Bash, use case statement to check if the word is in the array

The list in complete -W list is interpreted as a $IFS delimited list, and it's the $IFS at the time of completion that is taken into account.

So if you have:

complete -W 'a b,c d' do_work

do_work completion will offer a, b,c and d when $IFS contains space and a b and c d when $IFS contains , and not space.

So it's mostly broken by design. It also doesn't allow offering arbitrary strings as completions.

With these limitations, the best you can do is assume $IFS will never be modified (so will always contain space, tab and newline, characters that as a result can't be used in the completion words), and do:

choices='foo bar baz'
do_work() {
  case "$1" in
    (*' '*) echo 'no!';;
    (*)
      case " $choices " in 
        (*" $1 "*) echo 'yes!';;
        (*) echo 'no!';;
      esac;;
  esac
}
complete -W "$choices" do_work

You could add a readonly IFS to make sure $IFS is never modified, but that's likely to break things especially considering that bash doesn't let you declare a local variable that has been declared readonly in a parent scope, so even functions that do local IFS=, would break.

As for the more generic question of how to check whether a string is found amongst the elements of an array, bash (contrary to zsh) doesn't have an operator for that but, you could easily implement it with a loop:

amongst() {
  local string needle="$1"
  shift
  for string do
    [[ $needle = "$string" ]] && return
  done
  false
}

And then:

do_work {
  if amongst "$1" "${choices[@]}"; then
    echo 'yes!'
  else
    echo 'no!'
  fi
}

The more appropriate structure to look-up strings is to use hash tables or associative arrays:

typeset -A choices=( [foo]=1 [bar]=1 [baz]=1 )
do_work() {
  if [[ -n ${choices[+$1]} ]]; then
    echo 'yes!'
  else
    echo 'no!'
  fi
}
complete -W "${!choices[*]}" do_work

Here with "${!choices[*]}" joining the keys of the associative array with whichever is the first character of $IFS at that point (or with no separator if it's set but empty).

Note that bash associative arrays can't have an empty key, so the empty string can't be one of the choices, but anyway complete -W wouldn't support that either and completing an empty strings is not very useful anyway except maybe for the completion listing showing the user it's one of the accepted values.

Solved by user22565 (483,412 )
Sep 02, 2022 at 07:27