我使用核心数据与云工具包,因此要检查iCloud用户状态在应用程序启动。如果出现问题,我想向用户发出一个对话框,我使用UIApplication.shared.keyWindow?. rootviewcontroller ?.present(…)到目前为止。

在Xcode 11 beta 4中,现在有一个新的弃用消息,告诉我:

'keyWindow'在iOS 13.0中已弃用:不应该用于支持多个场景的应用程序,因为它在所有连接的场景中返回一个键窗口

我应该如何呈现对话呢?


当前回答

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 }
    }
}

其他回答

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;
}
- (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;
}

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)

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iOS 16,我使用了以下方法:

let keyWindow = UIApplication.shared.currentUIWindow()?.windowScene?.keyWindow