blocks|key|1252914|text|没有办法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1252915|我所做的是将UIViewController指针传递给UIView+(或适当的继承)。很抱歉，我不能用IB方法来解决这个问题，因为我不相信IB。|1252916|回答第一个评论者:有时你确实需要知道是谁给你打电话的，因为它决定了你能做什么。例如，对于数据库，您可能只有读访问权限或读/写...|1252917|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

There is no way.

What I do is pass the UIViewController pointer to the UIView (or an appropriate inheritance). I'm sorry I can't help with the IB approach to the problem because I don't believe in IB.

To answer the first commenter: sometimes you do need to know who called you because it determines what you can do. For example with a database you might have read access only or read/write ...

blocks|key|1252969|text|我的解决方案可能会被认为是虚假的，但我遇到了与mayoneez类似的情况(我想切换视图以响应EAGLView中的一个手势)，我以这种方式获得了EAGL的视图控制器：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1252970|EAGLViewController+*vc+=+((EAGLAppDelegate*)[[UIApplication+sharedApplication]+delegate]).viewController;|code-block|syntax|javascript|1252971|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|$]]

My solution would probably be considered kind of bogus but I had a similar situation as mayoneez (I wanted to switch views in response to a gesture in an EAGLView), and I got the EAGL's view controller this way:

<pre><code>EAGLViewController *vc = ((EAGLAppDelegate*)[[UIApplication sharedApplication] delegate]).viewController;
</code></pre>

blocks|key|1253018|text|UIView是UIResponder的一个子类。UIResponder通过一个返回nil的实现展示了-nextResponder方法。UIView覆盖了此方法，如UIResponder+(出于某种原因而不是在UIView中)中所述:如果视图具有视图控制器，则由-nextResponder返回它。如果没有视图控制器，该方法将返回superview。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1253019|将此代码添加到您的项目中，您就可以开始工作了。|1253020|@interface+UIView+(APIFix)
-+(UIViewController+*)viewController;
@end

@implementation+UIView+(APIFix)

-+(UIViewController+*)viewController+{
++++if+([self.nextResponder+isKindOfClass:UIViewController.class])
++++++++return+(UIViewController+*)self.nextResponder;
++++else
++++++++return+nil;
}
@end|code-block|syntax|javascript|1253021|现在UIView有了返回视图控制器的工作方法。|1253022|entityMap^0|0|6|7|B|O|B|15|3|1E|E|1V|6|29|B|2W|6|3M|E|0|0|0|2|6|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|19|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|P|$]]

<code>UIView</code> is a subclass of <code>UIResponder</code>. <code>UIResponder</code> lays out the method <code>-nextResponder</code> with an implementation that returns <code>nil</code>. <code>UIView</code> overrides this method, as documented in <code>UIResponder</code> (for some reason instead of in <code>UIView</code>) as follows: if the view has a view controller, it is returned by <code>-nextResponder</code>. If there is no view controller, the method will return the superview. 

Add this to your project and you're ready to roll. 

<pre><code>@interface UIView (APIFix)
- (UIViewController *)viewController;
@end

@implementation UIView (APIFix)

- (UIViewController *)viewController {
 if ([self.nextResponder isKindOfClass:UIViewController.class])
 return (UIViewController *)self.nextResponder;
 else
 return nil;
}
@end
</code></pre>

Now <code>UIView</code> has a working method for returning the view controller.

blocks|key|1467899|text|使用Brock发布的示例，我对其进行了修改，使其成为UIView而不是UIViewController的类别，并使其递归，以便任何子视图都可以(希望)找到父UIViewController。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1467900|@interface+UIView+(FindUIViewController)
-+(UIViewController+*)+firstAvailableUIViewController;
@end

@implementation+UIView+(FindUIViewController)
-+(UIViewController+*)+firstAvailableUIViewController+{
++++UIResponder+*responder+=+[self+nextResponder];
++++while+(responder+!=+nil)+{
++++++++if+([responder+isKindOfClass:[UIViewController+class]])+{
++++++++++++return+(UIViewController+*)responder;
++++++++}
++++++++responder+=+[responder+nextResponder];
++++}
++++return+nil;
}

