我正在制作一个简单的菜单栏应用程序，其中有2个项目-首选项&退出
我想在点击首选项时打开一个新窗口
目前我有

func applicationDidFinishLaunching(_ aNotification: Notification) {
        constructMenu()
    }

func constructMenu() {
        let menu = NSMenu()

        menu.addItem(NSMenuItem(
                       title: "Preferences...", 
                       action: #selector(AppDelegate.preferencesWindow(_:)), 
                       keyEquivalent: "P"))
        menu.addItem(NSMenuItem.separator())
        menu.addItem(NSMenuItem(
                       title: "Quit", 
                       action: #selector(NSApplication.terminate(_:)), 
                       keyEquivalent: "q"))

        statusItem.menu = menu
    }

@objc func preferencesWindow(_ sender: Any) {
        print("open preference window here")
        if let storyboard = NSStoryboard(name: NSStoryboard.Name(rawValue: "Main"), bundle: nil) {
            let controller = storyboard.instantiateControllerWithIdentifier("preferencesWindow")
 as NSViewController

            if let window = NSApplication.shared.mainWindow {
                window.contentViewController = controller // just swap
            }
        }
    }

但它在这一行抛出错误
如果let storyboard = NSStoryboard（名称：NSStoryboard.名称（原始值：“Main”），bundle：nil）{
陈述
条件绑定的初始化程序必须为Optional类型，而不是“NSStoryboard”
当我点击首选项时，print语句会被记录下来，但我不知道如何以编程方式打开Windows。
我刚刚从对象库中拖动了窗口控制器，并为StoryBoard ID指定了preferencesWindow的值
由于上述错误，我还尝试了以下方法

@objc func preferencesWindow(_ sender: Any) {
        print("open preference window here")
        let storyboard = NSStoryboard(name: NSStoryboard.Name(rawValue: "Main"), bundle: nil)
        let controller = storyboard.instantiateController(withIdentifier: NSStoryboard.SceneIdentifier(rawValue: "preferencesWindow")) as! NSViewController

        if let window = NSApplication.shared.mainWindow {
            window.contentViewController = controller // just swap
        }
    }

但它只记录&从不打开窗口。我该怎么办？