Remove specific word in variable

Try:

$ printf '%s\n' "${FOO//$WORDTOREMOVE/}"
CATS DOGS FISH

This also work in ksh93, mksh, zsh.

---

POSIXLY:

FOO="CATS DOGS FISH MICE"
WORDTOREMOVE="MICE"

remove_word() (
  set -f
  IFS=' '

  s=$1
  w=$2

  set -- $1
  for arg do
    shift
    [ "$arg" = "$w" ] && continue
    set -- "$@" "$arg"
  done

  printf '%s\n' "$*"
)

remove_word "$FOO" "$WORDTOREMOVE"

It assumes your words are space delimited and has side effect that remove spaces before and after "$WORDTOREMOVE".

Using bash substring replacement:

FOO=${FOO//$WORDTOREMOVE/}

The // replaces all occurences of the substring ($WORDTOREMOVE) with the content between / and }. In this case nothing.

For information on this and other ways to work with strings in bash, see the section 10.1. Manipulating Strings of the Advanced Bash-Scripting Guide.

echo $FOO | sed s/"$WORDTOREMOVE"//