bash completion: how to complete a list of hosts for a specific argument?

Question

My programm can have the following format:

<command_name> <arguments> <options> -hosts <host1> <host2> ... <hostn> (<more_arguments>) ...

I am trying to set bash auto-completion, and based on debian-administration's article, I have done host completion like this :

(...)
case "${prev}" in
    -hosts)
        _known_hosts_real -a "$cur"
        return 0
        ;;
    (...)
(...)

Of course after writing -hosts bash is able to complete the name of a single host, but not multiple hosts. Is there a beautiful way to achieve this hack without completing hosts when we deal with other arguments than -hosts?

Answer 1

If you ignore the (<more_arguments>) case then you can just check whether -hosts is one of the earlier words:

(...)
for ((i=1;i<COMP_CWORD;i++)); do 
  if [ "-hosts" == "${COMP_WORDS[i]}" ]; then 
    _known_hosts_real -a "$cur" 
    return 0 
  fi 
done 
case "${prev}" in 

Answer 2

I'm not sure if it's possible to create such autocompletion rule for space-separated hosts values. But if you don't mind using comma-separated lists then:

_myhost () {
  local cur prev opts
  _get_comp_words_by_ref cur prev

  opts='-hosts'
  servers='name1 name2 name3'

  if [[ ${cur} == -* ]] ; then
    COMPREPLY=( $(compgen -W "${opts}" -- ${cur}) )
  else
    case "${prev}" in
      -hosts)
        if [[ "$cur" == *,* ]]; then
          local realcur prefix
          realcur=${cur##*,}
          prefix=${cur%,*}
          COMPREPLY=( $(compgen -W "${servers}" -P "${prefix}," -- ${realcur}) )
        else
          COMPREPLY=( $(compgen -W "${servers}" -- ${cur}) )
        fi
        ;;
      *)
        # do nothing
        ;;
    esac
  fi
}
complete -F _myhost myhost