我遇到了一些情况,现在,它将是方便的,能够找到“最顶层”的视图控制器(一个负责当前视图),但还没有找到一种方法。
基本上,挑战是这样的:给定一个在一个类中执行,这个类不是一个视图控制器(或一个视图)[并且没有活动视图的地址],并且没有传递最顶层视图控制器的地址(或者,比如说,导航控制器的地址),是否有可能找到那个视图控制器?(如果是的话,是怎么做到的?)
或者,如果找不到,有没有可能找到最高处的风景?
我遇到了一些情况,现在,它将是方便的,能够找到“最顶层”的视图控制器(一个负责当前视图),但还没有找到一种方法。
基本上,挑战是这样的:给定一个在一个类中执行,这个类不是一个视图控制器(或一个视图)[并且没有活动视图的地址],并且没有传递最顶层视图控制器的地址(或者,比如说,导航控制器的地址),是否有可能找到那个视图控制器?(如果是的话,是怎么做到的?)
或者,如果找不到,有没有可能找到最高处的风景?
当前回答
我的问题有点不同,我在我的应用程序中使用SWRevealViewController。 我使用了Yuchen Zhong的答案,但它总是返回topViewController作为SWRevealViewController。对于那些使用SWRevealViewController或其他一些pod来开发侧菜单的人。以下是我对钟宇晨回答的延伸:
extension UIApplication {
class func topViewController() -> UIViewController? {
var topVC = shared.keyWindow!.rootViewController
while true {
if let presented = topVC?.presentedViewController {
topVC = presented
} else if let nav = topVC as? UINavigationController {
topVC = nav.visibleViewController
} else if let tab = topVC as? UITabBarController {
topVC = tab.selectedViewController
}else if let swRVC = topVC as? SWRevealViewController {
topVC = swRVC.frontViewController
} else {
break
}
}
return topVC
}
}
其他回答
iOS 4在UIWindow上引入了rootViewController属性:
[UIApplication sharedApplication].keyWindow.rootViewController;
你需要在创建视图控制器后自己设置。
使用下面的扩展抓取当前可见的UIViewController。适用于Swift 4.0及更高版本
Swift 4.0及以上版本:
extension UIApplication {
class func topViewController(_ viewController: UIViewController? = UIApplication.shared.keyWindow?.rootViewController) -> UIViewController? {
if let nav = viewController as? UINavigationController {
return topViewController(nav.visibleViewController)
}
if let tab = viewController as? UITabBarController {
if let selected = tab.selectedViewController {
return topViewController(selected)
}
}
if let presented = viewController?.presentedViewController {
return topViewController(presented)
}
return viewController
}
}
如何使用?
let objViewcontroller = UIApplication.topViewController()
下面两个函数可以帮助在视图控制器堆栈中找到topViewController。以后可能需要自定义,但是对于这段代码来说,理解topViewController或viewcontroller堆栈的概念非常棒。
- (UIViewController*)findTopViewController {
id topControler = [self topMostController];
UIViewController* topViewController;
if([topControler isKindOfClass:[UINavigationController class]]) {
topViewController = [[(UINavigationController*)topControler viewControllers] lastObject];
} else if ([topControler isKindOfClass:[UITabBarController class]]) {
//Here you can get reference of top viewcontroller from stack of viewcontrollers on UITabBarController
} else {
//topController is a preented viewController
topViewController = (UIViewController*)topControler;
}
//NSLog(@"Top ViewController is: %@",NSStringFromClass([topController class]));
return topViewController;
}
- (UIViewController*)topMostController
{
UIViewController *topController = [UIApplication sharedApplication].keyWindow.rootViewController;
while (topController.presentedViewController) {
topController = topController.presentedViewController;
}
//NSLog(@"Top View is: %@",NSStringFromClass([topController class]));
return topController;
}
你可以使用[viewController Class]方法找出一个viewController的类的类型。
I am thinking that perhaps one thing is being overlooked here. Perhaps it is better to pass the parent viewController into the function that is using the viewController. If you are fishing around in the view hierarchy to find the top view controller that it is probably violating separation of the Model layer and UI layer and is a code smell. Just pointing this out, I did the same, then realized it was much simpler just to pass it in to function, by having the model operation return to the UI layer where I have a reference to the view controller.
Swift替代解决方案:
static func topMostController() -> UIViewController {
var topController = UIApplication.sharedApplication().keyWindow?.rootViewController
while (topController?.presentedViewController != nil) {
topController = topController?.presentedViewController
}
return topController!
}