刚开始使用Xcode 4.5，我在控制台得到了这个错误:
警告:试图在< ViewController: 0x1ec3e000>上显示< finishViewController: 0x1e56e0a0 >，其视图不在窗口层次结构中!
视图仍在显示，应用程序中的一切都在正常工作。这是iOS 6的新功能吗?
这是我用来在视图之间更改的代码:
UIStoryboard *storyboard = self.storyboard;
finishViewController *finished =
[storyboard instantiateViewControllerWithIdentifier:@"finishViewController"];
[self presentViewController:finished animated:NO completion:NULL];
当前回答
let alert = UIAlertController(title: "", message: "YOU SUCCESSFULLY\nCREATED A NEW\nALERT CONTROLLER", preferredStyle: .alert)
func okAlert(alert: UIAlertAction!)
{
}
alert.addAction(UIAlertAction(title: "OK", style: .default, handler: okAlert))
let scenes = UIApplication.shared.connectedScenes
let windowScene = scenes.first as? UIWindowScene
let window = windowScene?.windows.first
var rootVC = window?.rootViewController
if var topController = rootVC
{
while let presentedViewController = topController.presentedViewController
{
topController = presentedViewController
}
rootVC = topController
}
rootVC?.present(alert, animated: true, completion: nil)
其他回答
如果你有播放视频的AVPlayer对象，你必须先暂停视频。
Swift 5 -后台线程
如果一个警报控制器在后台线程上执行，那么“Attempt to present…”“其视图不在窗口层次结构中”的错误可能会发生。
所以这个:
present(alert, animated: true, completion: nil)
是这样固定的:
DispatchQueue.main.async { [weak self] in
self?.present(alert, animated: true, completion: nil)
}
要将任何子视图显示到主视图，请使用以下代码
UIViewController *yourCurrentViewController = [UIApplication sharedApplication].keyWindow.rootViewController;
while (yourCurrentViewController.presentedViewController)
{
yourCurrentViewController = yourCurrentViewController.presentedViewController;
}
[yourCurrentViewController presentViewController:composeViewController animated:YES completion:nil];
对于从主视图中解散任何子视图，请使用以下代码
UIViewController *yourCurrentViewController = [UIApplication sharedApplication].keyWindow.rootViewController;
while (yourCurrentViewController.presentedViewController)
{
yourCurrentViewController = yourCurrentViewController.presentedViewController;
}
[yourCurrentViewController dismissViewControllerAnimated:YES completion:nil];
在我的情况下，我不能把我的类重写。所以，这是我得到的:
let viewController = self // I had viewController passed in as a function,
// but otherwise you can do this
// Present the view controller
let currentViewController = UIApplication.shared.keyWindow?.rootViewController
currentViewController?.dismiss(animated: true, completion: nil)
if viewController.presentedViewController == nil {
currentViewController?.present(alert, animated: true, completion: nil)
} else {
viewController.present(alert, animated: true, completion: nil)
}
另一个潜在原因:
当我不小心两次呈现同一个视图控制器时，我就遇到了这个问题。(第一次使用performSegueWithIdentifer:sender:当按钮被按下时调用，第二次使用直接连接到按钮的segue)。
实际上，两个segue同时被触发，我得到了错误:试图在Y上显示X，其视图不在窗口层次结构中!
