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
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
}
其他回答
选择要首先打开的视图控制器,然后转到属性检查器。 转到初始场景,检查初始视图控制器选项。
这是你的初始视图控制器在应用启动时首先打开。
打开mainstoryboard,首先选择要启动的视图,然后打开Utilities—> Attributes。在“视图控制器”下面,你会看到“Is initial View Controller”单选按钮。选择它。
——对修改后的问题:
也许你可以试试这个:在你的初始视图的ViewDidLoad部分写一个方法,当这个方法在应用程序启动时运行,方法触发一个到另一个视图的segue。
谢谢在AppDelegate中修改如下:
func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool {
//Some code to check value of pins
if pins! == "Verified"{
print(pins)
self.window = UIWindow(frame: UIScreen.mainScreen().bounds)
let mainStoryboard: UIStoryboard = UIStoryboard(name: "HomePage", bundle: nil)
let exampleViewController: UINavigationController = mainStoryboard.instantiateViewControllerWithIdentifier("SBHP") as! UINavigationController
self.window?.rootViewController = exampleViewController
self.window?.makeKeyAndVisible()
}else{
print(pins)
self.window = UIWindow(frame: UIScreen.mainScreen().bounds)
let mainStoryboard: UIStoryboard = UIStoryboard(name: "Main", bundle: nil)
let exampleViewController: UINavigationController = mainStoryboard.instantiateViewControllerWithIdentifier("SBUser") as! UINavigationController
self.window?.rootViewController = exampleViewController
self.window?.makeKeyAndVisible()
}
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
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
}
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