如何使用文件路径以编程方式从按钮打开应用程序？ Swift 4，MacOS 10.13+

Question

I am trying to create a menubar app that when a button in the menu is clicked, it will open an app already installed on the Mac.

@objc func kineticSelf(_ sender: Any){
     NSWorkspace.shared.open(URL(fileURLWithPath: "/Library/Addigy/macamanage/MacManage.app"))
    
}

menu.addItem(NSMenuItem(title: "Open Self Service", action:
        #selector(AppDelegate.kineticSelf(_:)), keyEquivalent: ""))

When I build the project I get no errors. The button shows up, but when I click it nothing happens. I have built a version in Xcode 11 and Swift 5, but I need it to run in Swift 4.

What I had in Xcode 11 and Swift 5:

@objc func kineticSelf(_ sender: Any){
    let path = "/bin"
    let selfServiceUrl = NSURL(fileURLWithPath: "/Library/Addigy/macmanage/MacManage.app", isDirectory: true) as URL
    let configuration = NSWorkspace.OpenConfiguration()
    configuration.arguments = [path]
    
    NSWorkspace.shared.openApplication(at: selfServiceUrl, configuration: configuration, completionHandler: nil)

Answer 1

I've figured out a solution:

@objc func kineticSelf(_ sender: Any){
    let kineticSelfUrl = URL(string: "/Library/Addigy/macmanage/MacManage.app")
    NSWorkspace.shared.openFile(kineticSelfUrl!.path)
    // NSWorkspace.shared.open(URL(fileURLWithPath: "/Library/Addigy/macmanage/MacManage.app"))
    
}

menu.addItem(NSMenuItem(title: "Open Self Service", action:
        #selector(AppDelegate.kineticSelf(_:)), keyEquivalent: ""))
    

Thanks!