是，这是不可能的，除非您子类化视图并为其提供一个实例属性或类似的属性，一旦视图添加到场景中，就会在其中存储视图控制器引用……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1462038|在大多数情况下-很容易解决这篇文章的原始问题，因为大多数视图控制器对于负责向ViewController的视图添加任何子视图的程序员来说都是众所周知的实体;-)这就是为什么我猜Apple从来没有费心添加那个属性的原因。|1462039|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Alas, this is impossible unless you subclass the view and provide it with an instance property or alike which stores the view controller reference inside it once the view is added to the scene...

In most cases - it is very easy to work around the original problem of this post as most view controllers are well-known entities to the programmer who was responsible for adding any subviews to the ViewController's View ;-) Which is why I guess that Apple never bothered to add that property.

blocks|key|1247422|text|有点晚了，但这里有一个扩展，可以让您找到任何类型的响应器，包括ViewController。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1247423|extension+NSObject{
func+findNext(type:+AnyClass)+->+Any{
++++var+resp+=+self+as!+UIResponder

++++while+!resp.isKind(of:+type.self)+&&+resp.next+!=+nil
++++{
++++++++resp+=+resp.next!
++++}

++++return+resp
++}+++++++++++++++++++++++
}|code-block|syntax|javascript|1247424|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

A little bit late, but here's an extension that enable you to find a responder of any type, including ViewController.

<pre><code>extension NSObject{
func findNext(type: AnyClass) -&gt; Any{
 var resp = self as! UIResponder

 while !resp.isKind(of: type.self) &amp;&amp; resp.next != nil
 {
 resp = resp.next!
 }

 return resp
 } 
}
</code></pre>

I have a pointer to a <code>UIView</code>. How do I access its <code>UIViewController</code>? <code>[self superview]</code> is another <code>UIView</code>, but not the <code>UIViewController</code>, right?

Given a view, how do I get its viewController?

调试器

Swift

我有一个指向UIView的指针。如何访问它的UIViewController？[self superview]是另一个UIView，但不是UIViewController，对吧？

问给定一个视图，我如何获取它的viewController？
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问给定一个视图，我如何获取它的viewController？
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