的XCode 6：我無法通過其故事板ID

Question

此選擇一個視圖是代碼：

let theBoard = self.storyboard! 
let vc = theBoard.instantiateViewControllerWithIdentifier("myViewId") as! UIViewController // this line causes the error 

第二行導致此錯誤：

unexpectedly found nil while unwrapping an Optional value 

*非常重要*：該錯誤僅在構建配置設置爲「debug」時觸發。當設置爲「release」時，它運行良好。

爲什麼？

Answer 1

您正在將UIViewController投射到let vc = theBoard.instantiateViewControllerWithIdentifier("myViewId") as! UIViewController這是錯誤的。 你應該嘗試：

var storyboard = UIStoryboard(name: "Main", bundle: nil) 

var vc = storyboard.instantiateViewControllerWithIdentifier("myViewId") as! ViewController 

Answer 2

使用這樣

var moveToNextVC:ViewController = self.storyboard?.instantiateViewControllerWithIdentifier("myViewId") as! ViewController 

self.presentViewController(moveToNextVC, animated: true, completion: nil) 

記住設置StoryboardID確定「myViewId」的文件檢查您的視圖 - 控制

Answer 3

這一切是新的語法，功能問題並沒有改變：

let storyboard = UIStoryboard(name: "MyStoryboardName", bundle: nil) 
let vc = storyboard.instantiateViewControllerWithIdentifier("someViewController") as! UIViewController 
self.presentViewController(vc, animated: true, completion: nil)