@end|code-block|syntax|javascript|1467901|要使用此代码，请将其添加到一个新的类文件中(我将我的类命名为"UIKitCategories")，并删除类数据……将@接口复制到标题中，并将@实现复制到.m文件中。然后在你的项目中，#import+"UIKitCategories.h“并在UIView代码中使用：|1467902|//+from+a+UIView+subclass...+returns+nil+if+UIViewController+not+available
UIViewController+*+myController+=+[self+firstAvailableUIViewController];|1467903|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Using the example posted by Brock, I modified it so that it is a category of UIView instead UIViewController and made it recursive so that any subview can (hopefully) find the parent UIViewController.
<pre><code>@interface UIView (FindUIViewController)
- (UIViewController *) firstAvailableUIViewController;
@end

@implementation UIView (FindUIViewController)
- (UIViewController *) firstAvailableUIViewController {
 UIResponder *responder = [self nextResponder];
 while (responder != nil) {
 if ([responder isKindOfClass:[UIViewController class]]) {
 return (UIViewController *)responder;
 }
 responder = [responder nextResponder];
 }
 return nil;
}

@end
</code></pre>
To use this code, add it into an new class file (I named mine &quot;UIKitCategories&quot;) and remove the class data... copy the @interface into the header, and the @implementation into the .m file. Then in your project, #import &quot;UIKitCategories.h&quot; and use within the UIView code:
<pre><code>// from a UIView subclass... returns nil if UIViewController not available
UIViewController * myController = [self firstAvailableUIViewController];
</code></pre>

blocks|key|1467972|text|不要忘记，您可以访问视图所在窗口的根视图控制器。在那里，如果你正在使用一个导航视图控制器，并且想要把一个新的视图推到它上面：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1467973|++++[[[[self+window]+rootViewController]+navigationController]+pushViewController:newController+animated:YES];|code-block|syntax|javascript|1467974|但是，您首先需要正确设置窗口的rootViewController属性。当你第一次创建控制器时，例如在你的应用委托中这样做：|1467975|-(void)+applicationDidFinishLaunching:(UIApplication+*)application+{
++++window+=+[[UIWindow+alloc]+initWithFrame:[[UIScreen+mainScreen]+bounds]];
++++RootViewController+*controller+=+[[YourRootViewController]+alloc]+init];
++++[window+setRootViewController:+controller];
++++navigationController+=+[[UINavigationController+alloc]+initWithRootViewController:rootViewController];
++++[controller+release];
++++[window+addSubview:[[self+navigationController]+view]];
++++[window+makeKeyAndVisible];
}|1467976|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Don't forget that you can get access to the root view controller for the window that the view is a subview of. From there, if you are e.g. using a navigation view controller and want to push a new view onto it:

<pre><code> [[[[self window] rootViewController] navigationController] pushViewController:newController animated:YES];
</code></pre>

You will need to set up the rootViewController property of the window properly first, however. Do this when you first create the controller e.g. in your app delegate:

<pre><code>-(void) applicationDidFinishLaunching:(UIApplication *)application {
 window = [[UIWindow alloc] initWithFrame:[[UIScreen mainScreen] bounds]];
 RootViewController *controller = [[YourRootViewController] alloc] init];
 [window setRootViewController: controller];
 navigationController = [[UINavigationController alloc] initWithRootViewController:rootViewController];
 [controller release];
 [window addSubview:[[self navigationController] view]];
 [window makeKeyAndVisible];
}
</code></pre>

blocks|key|1253240|text|我认为有这样一种情况，即被观察者需要通知观察者。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1253241|我看到了一个类似的问题，UIViewController中的UIView正在响应一种情况，它需要首先告诉它的父视图控制器隐藏back按钮，然后在完成时告诉父视图控制器它需要将自己弹出堆栈。|1253242|我一直在与代表们尝试这一点，但没有成功。|1253243|我不明白为什么这是个坏主意？|1253244|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

I think there is a case when the observed needs to inform the observer.

I see a similar problem where the UIView in a UIViewController is responding to a situation and it needs to first tell its parent view controller to hide the back button and then upon completion tell the parent view controller that it needs to pop itself off the stack.

I have been trying this with delegates with no success.

