How to extract the numbers from a filename

Question

How do I get only the numbers from a filename in another directory? Using ONLY sed awk or bash? Assume it can be any arbitrary length but the format will always be ABC1234.

If I have a file name ABC1311.crs then I only want 1311.
I've tried:

number=`awk -F '[]' '{print $2}' $data/ABC123.crs

Answer 1

This seems like an odd requirement, but to extract digits from arbitrary filenames, you could use parameter expansion to remove anything from the filename that's not a digit:

for f in *
do
  printf '%s\n' "${f//[![:digit:]]/}"
done

Answer 2

I would use grep:

grep -Eo '[0-9]+' file.txt

Answer 3

You can get the numbers by the following command:

echo * | grep -o '[[:digit:]]*'

You can replace echo * by ls if you like.

If you just want to use awk, then you can do the following:

\ls | awk 'match($0, /[[:digit:]]+/) {print substr($0, RSTART, RLENGTH)}'

\ls will list all the files in your directory but awk will print only the first set of numbers. If you have files with multiple numeric strings (such as 2043x924_fjnndkk.jpg), it will only output 2043.

Use \ls if your ls is aliased to something; otherwise, a plain ls will suffice. In my case, I have aliased ls to ls -s (among other options) and so, always get the size printed which interferes with this command.