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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
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3 Answers 3

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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
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I would use grep:

grep -Eo '[0-9]+' file.txt
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  • sorry I forgot to mention I can only use sed awk and bash
    – bull
    Oct 25, 2020 at 1:15
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    Why are you restricted to those commands? Is this homework?
    – waltinator
    Oct 25, 2020 at 3:54
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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.

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  • I can only use sed awk or bash
    – bull
    Oct 25, 2020 at 1:24
  • @kevo885 There is no compelling reason to impose that restriction as far as I can see. A Unix system always has both grep and ls.
    – Kusalananda
    Oct 25, 2020 at 9:32

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