Running sh/bash/python scripts with arguments using Go - linux

Question

I've been stuck on this one a few days, I'm trying to run a bash script which runs off of the first argument (maybe I should give up all hope, haha)
Syntax for running the script can be assumed to be:
sudo bash script argument or since it has og+x it can be ran as just sudo script argument
In go I'm running it using the following:
package main
import (
"os"
"os/exec"
"fmt"
)
func main() {
c := exec.Command("/bin/bash", "script " + argument)
if err := c.Run(); err != nil {
fmt.Println("Error: ", err)
}
os.Exit(0)
}
I have had absolutely no luck, I've tried loads of other variations as well for this...
exec.Command("/bin/sh", "-c", "sudo script", argument)
exec.Command("/bin/sh", "-c", "sudo script " + argument) (my first try)
exec.Command("/bin/bash", "-c", "sudo script" + argument)
exec.Command("/bin/bash", "sudo script", argument)
exec.Command("/bin/bash sudo script" + argument)
Most of these I am met with '/bin/bash sudo ect' no such file or directory, or Error: exit status 1 I have even gone as far as to write a Python wrapper looking for an argument and executing the bash script with subprocess. To rule out the path to the script not being defined I have tried all of the above with a direct route to the script rather than script name.
For the sake of my remaining hair, what am I doing wrong here? How can I better diagnose this problem so that I can get more information rather than exit status 1?

Answer 1

You don't need to call bash/sh at all, simply pass each argument alone, also to get the error you have to capture the command's stderr, here's a working example:
func main() {
c := exec.Command("sudo", "ls", "/tmp")
stderr := &bytes.Buffer{}
stdout := &bytes.Buffer{}
c.Stderr = stderr
c.Stdout = stdout
if err := c.Run(); err != nil {
fmt.Println("Error: ", err, "|", stderr.String())
} else {
fmt.Println(stdout.String())
}
os.Exit(0)
}