view中是不能进行UIViewController的push，pop等操作的，若进行跳转操作，一般是用代理，block,通知等实现，那如何实现在ViewController的subView中实现跳转操作呢，其实只要获取该view所在的ViewController即可。
获取view所在UIViewController
UIView+UIViewController.h
#import <UIKit/UIKit.h> @interface UIView (UIViewController) - (UIViewController *)viewController; @end
UIView+UIViewController.m
#import "UIView+UIViewController.h"
@implementation UIView (UIViewController)
- (UIViewController *)viewController {
//通过响应者链，获得view所在的视图控制器
UIResponder *next = self.nextResponder;
do {
//判断响应者对象是否是视图控制器类型
if ([next isKindOfClass:[UIViewController class]]) {
return (UIViewController *)next;
}
next = next.nextResponder;
}while(next != nil);
return nil;
}
@end
or
- (UIViewController *)viewController {
// Traverse responder chain. Return first found view controller, which will be the view‘s view controller.
UIResponder *responder = self;
while ((responder = [responder nextResponder]))
if ([responder isKindOfClass: [UIViewController class]])
return (UIViewController *)responder;
return nil;
}
- (UIViewController*)viewController {
for (UIView* next = [self superview]; next; next = next.superview) {
UIResponder* nextResponder = [next nextResponder];
if ([nextResponder isKindOfClass:[UIViewController class]]) {
return (UIViewController*)nextResponder;
}
}
return nil;
}
使用：
在subview中导入 #import "UIView+UIViewController.h"
-(void)viewDidSelect
{
WebViewController *vc = [[WebViewController alloc] init];
vc.webUrl =@"http://www.baidu.com"；
[self.viewController.navigationController pushViewController:vc animated:YES];
}
原文：http://www.cnblogs.com/sixindev/p/4856879.html