UIGestureRecognizer - 获取对触摸的UIViewController的引用，而不是其视图？

Question

如何获得触摸视图的UIViewController的引用？

我在UIViewController的视图上使用UIPanGestureRecognizer。以下是我初始化：

TaskUIViewController *thisTaskController = [[TaskUIViewController alloc]init]; 
    [[self view]addSubview:[thisTaskController view]]; 
    UIPanGestureRecognizer *panRec = [[UIPanGestureRecognizer alloc]initWithTarget:self action:@selector(handlePan:)]; 
    [[thisTaskController view] addGestureRecognizer:panRec]; 

在tiggered动作使用手势识别我能够得到从参数角度触发使用recognizer.view

- (IBAction)handlePan:(UIPanGestureRecognizer *)recognizer { 
    UIView *touchedView = [[UIView alloc]init]; 
    touchedView = (UIView*)[recognizer view]; 
    ... 
} 

但是我真正需要的是触摸视图的底层UIViewController。我如何获得对包含此视图而不是仅UIView的UIViewController的引用？

Answer 1

我要说的是，它更一个设计问题，而不仅仅是获得参考。所以我会遵循几个简单的建议：

1. 业主应该从它的视角捕捉事件。即TaskUIViewController应该是您添加到其视图的UIPanGestureRecognizer的目标。
2. 如果一个控制器有一个子控制器，并从其子控制器等待一些响应 - 将其作为委托实现。
3. 你的“handlePan：”方法有内存泄漏。

这是解决您的问题的骨架：

@protocol CallbackFromMySubcontroller <NSObject> 
- (void)calbackFromTaskUIViewControllerOnPanGesture:(UIViewController*)fromController; 
@end 

@interface OwnerController : UIViewController <CallbackFromMySubcontroller> 
@end 

@implementation OwnerController 

- (id)init 
{ 
    ... 
    TaskUIViewController *thisTaskController = [[TaskUIViewController alloc] init]; 
    ... 
} 

- (void)viewDidLoad 
{ 
    ... 
    [self.view addSubview:thisTaskController.view]; 
    ... 
} 

- (void)calbackFromTaskUIViewControllerOnPanGesture:(UIViewController*)fromController 
{ 
    NSLog(@"Yahoo. I got an event from my subController's view"); 
} 

@end 


@interface TaskUIViewController : UIViewController { 
    id <CallbackFromMySubcontroller> delegate; 
} 
@end 

@implementation TaskUIViewController 

- (id)initWithOwner:(id<CallbackFromMySubcontroller>)owner 
{ 
    ... 
    delegate = owner; 
    ... 
} 

- (void)viewDidLoad 
{ 
    UIPanGestureRecognizer *panRec = [[UIPanGestureRecognizer alloc] initWithTarget:self action:@selector(handlePan:)]; 
    [self.view addGestureRecognizer:panRec]; 
    [panRec release]; 
} 

- (void)handlePan:(UIPanGestureRecognizer *)recognizer { 
    ... 
    [delegate calbackFromTaskUIViewControllerOnPanGesture:self]; 
    ... 
} 

@end 

Answer 2

[touchedView nextResponder]将返回管理touchedView的UIViewController对象（如果有的话）或touchedView的上海华盈（如果它不有一个管理它UIViewController对象）。请参阅UIResponder Class Reference。 （UIViewController和UIView是UIResponder子类）。

在你的情况，因为你碰巧知道touchedView是你的viewController的观点（而不是，例如，您的viewController的视图的子视图），你可以使用：

TaskUIViewController *touchedController = (TaskUIViewController *)[touchedView nextResponder]; 

在更一般的情况下，你可以工作了响应链，直到找到一种UIViewController中的对象：

id aNextResponder = [touchedView nextResponder]; 

while (aNextResponder != nil) 
{ 
    if ([aNextResponder isKindOfClass:[UIViewController class]]) 
    { 
    // we have found the viewController that manages touchedView, 
    // so we break out of the while loop: 
    break; 
    } 
    else 
    { 
    // we have yet to find the managing viewController, 
    // so we examine the next responder in the responder chain 
    aNextResponder = [aNextResponder nextResponder]; 
    } 
} 

// outside the while loop. at this point aNextResponder points to 
// touchedView's managing viewController (or nil if it doesn't have one). 
UIViewController *eureka = (UIViewController *)aNextResponder;