I don't understand why this should be a bad idea?

blocks|key|1468097|text|我建议一种更轻量级的方法来遍历完整的响应器链，而不必在UIView上添加类别：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1468098|@implementation+MyUIViewSubclass

-+(UIViewController+*)viewController+{
++++UIResponder+*responder+=+self;
++++while+(![responder+isKindOfClass:[UIViewController+class]])+{
++++++++responder+=+[responder+nextResponder];
++++++++if+(nil+==+responder)+{
++++++++++++break;
++++++++}
++++}
++++return+(UIViewController+*)responder;
}

@end|code-block|syntax|javascript|1468099|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|$]]

I would suggest a more lightweight approach for traversing the complete responder chain without having to add a category on UIView:

<pre><code>@implementation MyUIViewSubclass

- (UIViewController *)viewController {
 UIResponder *responder = self;
 while (![responder isKindOfClass:[UIViewController class]]) {
 responder = [responder nextResponder];
 if (nil == responder) {
 break;
 }
 }
 return (UIViewController *)responder;
}

@end
</code></pre>

blocks|key|1253354|text|我不认为在某些情况下找出谁是视图控制器是个“坏”主意。一个坏主意是保存对这个控制器的引用，因为它可能会随着超级视图的改变而改变。在我的例子中，我有一个遍历响应器链的getter。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1253355|//.h|1253356|@property+(nonatomic,+readonly)+UIViewController+*+viewController;|code-block|syntax|javascript|1253357|//.m|1253358|-+(UIViewController+*)viewController
{
++++for+(UIResponder+*+nextResponder+=+self.nextResponder;
+++++++++nextResponder;
+++++++++nextResponder+=+nextResponder.nextResponder)
++++{
++++++++if+([nextResponder+isKindOfClass:[UIViewController+class]])
++++++++++++return+(UIViewController+*)nextResponder;
++++}

++++//+Not+found
++++NSLog(@"%25@+doesn't+seem+to+have+a+viewController".+self);
++++return+nil;
}|1253359|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I don't think it's "bad" idea to find out who is the view controller for some cases. What could be a bad idea is to save the reference to this controller as it could change just as superviews change.
In my case I have a getter that traverses the responder chain.

//.h

<pre><code>@property (nonatomic, readonly) UIViewController * viewController;
</code></pre>

//.m

<pre><code>- (UIViewController *)viewController
{
 for (UIResponder * nextResponder = self.nextResponder;
 nextResponder;
 nextResponder = nextResponder.nextResponder)
 {
 if ([nextResponder isKindOfClass:[UIViewController class]])
 return (UIViewController *)nextResponder;
 }

 // Not found
 NSLog(@"%@ doesn't seem to have a viewController". self);
 return nil;
}
</code></pre>

blocks|key|1468200|text|另一种简单的方法是拥有自己的视图类，并在视图类中添加视图控制器的属性。通常情况下，视图控制器创建视图，这是控制器可以将自己设置为属性的地方。基本上，它不是到处寻找控制器(通过一些破解)，而是让控制器将自己设置为视图-这很简单，但有意义，因为它是控制器“控制”视图。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1468201|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Another easy way is to have your own view class and add a property of the view controller in the view class. Usually the view controller creates the view and that is where the controller can set itself to the property. Basically it is instead of searching around (with a bit of hacking) for the controller, having the controller to set itself to the view - this is simple but makes sense because it is the controller that "controls" the view.

