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Swift:创建一个调用另一个class1中的方法的按钮

我有2个类,第一个类名为"store",我在其中创建一个调用方法的按钮:"storeSelected"位于第二个类中,名称为ExploreViewController.
该方法应该打印"完成"并带我到另一个视图控制器.如果没有segue,则会打印"done",但是当放置segue代码时,app会崩溃.

错误是:
Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: 'Receiver (<Glam.ExploreViewController: 0x14ed3be20>) has no segue with identifier 'ok'' .........

// ExploreViewController Class

let _sharedMonitor: ExploreViewController = { ExploreViewController() }()

class ExploreViewController: UIViewController, UIScrollViewDelegate {

    class func sharedMonitor() -> ExploreViewController {
        return _sharedMonitor
    }


    func storeSelected(sender: UIButton) {                        
        println("done") // it entered here and "done" is printed                                                                   
        self.performSegueWithIdentifier("ok", sender: self) //here is the problem                           
    }
}







// another class named "Store"  
// button is created …
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objective-c ios segue swift ios8

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