Find word in sentence with Bourne Shell (sh)

Question

I'm trying to determine if a particular word appears in a sentence using Bourne Shell (sh). For example:

#!/bin/bash
INPUT='Lorem ipsum dolor sit amet.'

if [[ "$INPUT" == *dolor* ]]
then
    echo "So true"
else
    echo "Not true"
fi

This works in bash, but I can't use bash, I need something that works in sh

Answer 1

I suspect you meant POSIX shell syntax instead of Bourne shell as the Bourne shell is becoming quite a rare dinosaur these days.

Anyway, the answer is the same in both cases:

case $INPUT in
  *dolor*) echo true;;
  *) echo false;;
esac

The last ;; before esac is not required. The POSIX syntax also allows (*dolor*).

Answer 2

You can use other tools:

• grep: if echo "$INPUT" | grep --quiet '\bdolor\b'

Answer 3

In case you want to match whole words portably, first, you need to define how words are delimited. If we go for GNU grep -w's definition, then a word is a sequence of letters, numbers or underscore (and a delimiter would be any character that is not any of those). Unfortunately, the definition of "letter" is locale-dependant. POSIX shells can specify letters (with [[:alpha:]], but I don't know of any variant of the Bourne shell that does. So with a POSIX shell, you could do:

word_delimiter='[![:alnum:]_]'
case +$INPUT+ in
 (*${word_delimiter}dolor${word_delimiter}*) echo true;;
 (*) echo false;;
esac

And in the Bourne shell, you would have to assume US letters:

word_delimiter='[!a-zA-Z0-9_]'
case +$INPUT+ in
  *${word_delimiter}dolor${word_delimiter}*) echo true;;
  *) echo false;;
esac

Other options:

if
  tr -cs '[:alnum:]_' '[\n*]' << EOF | grep -qx dolor
$INPUT
EOF
then
  echo true
else
  echo false
fi

While that syntax is POSIX, if you have to deal with as old a system as one having a Bourne shell, you may have issues with that "tr" syntax as there used to be two main variants of it in the old pre-POSIX days.