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
这个答案包含了childViewControllers,并维护了一个干净易读的实现。
+ (UIViewController *)topViewController
{
UIViewController *rootViewController = [UIApplication sharedApplication].keyWindow.rootViewController;
return [rootViewController topVisibleViewController];
}
- (UIViewController *)topVisibleViewController
{
if ([self isKindOfClass:[UITabBarController class]])
{
UITabBarController *tabBarController = (UITabBarController *)self;
return [tabBarController.selectedViewController topVisibleViewController];
}
else if ([self isKindOfClass:[UINavigationController class]])
{
UINavigationController *navigationController = (UINavigationController *)self;
return [navigationController.visibleViewController topVisibleViewController];
}
else if (self.presentedViewController)
{
return [self.presentedViewController topVisibleViewController];
}
else if (self.childViewControllers.count > 0)
{
return [self.childViewControllers.lastObject topVisibleViewController];
}
return self;
}
其他回答
伟大的解决方案在Swift,实现在AppDelegate
func getTopViewController()->UIViewController{
return topViewControllerWithRootViewController(UIApplication.sharedApplication().keyWindow!.rootViewController!)
}
func topViewControllerWithRootViewController(rootViewController:UIViewController)->UIViewController{
if rootViewController is UITabBarController{
let tabBarController = rootViewController as! UITabBarController
return topViewControllerWithRootViewController(tabBarController.selectedViewController!)
}
if rootViewController is UINavigationController{
let navBarController = rootViewController as! UINavigationController
return topViewControllerWithRootViewController(navBarController.visibleViewController)
}
if let presentedViewController = rootViewController.presentedViewController {
return topViewControllerWithRootViewController(presentedViewController)
}
return rootViewController
}
扩展@Eric的回答,你需要小心,keyWindow实际上是你想要的窗口。例如,如果您试图在点击警报视图中的某些内容后使用此方法,keyWindow实际上将是警报的窗口,这无疑会给您带来问题。这发生在我在野外通过警报处理深度链接时,并导致SIGABRTs没有堆栈跟踪。要调试的婊子。
下面是我现在使用的代码:
- (UIViewController *)getTopMostViewController {
UIWindow *topWindow = [UIApplication sharedApplication].keyWindow;
if (topWindow.windowLevel != UIWindowLevelNormal) {
NSArray *windows = [UIApplication sharedApplication].windows;
for(topWindow in windows)
{
if (topWindow.windowLevel == UIWindowLevelNormal)
break;
}
}
UIViewController *topViewController = topWindow.rootViewController;
while (topViewController.presentedViewController) {
topViewController = topViewController.presentedViewController;
}
return topViewController;
}
你可以把这个和你喜欢的从这个问题的其他答案中检索顶视图控制器的方法混合在一起。
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
}
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.
一个完整的非递归版本,照顾到不同的场景:
视图控制器正在呈现另一个视图 视图控制器是一个UINavigationController 视图控制器是一个UITabBarController
objective - c
UIViewController *topViewController = self.window.rootViewController;
while (true)
{
if (topViewController.presentedViewController) {
topViewController = topViewController.presentedViewController;
} else if ([topViewController isKindOfClass:[UINavigationController class]]) {
UINavigationController *nav = (UINavigationController *)topViewController;
topViewController = nav.topViewController;
} else if ([topViewController isKindOfClass:[UITabBarController class]]) {
UITabBarController *tab = (UITabBarController *)topViewController;
topViewController = tab.selectedViewController;
} else {
break;
}
}
斯威夫特 4+
extension UIWindow {
func topViewController() -> UIViewController? {
var top = self.rootViewController
while true {
if let presented = top?.presentedViewController {
top = presented
} else if let nav = top as? UINavigationController {
top = nav.visibleViewController
} else if let tab = top as? UITabBarController {
top = tab.selectedViewController
} else {
break
}
}
return top
}
}