Swift, macOS: Open terminal window at the location

Question

I'm trying to lunch terminal window and go to some local path in it:

my code:

public class func openShell(at url: URL?) {
    guard let url = url else { return }
    
    let shellProcess = Process();

    shellProcess.launchPath = url.path;

    //shellProcess.arguments = [
    //  "osascript -e 'tell application \"terminal\" to do script \"cd \(url)\"'"
    //];

    shellProcess.launch();
}

but as result there is some output into debug console inside of XCode but no shell window is opened.

commented code - alternative way that I have tried

Answer 1

2021, Swift 5.0

Answer is simple:

@available(OSX 10.15, *)
public class func openTerminal(at url: URL?){
    guard let url = url,
          let appUrl = NSWorkspace.shared.urlForApplication(withBundleIdentifier: "com.apple.Terminal")
    else { return }
    
    NSWorkspace.shared.open([url], withApplicationAt: appUrl, configuration: NSWorkspace.OpenConfiguration() )
}

on each func call will be opened the new instance of the Terminal with selected location.

And no need to make XPC service or turn off sandbox for this.