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
}
其他回答
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Rajesh的解决方案几乎是完美的,但我认为从上到下遍历子视图更好,我改为如下:
+ (UIViewController *)topViewController:(UIViewController *)viewController{
if (viewController.presentedViewController)
{
UIViewController *presentedViewController = viewController.presentedViewController;
return [self topViewController:presentedViewController];
}
else if ([viewController isKindOfClass:[UITabBarController class]])
{
UITabBarController *tabBarController = (UITabBarController *)viewController;
return [self topViewController:tabBarController.selectedViewController];
}
else if ([viewController isKindOfClass:[UINavigationController class]])
{
UINavigationController *navController = (UINavigationController *)viewController;
return [self topViewController:navController.visibleViewController];
}
// Handling UIViewController's added as subviews to some other views.
else {
NSInteger subCount = [viewController.view subviews].count - 1;
for (NSInteger index = subCount; index >=0 ; --index)
{
UIView *view = [[viewController.view subviews] objectAtIndex:index];
id subViewController = [view nextResponder]; // Key property which most of us are unaware of / rarely use.
if ( subViewController && [subViewController isKindOfClass:[UIViewController class]])
{
return [self topViewController:subViewController];
}
}
return viewController;
}
}
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 4.2中一个简洁而全面的解决方案,考虑了UINavigationControllers, UITabBarControllers, presenting和子视图控制器:
extension UIViewController {
func topmostViewController() -> UIViewController {
if let navigationVC = self as? UINavigationController,
let topVC = navigationVC.topViewController {
return topVC.topmostViewController()
}
if let tabBarVC = self as? UITabBarController,
let selectedVC = tabBarVC.selectedViewController {
return selectedVC.topmostViewController()
}
if let presentedVC = presentedViewController {
return presentedVC.topmostViewController()
}
if let childVC = children.last {
return childVC.topmostViewController()
}
return self
}
}
extension UIApplication {
func topmostViewController() -> UIViewController? {
return keyWindow?.rootViewController?.topmostViewController()
}
}
用法:
let viewController = UIApplication.shared.topmostViewController()
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;
}