Pass on arguments with double quotation marks from one bash script to another

Question

A bash scrip is invoked like this:

$./script 25 "str1 str2"

and it is supposed to launch a terminal, that runs another script that receives both arguments, exactly as they are above (including quotation marks). I've tried this:

lxterminal --command=$"./script2 "$"$@"

but this seems to omit the quotations marks, so the call would end up as ./script2 25 str1 str2. What is the correct notation to replicate the arguments as they were in the original command line?

Answer 1

The problem is that the argument to lxterminal's --command is just one string, it can't take a command and its arguments like some other terminals like xterm do.

lxterminal parses that string using its own rules to determine the command to run an its arguments. That's similar but not identical to the Bourne shell parsing.

It does recognise '...' as strong quotes and space as argument separator, so you can implement quoting for it as:

lxquote() {
  awk -v q="'" '
    function lxquote(s) {
      gsub(q, q "\\" q q, s)
      return q s q
    }
    BEGIN {
      for (i = 1; i < ARGC; i++) {
        printf sep "%s", lxquote(ARGV[i])
        sep = " "
      }
    }' "$@"
}

And call lxterminal as:

lxterminal --command="$(lxquote ./script2 "$@")"

Alternatively, if script's interpreter is bash, you can do:

printf -v code '%q ' ./script2 "$@"
CODE=$code lxterminal --command="bash -c 'eval \"\$CODE\"'"