blocks|key|1253528|text|这并没有直接回答问题，而是对问题的意图进行了假设。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1253529|如果您有一个视图，且在该视图中需要调用另一个对象上的方法，例如视图控制器，则可以使用NSNotificationCenter代替。|1253530|首先在头文件中创建通知字符串|1253531|#define+SLCopyStringNotification+@"ShaoloCopyStringNotification"|code-block|syntax|javascript|1253532|在您的视图中调用postNotificationName：|1253533|-+(IBAction)+copyString:(id)sender
{
++++[[NSNotificationCenter+defaultCenter]+postNotificationName:SLCopyStringNotification+object:nil];
}|1253534|然后在视图控制器中添加一个观察者。我在viewDidLoad中做到了这一点|1253535|-+(void)viewDidLoad
{
++++[[NSNotificationCenter+defaultCenter]+addObserver:self
+++++++++++++++++++++++++++++++++++++++++++++selector:@selector(copyString:)
+++++++++++++++++++++++++++++++++++++++++++++++++name:SLCopyStringNotification
+++++++++++++++++++++++++++++++++++++++++++++++object:nil];
}|1253536|现在(同样在同一个视图控制器中)实现您的方法copyString:如上面的@selector所示。|1253537|-+(IBAction)+copyString:(id)sender
{
++++CalculatorResult*+result+=+(CalculatorResult*)[[PercentCalculator+sharedInstance].arrayTableDS+objectAtIndex:([self.viewTableResults+indexPathForSelectedRow].row)];
++++UIPasteboard+*gpBoard+=+[UIPasteboard+generalPasteboard];
++++[gpBoard+setString:result.stringResult];
}|1253538|我并不是说这是做这件事的正确方式，只是它看起来比运行第一响应器链更干净。我使用这段代码在UITableView上实现了一个UIMenuController，并将事件传递回UIViewController，这样我就可以对数据执行一些操作。|1253539|entityMap^0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|12|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|13|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|15|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|R|5|H|7|17|8|@]|9|@]|A|$I|J]]|$1|S|3|T|5|6|7|18|8|@]|9|@]|A|$]]|$1|U|3|V|5|H|7|19|8|@]|9|@]|A|$I|J]]|$1|W|3|X|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

This doesn't answer the question directly, but rather makes an assumption about the intent of the question.

If you have a view and in that view you need to call a method on another object, like say the view controller, you can use the NSNotificationCenter instead.

First create your notification string in a header file

<pre><code>#define SLCopyStringNotification @"ShaoloCopyStringNotification"
</code></pre>

In your view call postNotificationName:

<pre><code>- (IBAction) copyString:(id)sender
{
 [[NSNotificationCenter defaultCenter] postNotificationName:SLCopyStringNotification object:nil];
}
</code></pre>

Then in your view controller you add an observer. I do this in viewDidLoad

<pre><code>- (void)viewDidLoad
{
 [[NSNotificationCenter defaultCenter] addObserver:self
 selector:@selector(copyString:)
 name:SLCopyStringNotification
 object:nil];
}
</code></pre>

Now (also in the same view controller) implement your method copyString: as depicted in the @selector above.

<pre><code>- (IBAction) copyString:(id)sender
{
 CalculatorResult* result = (CalculatorResult*)[[PercentCalculator sharedInstance].arrayTableDS objectAtIndex:([self.viewTableResults indexPathForSelectedRow].row)];
 UIPasteboard *gpBoard = [UIPasteboard generalPasteboard];
 [gpBoard setString:result.stringResult];
}
</code></pre>

I'm not saying this is the right way to do this, it just seems cleaner than running up the first responder chain. I used this code to implement a UIMenuController on a UITableView and pass the event back up to the UIViewController so I can do something with the data.

blocks|key|1468373|text|对于Phil的回答：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1468374|行：id+nextResponder+=+[self+nextResponder];+if+self(UIView)不是视图控制器视图的子视图，如果你知道自己的层次结构(UIView)，你也可以使用：id+nextResponder+=+[[self+superview]+nextResponder];...|offset|length|style|CODE|1468375|entityMap^0|0|2|14|2S|1G|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

To Phil's answer:

In line: <code>id nextResponder = [self nextResponder];</code> if self(UIView) is not a subview of ViewController's view, if you know hierarchy of self(UIView) you can use also: <code>id nextResponder = [[self superview] nextResponder];</code>...

blocks|key|1468451|text|用于查找viewController的最简单的do+while循环。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1468452|-(UIViewController*)viewController
{
++++UIResponder+*nextResponder+=++self;

++++do
++++{
++++++++nextResponder+=+[nextResponder+nextResponder];

++++++++if+([nextResponder+isKindOfClass:[UIViewController+class]])
++++++++++++return+(UIViewController*)nextResponder;

++++}+while+(nextResponder+!=+nil);

++++return+nil;
}|code-block|syntax|javascript|1468453|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 simplest do while loop for finding the viewController.

