Finding strings with a specific number of letters, extensions included with grep

Question

If I pipe ls to grep, how would I be able to return a list of files with the condition that they only have an x amount of letters, extension included?

So for example, if ls gives me:

abcde.jav a156e.exc test.c prog1.c qwert r.c 

and I'm looking for all files that contain strictly 5 letters, extensions included:

a156e.exc test.c prog1.c qwert

I've tried:

ls | grep '^[a-z]${5}'
ls | grep "^[a-z]|[a-z]$"

and other things like that, but I can't seem to get it. It seems like the solution should be really simple but I can't figure it out. Any help would be appreciated.

Answer 1

You could use the following expression:

Live Example

^([^a-z]*[a-z][^a-z]*){5}$

Explanation:

• ^ - Anchor denoting the start of the string
• ([^a-z]*[a-z][^a-z]*) - Group that matches a single letter between zero or more non-letter characters
• {5} - Match the previous group 5 times
• $ - Anchor denoting the end of the string.

Usage:

ls | grep -E '^([^a-z]*[a-z][^a-z]*){5}$'

Or without ls:

for f in *; do
  if [[ $f =~ ^([^a-z]*[a-z][^a-z]*){5}$ ]]
    then
      echo $f
  fi
done