UIStoryBoard从ApplicationDelegate获取第一个视图控制器

Question

我有一个我的ApplicationDelegate实现的协议。我想把它传递给故事板中定义的第一个视图控制器。如何从该方法访问它？

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

Answer 1

我不知道如果我理解正确的问题，我认为你是问你如何让在故事板第一的viewController。为了得到这个电话，

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

更改故事板的名称以适合您的名字，MainStoryboard只是默认名称。希望这是你正在寻找的。

编辑：

- (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

而是然后用

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

您可以在您的应用程序委托的头文件设置该属性

@property (nonatomic, weak) UIViewController* initialViewController; 

和创造现有故事板的新副本此方法只需将self.window.rootViewController分配给属性

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

    return YES; 
} 

这应该工作，因为在窗口的应用程序委托协议的属性是访问rootViewConroller这个控制器initialViewController如果用故事板。

Answer 3

Swift 2.0为那些想要的实现。

您可能需要修改STROYBAORD和控制器视图标识符。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 
    }