<pre><code>-(UIViewController*)viewController
{
 UIResponder *nextResponder = self;

 do
 {
 nextResponder = [nextResponder nextResponder];

 if ([nextResponder isKindOfClass:[UIViewController class]])
 return (UIViewController*)nextResponder;

 } while (nextResponder != nil);

 return nil;
}
</code></pre>

blocks|key|1253768|text|如果您的rootViewController是在AppDelegate类中设置的UINavigationViewController，那么|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1253769|++++%2B+(UIViewController+*)+getNearestViewController:(Class)+c+{
NSArray+*arrVc+=+[[[[UIApplication+sharedApplication]+keyWindow]+rootViewController]+childViewControllers];

for+(UIViewController+*v+in+arrVc)
{
++++if+([v+isKindOfClass:c])
++++{
++++++++return+v;
++++}
}

return+nil;}|code-block|syntax|javascript|1253770|其中c需要视图控制器类。|1253771|用法：|1253772|+++++RequiredViewController*+rvc+=+[Utilities+getNearestViewController:[RequiredViewController+class]];|1253773|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If your rootViewController is UINavigationViewController, which was set up in AppDelegate class, then

<pre><code> + (UIViewController *) getNearestViewController:(Class) c {
NSArray *arrVc = [[[[UIApplication sharedApplication] keyWindow] rootViewController] childViewControllers];

for (UIViewController *v in arrVc)
{
 if ([v isKindOfClass:c])
 {
 return v;
 }
}

return nil;}
</code></pre>

Where c required view controllers class.

USAGE:

<pre><code> RequiredViewController* rvc = [Utilities getNearestViewController:[RequiredViewController class]];
</code></pre>

blocks|key|1468566|text|这肯定是一个糟糕的想法和错误的设计，但我相信我们都可以享受@Phil_M提出的最佳答案的快速解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1468567|static+func+firstAvailableUIViewController(fromResponder+responder:+UIResponder)+->+UIViewController?+{
++++func+traverseResponderChainForUIViewController(responder:+UIResponder)+->+UIViewController?+{
++++++++if+let+nextResponder+=+responder.nextResponder()+{
++++++++++++if+let+nextResp+=+nextResponder+as?+UIViewController+{
++++++++++++++++return+nextResp
++++++++++++}+else+{
++++++++++++++++return+traverseResponderChainForUIViewController(nextResponder)
++++++++++++}
++++++++}
++++++++return+nil
++++}

++++return+traverseResponderChainForUIViewController(responder)
}|code-block|syntax|javascript|1468568|如果您的目的是做一些简单的事情，比如显示一个模式对话框或跟踪数据，这并不能证明使用协议是合理的。我个人将这个函数存储在一个实用程序对象中，您可以在实现UIResponder协议的任何东西中使用它，如下所示：|1468569|if+let+viewController+=+MyUtilityClass.firstAvailableUIViewController(self)+{}|1468570|所有功劳都归于@Phil_M|1468571|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

It's surely a bad idea and a wrong design, but I'm sure we can all enjoy a Swift solution of the best answer proposed by @Phil_M:

<pre><code>static func firstAvailableUIViewController(fromResponder responder: UIResponder) -&gt; UIViewController? {
 func traverseResponderChainForUIViewController(responder: UIResponder) -&gt; UIViewController? {
 if let nextResponder = responder.nextResponder() {
 if let nextResp = nextResponder as? UIViewController {
 return nextResp
 } else {
 return traverseResponderChainForUIViewController(nextResponder)
 }
 }
 return nil
 }

 return traverseResponderChainForUIViewController(responder)
}
</code></pre>

If your intention is to do simple things, as showing a modal dialog or tracking data, that doesn't justify the use of a protocol. I personally store this function in an utility object, you can use it from anything that implement the UIResponder protocol as:

<pre><code>if let viewController = MyUtilityClass.firstAvailableUIViewController(self) {}
</code></pre>

All credit to @Phil_M

blocks|key|1468622|text|我偶然发现我有一个想要重用的小组件，并在一个可重用的视图本身中添加了一些代码(它实际上只是一个打开PopoverController的按钮)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1468623|虽然这在iPad中工作得很好(+UIPopoverController显示自己，因此不需要引用UIViewController)，但让相同的代码工作意味着突然从UIViewController引用您的presentViewController。有点不一致，对吧？|1468624|正如前面提到的，在UIView中加入逻辑并不是最好的方法。但是，将所需的几行代码包装在一个单独的控制器中，感觉真的没有用。|1468625|无论哪种方式，这里都有一个快速的解决方案，它为任何UIView添加了一个新属性：|1468626|extension+UIView+{

