我如何以编程方式设置一个故事板的InitialViewController ?我想打开我的故事板到一个不同的视图这取决于不同的启动条件。


当前回答

Swift 4, Xcode 9

在AppDelegate.swift文件中

func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {
    let storyboard = UIStoryboard(name: "Main", bundle: nil)
    let firstVC = storyboard.instantiateViewController(withIdentifier: "firstViewController") as! firstViewController
    self.window?.rootViewController = firstVC
}

其他回答

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

找到了简单的解决方案-不需要从故事板和编辑项目信息选项卡中删除“初始视图控制器检查”,并使用makeKeyAndVisible,只是放置

self.window.rootViewController = rootVC;

in

- (BOOL) application:didFinishLaunchingWithOptions:

使用Swift 3和Swift 4避免强制铸造的另一个解决方案是这样的

func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {
    self.window = UIWindow(frame: UIScreen.main.bounds)
    let storyboard = UIStoryboard(name: "Main", bundle: nil)
    guard let viewController = storyboard.instantiateViewController(withIdentifier: "YourViewController") as? YourViewController else {
        return false
    }
    self.window?.rootViewController = viewController
    self.window?.makeKeyAndVisible()
    return true
}

下面是使用UINavigationController

func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {
    self.window = UIWindow(frame: UIScreen.main.bounds)
    let storyboard = UIStoryboard(name: "Main", bundle: nil)
    guard let viewController = storyboard.instantiateViewController(withIdentifier: "YourViewController") as? YourViewController else {
        return false
    }
    let navigationController = UINavigationController(rootViewController: viewController)
    self.window?.rootViewController = navigationController
    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。

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
}