我的解决方案如下，适用于iOS 15

let window = (UIApplication.shared.connectedScenes.first as? UIWindowScene)?.windows.first

灵感来自berni的回答

let keyWindow = Array(UIApplication.shared.connectedScenes)
        .compactMap { $0 as? UIWindowScene }
        .flatMap { $0.windows }
        .first(where: { $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

对于Objective-C解决方案

+ (UIWindow *)keyWindow
{
    NSArray<UIWindow *> *windows = [[UIApplication sharedApplication] windows];
    for (UIWindow *window in windows) {
        if (window.isKeyWindow) {
            return window;
        }
    }
    return nil;
}

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

- (UIWindow *)mainWindow {
    NSEnumerator *frontToBackWindows = [UIApplication.sharedApplication.windows reverseObjectEnumerator];
    for (UIWindow *window in frontToBackWindows) {
        BOOL windowOnMainScreen = window.screen == UIScreen.mainScreen;
        BOOL windowIsVisible = !window.hidden && window.alpha > 0;
        BOOL windowLevelSupported = (window.windowLevel >= UIWindowLevelNormal);
        BOOL windowKeyWindow = window.isKeyWindow;
        if(windowOnMainScreen && windowIsVisible && windowLevelSupported && windowKeyWindow) {
            return window;
        }
    }
    return nil;
}