++++var+viewController:+UIViewController?+{

++++++++var+responder:+UIResponder?+=+self

++++++++while+responder+!=+nil+{

++++++++++++if+let+responder+=+responder+as?+UIViewController+{
++++++++++++++++return+responder
++++++++++++}
++++++++++++responder+=+responder?.nextResponder()
++++++++}
++++++++return+nil
++++}
}|code-block|syntax|javascript|1468627|entityMap^0|1D|H|0|G|J|1B|G|28|G|2S|L|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|14|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|15|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|16|8|@]|D|@]|E|$O|P]]|$1|Q|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|R|$]]

I stumbled upon a situation where I have a small component I want to reuse, and added some code in a reusable view itself(it's really not much more than a button that opens a <code>PopoverController</code>).

While this works fine in the iPad (the <code>UIPopoverController</code> presents itself, therefor needs no reference to a <code>UIViewController</code>), getting the same code to work means suddenly referencing your <code>presentViewController</code> from your <code>UIViewController</code>. Kinda inconsistent right?

Like mentioned before, it's not the best approach to have logic in your UIView. But it felt really useless to wrap the few lines of code needed in a separate controller.

Either way, here's a swift solution, which adds a new property to any UIView:

<pre><code>extension UIView {

 var viewController: UIViewController? {

 var responder: UIResponder? = self

 while responder != nil {

 if let responder = responder as? UIViewController {
 return responder
 }
 responder = responder?.nextResponder()
 }
 return nil
 }
}
</code></pre>

blocks|key|1253928|text|也许我来晚了。但在这种情况下，我不喜欢分类(污染)。我喜欢这样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1253929|#define+UIViewParentController(__view)+({+\
UIResponder+*__responder+=+__view;+\
while+([__responder+isKindOfClass:[UIView+class]])+\
__responder+=+[__responder+nextResponder];+\
(UIViewController+*)__responder;+\
})|code-block|syntax|javascript|1253930|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|$]]

Maybe I'm late here. But in this situation I don't like category (pollution). I love this way:

<pre><code>#define UIViewParentController(__view) ({ \
UIResponder *__responder = __view; \
while ([__responder isKindOfClass:[UIView class]]) \
__responder = [__responder nextResponder]; \
(UIViewController *)__responder; \
})
</code></pre>

blocks|key|1468701|text|swift+4的更新版:感谢@Phil_M和@paul-slm|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1468702|static+func+firstAvailableUIViewController(fromResponder+responder:+UIResponder)+->+UIViewController?+{
++++func+traverseResponderChainForUIViewController(responder:+UIResponder)+->+UIViewController?+{
++++++++if+let+nextResponder+=+responder.next+{
++++++++++++if+let+nextResp+=+nextResponder+as?+UIViewController+{
++++++++++++++++return+nextResp
++++++++++++}+else+{
++++++++++++++++return+traverseResponderChainForUIViewController(responder:+nextResponder)
++++++++++++}
++++++++}
++++++++return+nil
++++}

++++return+traverseResponderChainForUIViewController(responder:+responder)
}|code-block|syntax|javascript|1468703|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|$]]

Updated version for swift 4 : Thanks for @Phil_M and @paul-slm

<pre><code>static func firstAvailableUIViewController(fromResponder responder: UIResponder) -&gt; UIViewController? {
 func traverseResponderChainForUIViewController(responder: UIResponder) -&gt; UIViewController? {
 if let nextResponder = responder.next {
 if let nextResp = nextResponder as? UIViewController {
 return nextResp
 } else {
 return traverseResponderChainForUIViewController(responder: nextResponder)
 }
 }
 return nil
 }

 return traverseResponderChainForUIViewController(responder: responder)
}
</code></pre>

blocks|key|1254029|text|Swiftier解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1254030|extension+UIView+{
++++var+parentViewController:+UIViewController?+{
++++++++for+responder+in+sequence(first:+self,+next:+{+$0.next+})+{
++++++++++++if+let+viewController+=+responder+as?+UIViewController+{
++++++++++++++++return+viewController
++++++++++++}
++++++++}
++++++++return+nil
++++}
}|code-block|syntax|javascript|1254031|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|$]]

