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
- (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;
}
其他回答
(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)
}
}
}
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
}
}
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;
}
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 }
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 })