Question

I’m trying to write a script with a file as an argument that greps the text file to find any word that starts with a capital and has 8 letters following it. I’m bad with syntax so I’ll show you my code, I’m sure it’s an easy fix.

grep -o '[A-Z][^ ]*' $1

I’m not sure how to specify that:

a) it starts with a capital letter, and

b)that it’s a 9 letter word.

Cheers

EDIT:

As an edit I’d like to add my new code:

while read p
do
echo $p | grep -Eo '^[A-Z][[:alpha:]]{8}'
done < $1

I still can’t get it to work, any help on my new code?

Answer 1

'[A-Z][^ ]*' will match one character between A and Z, followed by zero or more non-space characters. So it would match any A-Z character on its own.

Use \b to indicate a word boundary, and a quantifier inside braces, for example:

grep '\b[A-Z][a-z]\{8\}\b'

If you just did grep '[A-Z][a-z]\{8\}' that would match (for example) “aaaaHellosailor”.

I use \{8\}, the braces need to be escaped unless you use grep -E, also known as egrep, which uses Extended Regular Expressions. Vanilla grep, that you are using, uses Basic Regular Expressions. Also note that \b is not part of the standard, but commonly supported.

If you use ^ at the beginning and $ at the end then it will not find “Wiltshire” in “A Wiltshire pig makes great sausages”, it will only find lines which just consist of a 9 character pronoun and nothing else.