Swiftier solution

<pre><code>extension UIView {
 var parentViewController: UIViewController? {
 for responder in sequence(first: self, next: { $0.next }) {
 if let viewController = responder as? UIViewController {
 return viewController
 }
 }
 return nil
 }
}
</code></pre>

blocks|key|1468798|text|Swift+4|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1468799|(比其他答案更简洁)|1468800|fileprivate+extension+UIView+{

++var+firstViewController:+UIViewController?+{
++++let+firstViewController+=+sequence(first:+self,+next:+{+$0.next+}).first(where:+{+$0+is+UIViewController+})
++++return+firstViewController+as?+UIViewController
++}

}|code-block|syntax|javascript|1468801|我需要首先访问视图UIViewController的用例:我有一个包装在AVPlayer+/+AVPlayerViewController周围的对象，我想提供一个简单的show(in+view:+UIView)方法，将AVPlayerViewController嵌入到view中。为此，我需要访问view的UIViewController。|offset|length|style|CODE|1468802|entityMap^0|0|0|0|9|G|10|8|1B|M|2C|L|31|M|3Q|4|44|4|49|G|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]|$K|W|L|X|M|N]|$K|Y|L|Z|M|N]|$K|10|L|11|M|N]|$K|12|L|13|M|N]|$K|14|L|15|M|N]|$K|16|L|17|M|N]|$K|18|L|19|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|P|$]]

<h1>Swift 4</h1>

(more concise than the other answers)

<pre><code>fileprivate extension UIView {

 var firstViewController: UIViewController? {
 let firstViewController = sequence(first: self, next: { $0.next }).first(where: { $0 is UIViewController })
 return firstViewController as? UIViewController
 }

}
</code></pre>

My use case for which I need to access the view first <code>UIViewController</code>: I have an object that wraps around <code>AVPlayer</code> / <code>AVPlayerViewController</code> and I want to provide a simple <code>show(in view: UIView)</code> method that will embed <code>AVPlayerViewController</code> into <code>view</code>. For that, I need to access <code>view</code>'s <code>UIViewController</code>.

blocks|key|1254101|text|如果您不打算将其上传到App+Store，您也可以使用UIView的私有方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1254102|@interface+UIView(Private)
-+(UIViewController+*)_viewControllerForAncestor;
@end

//+Later+in+the+code
UIViewController+*vc+=+[myView+_viewControllerForAncestor];|code-block|syntax|javascript|1254103|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|$]]

If you aren't going to upload this to the App Store, you can also use a private method of UIView.

<pre><code>@interface UIView(Private)
- (UIViewController *)_viewControllerForAncestor;
@end

// Later in the code
UIViewController *vc = [myView _viewControllerForAncestor];
</code></pre>

blocks|key|1468863|text|var+parentViewController:+UIViewController?+{
++++let+s+=+sequence(first:+self)+{+$0.next+}
++++return+s.compactMap+{+$0+as?+UIViewController+}.first
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1468864|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>var parentViewController: UIViewController? {
 let s = sequence(first: self) { $0.next }
 return s.compactMap { $0 as? UIViewController }.first
}
</code></pre>

blocks|key|1468880|text|要获得给定视图的控制器，可以使用UIFirstResponder链。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1468881|customView.target(forAction:+Selector("viewDidLoad"),+withSender:+nil)|code-block|syntax|javascript|1468882|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 the controller of a given view, one can use UIFirstResponder chain.

<pre><code>customView.target(forAction: Selector("viewDidLoad"), withSender: nil)
</code></pre>

Is there a built-in way to get from a <code>UIView</code> to its <code>UIViewController</code>? I know you can get from <code>UIViewController</code> to its <code>UIView</code> via <code>[self view]</code> but I was wondering if there is a reverse reference?

Get to UIViewController from UIView?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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有没有一种从UIView到UIViewController的内置方法？我知道你可以通过[self view]从UIViewController转到它的UIView，但是我想知道是否有反向引用？

问从UIView转到UIViewController？
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问从UIView转到UIViewController？EN