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
NSSet *connectedScenes = [UIApplication sharedApplication].connectedScenes;
for (UIScene *scene in connectedScenes) {
if (scene.activationState == UISceneActivationStateForegroundActive && [scene isKindOfClass:[UIWindowScene class]]) {
UIWindowScene *windowScene = (UIWindowScene *)scene;
for (UIWindow *window in windowScene.windows) {
UIViewController *viewController = window.rootViewController;
// Get the instance of your view controller
if ([viewController isKindOfClass:[YOUR_VIEW_CONTROLLER class]]) {
// Your code here...
break;
}
}
}
}
其他回答
Berni的代码很好，但它不工作时，应用程序从后台回来。
这是我的代码:
class var safeArea : UIEdgeInsets
{
if #available(iOS 13, *) {
var keyWindow = UIApplication.shared.connectedScenes
.filter({$0.activationState == .foregroundActive})
.map({$0 as? UIWindowScene})
.compactMap({$0})
.first?.windows
.filter({$0.isKeyWindow}).first
// <FIX> the above code doesn't work if the app comes back from background!
if (keyWindow == nil) {
keyWindow = UIApplication.shared.windows.first { $0.isKeyWindow }
}
return keyWindow?.safeAreaInsets ?? UIEdgeInsets()
}
else {
guard let keyWindow = UIApplication.shared.keyWindow else { return UIEdgeInsets() }
return keyWindow.safeAreaInsets
}
}
一个UIApplication扩展:
extension UIApplication {
/// The app's key window taking into consideration apps that support multiple scenes.
var keyWindowInConnectedScenes: UIWindow? {
return windows.first(where: { $0.isKeyWindow })
}
}
用法:
let myKeyWindow: UIWindow? = UIApplication.shared.keyWindowInConnectedScenes
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;
}
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 })
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
