iOS 16，兼容至iOS 15

由于这个帖子在三年后仍有流量，我想分享我认为目前功能最优雅的解决方案。它也适用于SwiftUI。

UIApplication
    .shared
    .connectedScenes
    .compactMap { ($0 as? UIWindowScene)?.keyWindow }
    .first

iOS 15和16，兼容至iOS 13

UIApplication
    .shared
    .connectedScenes
    .flatMap { ($0 as? UIWindowScene)?.windows ?? [] }
    .first { $0.isKeyWindow }

注意，connectedScenes只在iOS 13之后才可用。如果你需要支持早期版本的iOS，你必须把这个放在If #available(iOS 13， *)语句中。

变体:更长，但更容易理解的变体:

UIApplication
    .shared
    .connectedScenes
    .compactMap { $0 as? UIWindowScene }
    .flatMap { $0.windows }
    .first { $0.isKeyWindow }

iOS 13和14

下面的历史答案在iOS 15上仍然有效，但应该被替换，因为UIApplication.shared.windows已弃用。感谢@matt指出这一点!

最初的回答:

在matt的精彩回答基础上稍作改进，这个回答更简单、更简短、更优雅:

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

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;
}

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

当.foregroundActive场景为空时，我遇到了这个问题

这是我的变通办法

public extension UIWindow {
    @objc
    static var main: UIWindow {
        // Here we sort all the scenes in order to work around the case
        // when no .foregroundActive scenes available and we need to look through
        // all connectedScenes in order to find the most suitable one
        let connectedScenes = UIApplication.shared.connectedScenes
            .sorted { lhs, rhs in
                let lhs = lhs.activationState
                let rhs = rhs.activationState
                switch lhs {
                case .foregroundActive:
                    return true
                case .foregroundInactive:
                    return rhs == .background || rhs == .unattached
                case .background:
                    return rhs == .unattached
                case .unattached:
                    return false
                @unknown default:
                    return false
                }
            }
            .compactMap { $0 as? UIWindowScene }

        guard connectedScenes.isEmpty == false else {
            fatalError("Connected scenes is empty")
        }
        let mainWindow = connectedScenes
            .flatMap { $0.windows }
            .first(where: \.isKeyWindow)

        guard let window = mainWindow else {
            fatalError("Couldn't get main window")
        }
        return window
    }
}

我已经解决了:

let scenes = UIApplication.shared.connectedScenes
let windowScene = scenes.first as? UIWindowScene
let window = windowScene?.windows.first

如果你的应用还没有更新到采用基于场景的应用生命周期，另一种获得活动窗口对象的简单方法是通过UIApplicationDelegate:

let window = UIApplication.shared.delegate?.window
let rootViewController = window??.rootViewController