Ios 当我想从推送通知启动时，如何处理RootView

我的应用程序有一个嵌入式导航控制器。我想启动一个特定的viewController，它不是初始视图或rootViewController

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions: (NSDictionary *)launchOptions
{
NSDictionary *dictionary = [launchOptions objectForKey:UIApplicationLaunchOptionsRemoteNotificationKey];

if (dictionary != nil)
{
    UIViewController *rootController = (UIViewController *)self.window.rootViewController;
UIViewController *notificationController = [rootController.storyboard instantiateViewControllerWithIdentifier:@"NotificationsViewController"];
[rootController presentViewController:notificationController animated:NO completion:^{

}];
 }
}

我知道我必须在这里输入代码，但我不确定如何准确地编码它。有人能帮忙吗？我得到了错误代码：

Warning: Attempt to present <NotificationsViewController: 0x176883a0> on <SWRevealViewController: 0x17683750> whose view is not in the window hierarchy!

警告：尝试显示其视图不在窗口层次结构中的对象！

SWRevealViewController是我的边栏视图库。我猜我的“根”不是我想的那样

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如果我想将viewControllerX设置为我的根视图，我该如何实现这一点？

您应该首先初始化您的根视图控制器

self.window = [[UIWindow alloc] initWithFrame:[[UIScreen mainScreen] bounds]];    
YourViewControllerX * viewControllerX = [[YourViewControllerX alloc] init];
self.window.rootViewController = viewControllerX;
[self.window makeKeyAndVisible];

试一试

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有一件事需要问。当应用程序在没有选项的情况下启动时，它是如何工作的？？？据我所见，当launchOptions为nil时，没有代码可以运行。接下来，您需要在调用self.window.rootViewController之前首先创建一个UIWindow实例。

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions: (NSDictionary *)launchOptions{
    [self.window addSubview:self.window.rootViewController.view];
    [self.window makeKeyAndVisible];
    NSDictionary *dictionary = [launchOptions objectForKey:UIApplicationLaunchOptionsRemoteNotificationKey];
    if (dictionary != nil){
        ...
    }
}