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
很多答案都是不完整的。虽然这是在Objective-C中,但这是我现在能把它们放在一起的最好的编译,作为一个非递归的块:
链接到Gist,以防修改:https://gist.github.com/benguild/0d149bb3caaabea2dac3d2dca58c0816 供参考/比较的代码:
UIViewController *(^topmostViewControllerForFrontmostNormalLevelWindow)(void) = ^UIViewController *{
// NOTE: Adapted from various stray answers here:
// https://stackoverflow.com/questions/6131205/iphone-how-to-find-topmost-view-controller/20515681
UIViewController *viewController;
for (UIWindow *window in UIApplication.sharedApplication.windows.reverseObjectEnumerator.allObjects) {
if (window.windowLevel == UIWindowLevelNormal) {
viewController = window.rootViewController;
break;
}
}
while (viewController != nil) {
if ([viewController isKindOfClass:[UITabBarController class]]) {
viewController = ((UITabBarController *)viewController).selectedViewController;
} else if ([viewController isKindOfClass:[UINavigationController class]]) {
viewController = ((UINavigationController *)viewController).visibleViewController;
} else if (viewController.presentedViewController != nil && !viewController.presentedViewController.isBeingDismissed) {
viewController = viewController.presentedViewController;
} else if (viewController.childViewControllers.count > 0) {
viewController = viewController.childViewControllers.lastObject;
} else {
BOOL repeat = NO;
for (UIView *view in viewController.view.subviews.reverseObjectEnumerator.allObjects) {
if ([view.nextResponder isKindOfClass:[UIViewController class]]) {
viewController = (UIViewController *)view.nextResponder;
repeat = YES;
break;
}
}
if (!repeat) {
break;
}
}
}
return viewController;
};
另一个解决方案依赖于responder链,它可能工作,也可能不工作,这取决于第一个responder是什么:
获取第一个响应器。 获取与第一个responder相关联的UIViewController。
示例伪代码:
+ (UIViewController *)currentViewController {
UIView *firstResponder = [self firstResponder]; // from the first link above, but not guaranteed to return a UIView, so this should be handled more appropriately.
UIViewController *viewController = [firstResponder viewController]; // from the second link above
return viewController;
}
这是对Eric的回答的改进:
UIViewController *_topMostController(UIViewController *cont) {
UIViewController *topController = cont;
while (topController.presentedViewController) {
topController = topController.presentedViewController;
}
if ([topController isKindOfClass:[UINavigationController class]]) {
UIViewController *visible = ((UINavigationController *)topController).visibleViewController;
if (visible) {
topController = visible;
}
}
return (topController != cont ? topController : nil);
}
UIViewController *topMostController() {
UIViewController *topController = [UIApplication sharedApplication].keyWindow.rootViewController;
UIViewController *next = nil;
while ((next = _topMostController(topController)) != nil) {
topController = next;
}
return topController;
}
UIViewController *cont是一个辅助函数。
现在你所需要做的就是调用topMostController()和最顶端的UIViewController应该被返回!
我认为你需要一个公认的答案和@fishstix的组合
+ (UIViewController*) topMostController
{
UIViewController *topController = [UIApplication sharedApplication].keyWindow.rootViewController;
while (topController.presentedViewController) {
topController = topController.presentedViewController;
}
return topController;
}
Swift 3.0 +
func topMostController() -> UIViewController? {
guard let window = UIApplication.shared.keyWindow, let rootViewController = window.rootViewController else {
return nil
}
var topController = rootViewController
while let newTopController = topController.presentedViewController {
topController = newTopController
}
return topController
}
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!
}