Question

From /etc/passwd, I need just the “user” part.

Example:

backup:x:34:34:backup:/var/backups:/bin/sh

has backup as the user.

To extract the user part I am using this now:

perl -pe 's/^(.*?):.*/\1/g' < /etc/passwd

which is a few characters less than my old method:

while read line; do echo ${line%%:*}; done < /etc/passwd

but it sounds like an overkill. It’s a lot of text to write.

A Google search gives me:

awk -F  ':' '{print $1}' < /etc/passwd

which is equivalent, I suppose? Is it?

cmp <(perl -pe 's/^(.*?):.*/\1/g' < /etc/passwd) <(awk -F  ':' '{print $1}' < /etc/passwd)

Is there a standard UNIX tool to do this or an easier method using perl? Or cut?

I am a beginner.

BTW, I tried my hand at Python as follows but I suck at it, there may be better ways to do this in Python :(

python -c 'print "\n".join([u[:u.find(":")] for u in open("/etc/passwd")])'

EDIT: actually, maybe more like this for Python:

python -c 'print "\n".join([u.split(":")[0] for u in open("/etc/passwd")])'

Hmm… still very verbose.

Answer 1

perl -F: -lane 'print $F[0]' < /etc/passwd

does nore-or-less what the awk solution does. -a means split each line of input into fields, -F: means that : is the field delimiter. And the -l outputs a newline for each line of input so you don’t have to say print "$F[0]\n".

Since you requested it, the cut solution is even more straightforward:

cut -d: -f1 < /etc/passwd

Meaning: split on :, output the first field.