Question

我正在制作一个简单的菜单栏应用程序，其中包含2个项目-首选项和退出

当我单击首选项

时，我想打开一个新窗口

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
            }
        }
    }

但是它会在这行上引发错误

如果让Storyboard = NSStoryboard（name：NSStoryboard.Name（rawValue：“ Main”），bundle：nil）{

状态

用于条件绑定的初始化程序必须具有可选类型，而不是“ NSStoryboard”

当我单击首选项时，会记录 print 语句，但是我不知道如何以编程方式打开Window。

我刚刚从对象库中拖动了 Window Controller ，并为 StoryBoard ID 指定了 preferencesWindow

由于上述错误，我也尝试了以下方法

@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
        }
    }

，但它仅记录日志且从不打开窗口。我该怎么办？

Answer 1

当我在SO上发布问题时，我会很快找到答案。这是对我有用的东西

@objc func preferencesWindow(_ sender: Any) {
        var myWindow: NSWindow? = nil
        let storyboard = NSStoryboard(name: NSStoryboard.Name(rawValue: "Main"),bundle: nil)
        let controller = storyboard.instantiateController(withIdentifier: NSStoryboard.SceneIdentifier(rawValue: "preferencesWindow")) as! NSViewController
        myWindow = NSWindow(contentViewController: controller)
        NSApp.activate(ignoringOtherApps: true)
        myWindow?.makeKeyAndOrderFront(self)
        let vc = NSWindowController(window: myWindow)
        vc.showWindow(self)
    }

Cocoa Xcode：使用Mac的Swift 4以编程方式从AppDelegate打开窗口控制器？

1 个答案: