Matching files of certain length that only contain digits?

Question

I've been trying to match all file names in a directory that are 6 characters long and only contain digits. I've been banging my head trying to get it to work but with no luck. This is what I have:

for f in "$path"*
do
        if echo "$f" | tail -c 7 | grep "^[[:digit:]][:digit:]]*$"; then
                echo "$f is good"
        else
                echo "$f" | tail -c 7
        fi
done

Every time I run the script the else gets ran and I just get the last 6 characters from each file in the directory. I don't want to have to cd into the directory to check the files so I thought taking the last 6 and checking would be a good idea (maybe not). I want to eventually redirect all files in $path that are 6 characters long and all digits. I just cant figure out how to match this. Thank you!

Answer 1

here is a one liner to just list the filenames:

for f in "$path"[0-9][0-9][0-9][0-9][0-9][0-9];do echo ${f##*/};done

if you want the full path just omit the ##*/

Answer 2

ls [0-9][0-9][0-9][0-9][0-9][0-9]

Or, if you simply want to edit your script:

for f in "$path"*
do
    if echo "$f" | tail -c 7 | grep "^[[:digit:]][[:digit:]][[:digit:]][[:digit:]][[:digit:]][[:digit:]]$"; then
            echo "$f is good"
    else
            echo "$f" | tail -c 7
    fi
done

Or, just use the regex pattern:

'^[[:digit:]]{6}$'

Answer 3

Use basename to get the filename part of a pathname, and test that:

for f in "$path"*
do
    base=$(basename "$f")
    case "$base" in
        [0-9][0-9][0-9][0-9][0-9][0-9])
            echo "$f is good" ;;
        *) echo "$base" ;;
    esac
done