UIStoryBoard從ApplicationDelegate獲取第一個視圖控制器

Question

我有一個我的ApplicationDelegate實現的協議。我想把它傳遞給故事板中定義的第一個視圖控制器。如何從該方法訪問它？

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions 

Answer 1

我不知道如果我理解正確的問題，我認爲你是問你如何讓在故事板第一的viewController。爲了得到這個電話，

UIStoryboard *storybord = [UIStoryboard storyboardWithName:@"MainStoryboard" bundle:nil]; 
UIViewController *vc =[storybord instantiateInitialViewController]; 

更改故事板的名稱以適合您的名字，MainStoryboard只是默認名稱。希望這是你正在尋找的。

編輯：

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions 
{ 
    UIStoryboard *storybord = [UIStoryboard storyboardWithName:@"MainStoryboard" bundle:nil]; 
    UIViewController *vc =[storybord instantiateInitialViewController]; 
    //set the delegate on the view controller that you have loaded 
    // Override point for customization after application launch. 
    return YES; 
} 

Answer 2

而是然後用

UIStoryboard *storybord = [UIStoryboard storyboardWithName:@"MainStoryboard" bundle:nil]; 

您可以在您的應用程序委託的頭文件設置該屬性

@property (nonatomic, weak) UIViewController* initialViewController; 

和創造現有故事板的新副本此方法只需將self.window.rootViewController分配給屬性

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions 
{ 
    self.initialViewController = self.window.rootViewController; 

    return YES; 
} 

這應該工作，因爲在窗口的應用程序委託協議的屬性是訪問rootViewConroller這個控制器initialViewController如果用故事板。

Answer 3

Swift 2.0爲那些想要的實現。

您可能需要修改STROYBAORD和控制器視圖標識符。另外o

func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool { 

     let storyboard: UIStoryboard = UIStoryboard(name: "[STORYBOARD NAME]", bundle: nil) 
     let viewController:UIViewController = storyboard.instantiateViewControllerWithIdentifier("[CONTROLLER IDENTIFIER]") // You have replace UIViewController with you customer controll type 


     self.window?.rootViewController = viewController 
     self.window?.makeKeyAndVisible() 

     return true 
    }