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

我修复了这个错误,将最顶部的视图控制器存储为常量,这是在rootViewController的while循环中发现的:

if var topController = UIApplication.shared.keyWindow?.rootViewController {
    while let presentedViewController = topController.presentedViewController {
        topController = presentedViewController
    }
    topController.present(controller, animated: false, completion: nil)
    // topController should now be your topmost view controller
}

其他回答

在主窗口中,可能总是会出现与显示警报不兼容的过渡。为了允许在应用程序生命周期的任何时候显示警报,您应该有一个单独的窗口来完成这项工作。

/// independant window for alerts
@interface AlertWindow: UIWindow

+ (void)presentAlertWithTitle:(NSString *)title message:(NSString *)message;

@end

@implementation AlertWindow

+ (AlertWindow *)sharedInstance
{
    static AlertWindow *sharedInstance;
    static dispatch_once_t onceToken;
    dispatch_once(&onceToken, ^{
        sharedInstance = [[AlertWindow alloc] initWithFrame:UIScreen.mainScreen.bounds];
    });
    return sharedInstance;
}

+ (void)presentAlertWithTitle:(NSString *)title message:(NSString *)message
{
    // Using a separate window to solve "Warning: Attempt to present <UIAlertController> on <UIViewController> whose view is not in the window hierarchy!"
    UIWindow *shared = AlertWindow.sharedInstance;
    shared.userInteractionEnabled = YES;
    UIViewController *root = shared.rootViewController;
    UIAlertController *alert = [UIAlertController alertControllerWithTitle:title message:message preferredStyle:UIAlertControllerStyleAlert];
    alert.modalInPopover = true;
    [alert addAction:[UIAlertAction actionWithTitle:@"OK" style:UIAlertActionStyleCancel handler:^(UIAlertAction *action) {
        shared.userInteractionEnabled = NO;
        [root dismissViewControllerAnimated:YES completion:nil];
    }]];
    [root presentViewController:alert animated:YES completion:nil];
}

- (instancetype)initWithFrame:(CGRect)frame
{
    self = [super initWithFrame:frame];

    self.userInteractionEnabled = NO;
    self.windowLevel = CGFLOAT_MAX;
    self.backgroundColor = UIColor.clearColor;
    self.hidden = NO;
    self.rootViewController = UIViewController.new;

    [NSNotificationCenter.defaultCenter addObserver:self
                                           selector:@selector(bringWindowToTop:)
                                               name:UIWindowDidBecomeVisibleNotification
                                             object:nil];

    return self;
}

/// Bring AlertWindow to top when another window is being shown.
- (void)bringWindowToTop:(NSNotification *)notification {
    if (![notification.object isKindOfClass:[AlertWindow class]]) {
        self.hidden = YES;
        self.hidden = NO;
    }
}

@end

在设计上总是成功的基本用法:

[AlertWindow presentAlertWithTitle:@"My title" message:@"My message"];

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。

我有这个问题,根本原因是多次订阅按钮单击处理程序(TouchUpInside)。

它在ViewWillAppear中订阅,它被多次调用因为我们添加了导航去到另一个控制器,然后unwind回它。

我发现故事板中的segue有点坏了。删除segue(并再次创建完全相同的segue)解决了这个问题。

当从嵌入在容器中的视图控制器执行segue时,你也可以得到这个警告。正确的解决方案是从容器的父容器使用segue,而不是从容器的视图控制器。