Print a specific letter of “ls -1” command [bash, sed, awk, grep]

Question

Good day!

I was wondering how to print a specific letter of the following output:

ls -1 *Mcmm*txt *Mmmm*txt

It gives:

Mcmm_E.txt
Mcmm_M.txt
Mmmm_E.txt
Mmmm_M.txt

And I want to obtain:

E
M
E
M

Thanks in advance for any suggestion

Answer 1

You could try this. It prints the last-but-one field based on _ and . as separators(assuming that you need to print everything between the last _ and . and assuming that there is one instance of . in each file name)

ls -1 *Mcmm*txt *Mmmm*txt | awk -F'_|\\.' '{print $(NF-1)}'

Answer 2

ls -1 *Mcmm*txt *Mmmm*txt | cut -b6

Is a lot less typing than using awk...

cut -b6 selects byte 6 of each line.

Answer 3

You should always avoid parsing ls ; see here . Use find instead:

find . -type f -name '*M[cm]mm*.txt' -print0

And if you have gnu grep , you could pipe into:

grep -oPz '[^_](?=.txt)'

Results:

E
E
M
M

Answer 4

Another awk

ls -1 *Mcmm*txt *Mmmm*txt | awk -F"[_.]" '{print $2}'
E
M
E
M

Answer 5

Do not parse the output of ls . In your specific case, if you only want to print the sixth character of each file name matching your pattern, do:

files_ary=( *Mcmm*txt *Mmmm*txt )
for f in "${files_ary[@]}"; do
    printf '%s\n' "${f:5:1}"
done

To only print the character in front of .txt , a funny possibility is:

files_ary=( *Mcmm*.txt *Mmmm*.txt )
for f in "${files_ary[@]%.txt}"; do
    printf '%s\n' "${f: -1}"
done

These methods are 100% pure bash, and they will work even if you have funny symbols in the file names (spaces, newlines, etc.). Oh, and to be 100% safe, don't forget to use globs with shopt -s nullglob .

Answer 6

ls -1 *Mcmm*txt *Mmmm*txt | sed -u "s/^.\{5\}\(.\).*/\1/"

-u for stream treatment (avoir buffering on huge folder)

I think cut will be faster

In this case, you don't take care of the first char even with a * in the ls