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
In case it helps anyone, my issue was extremely silly. Totally my fault of course. A notification was triggering a method that was calling the modal. But I wasn't removing the notification correctly, so at some point, I would have more than one notification, so the modal would get called multiple times. Of course, after you call the modal once, the viewcontroller that calls it it's not longer in the view hierarchy, that's why we see this issue. My situation caused a bunch of other issue too, as you would expect.
总之，无论你做什么，都要确保模态不会被调用超过一次。
其他回答
viewWillLayoutSubviews和viewDidLayoutSubviews (iOS 5.0+)可以用于此目的。它们在viewDidAppear之前被调用。
必须写在线下。
self.searchController.definesPresentationContext = true
而不是
self.definesPresentationContext = true
在ui
这适用于呈现任何视图控制器，如果你有可用的导航控制器。 self.navigationController ?。present(MyViewController, animated: true, completion: nil) 此外，我可以能够呈现警报和邮件控制器也。
斯威夫特5
我在viewDidLayoutSubviews中调用present，因为在viewDidAppear中呈现会在模态加载之前导致视图控制器的瞬间显示，这看起来像一个丑陋的故障
确保检查窗口是否存在并只执行一次代码
var alreadyPresentedVCOnDisplay = false
override func viewDidLayoutSubviews() {
super.viewDidLayoutSubviews()
// we call present in viewDidLayoutSubviews as
// presenting in viewDidAppear causes a split second showing
// of the view controller before the modal is loaded
guard let _ = view?.window else {
// window must be assigned
return
}
if !alreadyPresentedVCOnDisplay {
alreadyPresentedVCOnDisplay = true
present(...)
}
}
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)
}
