using awk on only files which START with 'xyz'

Question

I have 3 files in a directory: xyz1.txt, xyz2.txt, rst_xyz.txt and I want to work with ONLY the first two (those which start with xyz). The command ls | awk '/xyz/{print $1}' will return all three of them. What alterations can I make to the command to specify ONLY the xyz* files? I tried ls | awk '/^xyz/{print $1}' thinking that it would make sure the file started with xyz, but that did not exclude rst_xyz.txt.

Answer 1

Don't parse ls!

You want

awk '...' xyz*

If, for some very odd reason, you want to do this entirely within awk:

awk 'FILENAME ~ /^xyz/ { .... }' *

Answer 2

no need for awk. just use filename glob:

ls xyz*

most likely you will want to pipe the filenames to work on them? like this maybe?

ls xyz* | some other commands

in that case it is better to use a for loop.

for filename in xyz*
do
  some other commands "$filename"
done

note that in the first variant the filenames will come as a stream to stdin of some other commands while in the for loop the filenames will come one by one as a command line argument for some other command.

Answer 3

ls | awk '/^xyz/'

Your original solution was just missing the ^ in the regep, which anchors the match to the beginning of the line. You don't need a print block because the default action when finding a match is to print the line. If you want to be explicit, you could write:

ls | awk '/^xyz/ {print $0}'

You should use $0 rather than $1 in case the filename contains spaces (although if the filename contains newline, this will still treat it as multiple names).