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?
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
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
}
}
其他回答
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
如果你使用SwiftLint的'first_where'规则,并想静音交战:
UIApplication.shared.windows.first(where: { $0.isKeyWindow })
理想情况下,由于它已经被弃用,我建议你将窗口存储在SceneDelegate中。但是,如果您确实想要一个临时的解决方案,您可以创建一个过滤器并像这样检索keyWindow。
let window = UIApplication.shared.windows.filter {$0.isKeyWindow}.first
试试这个:
UIApplication.shared.windows.filter { $0.isKeyWindow }.first?.rootViewController!.present(alert, animated: true, completion: nil)
(在运行于Xcode 13.2.1的iOS 15.2上测试)
extension UIApplication {
var keyWindow: UIWindow? {
// Get connected scenes
return UIApplication.shared.connectedScenes
// Keep only active scenes, onscreen and visible to the user
.filter { $0.activationState == .foregroundActive }
// Keep only the first `UIWindowScene`
.first(where: { $0 is UIWindowScene })
// Get its associated windows
.flatMap({ $0 as? UIWindowScene })?.windows
// Finally, keep only the key window
.first(where: \.isKeyWindow)
}
}
如果你想在关键的UIWindow中找到呈现的UIViewController,这是另一个你可以发现有用的扩展:
extension UIApplication {
var keyWindowPresentedController: UIViewController? {
var viewController = self.keyWindow?.rootViewController
// If root `UIViewController` is a `UITabBarController`
if let presentedController = viewController as? UITabBarController {
// Move to selected `UIViewController`
viewController = presentedController.selectedViewController
}
// Go deeper to find the last presented `UIViewController`
while let presentedController = viewController?.presentedViewController {
// If root `UIViewController` is a `UITabBarController`
if let presentedController = presentedController as? UITabBarController {
// Move to selected `UIViewController`
viewController = presentedController.selectedViewController
} else {
// Otherwise, go deeper
viewController = presentedController
}
}
return viewController
}
}
你可以把它放在任何你想要的地方,但我个人把它作为UIViewController的扩展。
这让我可以添加更多有用的扩展,比如更容易地呈现UIViewControllers:
extension UIViewController {
func presentInKeyWindow(animated: Bool = true, completion: (() -> Void)? = nil) {
DispatchQueue.main.async {
UIApplication.shared.keyWindow?.rootViewController?
.present(self, animated: animated, completion: completion)
}
}
func presentInKeyWindowPresentedController(animated: Bool = true, completion: (() -> Void)? = nil) {
DispatchQueue.main.async {
UIApplication.shared.keyWindowPresentedController?
.present(self, animated: animated, completion: completion)
}
}
}