通常使用

斯威夫特5

UIApplication.shared.windows.filter {$0.isKeyWindow}.first

另外，在UIViewController中:

self.view.window

视图。窗口是当前场景的窗口

WWDC 2019:

关键的窗口 手动跟踪窗口

在iPad上介绍多个Windows - WWDC 2019 -视频-苹果开发人员 支持iPad上的多个Windows |苹果开发者文档

Objective C解决方案:

UIWindow *foundWindow = nil;
NSSet *scenes=[[UIApplication sharedApplication] connectedScenes];
NSArray *windows;
for(id aScene in scenes){  // it's an NSSet so you can't use the first object
    windows=[aScene windows];
    if([aScene activationState]==UISceneActivationStateForegroundActive)
         break;
}
for (UIWindow  *window in windows) {
    if (window.isKeyWindow) {
        foundWindow = window;
        break;
    }
}
 // and to find the parent viewController:
UIViewController* parentController = foundWindow.rootViewController;
while( parentController.presentedViewController &&
      parentController != parentController.presentedViewController ){
    parentController = parentController.presentedViewController;
}

支持iOS 13及以上版本。

为了继续使用与旧iOS版本相似的语法，UIApplication.shared.keyWindow创建以下扩展:

extension UIApplication {
    var mainKeyWindow: UIWindow? {
        get {
            if #available(iOS 13, *) {
                return connectedScenes
                    .flatMap { ($0 as? UIWindowScene)?.windows ?? [] }
                    .first { $0.isKeyWindow }
            } else {
                return keyWindow
            }
        }
    }
}

使用

if let keyWindow = UIApplication.shared.mainKeyWindow {
    // Do Stuff
}

这是我的解决方案:

let keyWindow = UIApplication.shared.connectedScenes
        .filter({$0.activationState == .foregroundActive})
        .compactMap({$0 as? UIWindowScene})
        .first?.windows
        .filter({$0.isKeyWindow}).first

使用如:

keyWindow?.endEditing(true)

您可能知道，由于可能存在多个场景，因此不建议使用键窗口。最方便的解决方案是提供一个currentWindow作为扩展，然后进行搜索和替换。

extension UIApplication {
    var currentWindow: UIWindow? {
        connectedScenes
            .compactMap { $0 as? UIWindowScene }
            .flatMap { $0.windows }
            .first { $0.isKeyWindow }
    }
}

对于Objective-C解决方案也是如此

@implementation UIWindow (iOS13)

+ (UIWindow*) keyWindow {
   NSPredicate *isKeyWindow = [NSPredicate predicateWithFormat:@"isKeyWindow == YES"];
   return [[[UIApplication sharedApplication] windows] filteredArrayUsingPredicate:isKeyWindow].firstObject;
}

@end