bash: test if $WORD is in set

case $word in
    dog|cat|horse) echo yes;;
    *)             echo no;;
esac

This is a Bash-only (>= version 3) solution that uses regular expressions:

if [[ "$WORD" =~ ^(cat|dog|horse)$ ]]; then
    echo "$WORD is in the list"
else
    echo "$WORD is not in the list"
fi

If your word list is long, you can store it in a file (one word per line) and do this:

if [[ "$WORD" =~ $(echo ^\($(paste -sd'|' /your/file)\)$) ]]; then
    echo "$WORD is in the list"
else
    echo "$WORD is not in the list"
fi

One caveat with the file approach:

• It will break if the file has whitespace. This can be remedied by something like:

  sed 's/[[:blank:]]//g' /your/file | paste -sd '|' /dev/stdin
  

Thanks to @terdon for reminding me to properly anchor the pattern with ^ and $.

How about:

#!/usr/bin/env bash

WORD="$1"
for w in dog cat horse
do
  if [ "$w" == "$WORD" ] 
  then
      yes=1
      break
  fi
done;
[ "$yes" == "1" ] && echo "$WORD is in the list" || 
                     echo "$WORD is not in the list"

Then:

$ a.sh cat
cat is in the list
$ a.sh bob
bob is not in the list