您可以使用接口生成器以及编程方式设置初始视图控制器。

下面是编程使用的方法。

objective - c:

        self.window = [[UIWindow alloc] initWithFrame:UIScreen.mainScreen.bounds];
        UIStoryboard *storyboard = [UIStoryboard storyboardWithName:@"MainStoryboard" bundle:nil];

        UIViewController *viewController = [storyboard instantiateViewControllerWithIdentifier:@"HomeViewController"]; // <storyboard id>

        self.window.rootViewController = viewController;
        [self.window makeKeyAndVisible];

        return YES;

迅速:

        self.window = UIWindow(frame: UIScreen.mainScreen().bounds)
        let mainStoryboard: UIStoryboard = UIStoryboard(name: "Main", bundle: nil)

        var objMainViewController: MainViewController = mainStoryboard.instantiateViewControllerWithIdentifier("MainController") as! MainViewController

        self.window?.rootViewController = objMainViewController

        self.window?.makeKeyAndVisible()

        return true

如果你在iOS 13+中使用场景委托:

在你的信息中确认。Plist文件你发现行在:

Application Scene Manifest > Scene Configuration > Application Session Role > Item 0

并在那里删除对Main Storyboard的引用。否则，您将得到关于从storyboard实例化失败的相同警告。

同样，将代码从应用程序委托移动到场景委托方法scene(_:willConnectTo:options:)，因为这是现在处理生命周期事件的地方。

几天前我也遇到了同样的情况。一个非常简单的技巧解决了这个问题。 我设置在launch2之前隐藏我的初始视图控制器。如果初始视图控制器是正确的控制器，它在viewDidLoad中被设置为可见。否则，执行一个segue到所需的视图控制器。它在iOS 6.1及以上版本运行良好。我确信它适用于早期版本的iOS。

Xcode 12.4 Swift 5

在SceneDelegate.Swift

func scene(_ scene: UIScene, willConnectTo session: UISceneSession, options connectionOptions: UIScene.ConnectionOptions) {
    guard let windowScene = (scene as? UIWindowScene) else { return }
    let window = UIWindow(windowScene: windowScene)
    let storyboard = UIStoryboard(name: "Main", bundle: nil)
    window.rootViewController = storyboard.instantiateViewController(withIdentifier: "UserViewController") as! UserViewController
    
    self.window = window
    window.makeKeyAndVisible()
}

你可以添加视图控制器场景将要显示的条件

Swift 3:更新到@victor-sigler的代码

func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {
    self.window = UIWindow(frame: UIScreen.main.bounds)

    // Assuming your storyboard is named "Main"
    let mainStoryboard: UIStoryboard = UIStoryboard(name: "Main", bundle: nil)

    // Add code here (e.g. if/else) to determine which view controller class (chooseViewControllerA or chooseViewControllerB) and storyboard ID (chooseStoryboardA or chooseStoryboardB) to send the user to

    if(condition){
        let initialViewController: chooseViewControllerA = mainStoryboard.instantiateViewController(withIdentifier: "chooseStoryboardA") as! chooseViewControllerA
        self.window?.rootViewController = initialViewController
    )
    }else{
        let initialViewController: chooseViewControllerB = mainStoryboard.instantiateViewController(withIdentifier: "chooseStoryboardB") as! chooseViewControllerB
        self.window?.rootViewController = initialViewController
    )

    self.window?.makeKeyAndVisible(

    return true
}

我认为这是不可能的。 相反，你可以有一个初始控制器它有segue到不同的视图控制器。在启动时，您可以决定以编程方式执行哪个segue。