How to open new app with arguments from swift app using terminal command

Question

Hello I want to open new app with arguments using terminal command with in swift code of my application MacOS

I have terminal command open -n /Applications/test.app -- args arg1 this work fine when I run it in terminal

but when I am trying to run it using swift code

 static func shellCommand () {
        
        let task = Process()

        task.launchPath = "/bin/zsh"

        let args:[String] = ["-c","open -n /Applications/test.app","--args aaaa"]

        task.arguments = args
        
        let pipe = Pipe()
        let errorPipe = Pipe()
        task.standardOutput = pipe
        task.standardError = errorPipe
        task.launch()
        task.waitUntilExit()
        
        let data = pipe.fileHandleForReading.readDataToEndOfFile()
        let output = String(data: data, encoding: String.Encoding.utf8)!
        let errorData = errorPipe.fileHandleForReading.readDataToEndOfFile()
        let error = String(decoding: errorData, as: UTF8.self)

        print("out put from shell command \(output) error \(error)")
    }

It doesn't work I also tried let args:[String] = ["-c","open -n /Applications/test.app --args arg1"]

Thank you for any kind of hint or help

Answer 1

I'm pretty sure that instead of this

let args = ["-c","open -n /Applications/test.app","--args aaaa"]

it should be this

let args = ["-c", "open", "-n", "/Applications/test.app", "--args", "aaaa"]

Also, you don't really need to go through zsh. You could just call open directly.

task.launchPath = "/usr/bin/open"
let args = ["-n", "/Applications/test.app", "--args", "aaaa"]