On 02/15/2018 06:09 AM, Dave Howorth wrote:

On Thu, 15 Feb 2018 11:07:55 +0100 Alex Naumov wrote:

...

Hey,

can someone tell me how to get users list of some primary group. For example, I know only the name of this group: users. This is primary group and I can see it in id(1) output. How to get list of users who has the same prrmary group? Without needing to parse files.

I don't think there's a simple answer. Search unix.stackexchange to see what's possible.

A stupidly simple but not terribly accurate answer is: #> less /etc/passwd |grep <users group number> I say not terribly accurate because in my system, users group is 100 and grep returns not only 100 but anything with 100 in it. You did say without parsing files. That is as close as I can get without more coffee. -- To unsubscribe, e-mail: opensuse+unsubscribe@opensuse.org To contact the owner, e-mail: opensuse+owner@opensuse.org

On 02/15/2018 04:08 PM, Alex Naumov wrote:

The problem in your example is not parsing. My GID can come from, for example, LDAP. In that case you will get nothing.

I found this solution:

...

getent passwd | awk -F: '{if($4=='$gid')print $1}'

But I still have some problems. One of these is - how to convert name to number clear. I don't want to parse /etc/group for that.

That wouldn't help because also the groups may come from LDAP. What about: $ getent passwd | awk -F: -v gid="$(getent group users | cut -d: -f3)" '$4 == gid' ? Or below the longer variant as script. Have a nice day, Berny #!/bin/sh # Get group name from parameter. gname="$1" \ && test "$gname" \ || { echo "Usage: $0 <group-name>" >&2; exit 1; } # Determine GID of given group. gid="$(getent group "$gname" | cut -d: -f3)" \ && test "$gid" \ || { echo "$0: error: getting GID of group failed: '$gname'" >&2; exit 1; } set -o pipefail getent passwd | awk -F: -v gid="$gid" '$4 == gid' #-eof -- To unsubscribe, e-mail: opensuse+unsubscribe@opensuse.org To contact the owner, e-mail: opensuse+owner@opensuse.org

On 15.02.2018 16:08, Alex Naumov wrote:

The problem in your example is not parsing. My GID can come from, for example, LDAP. In that case you will get nothing.

I did not try, but if you use sssd, you should be possible to use it for that. You can use sss_groupshow to show information about a group, containing users. But you have to configure sssd correctly: - you have to include the "local" domain in sssd.conf - you have to enable "enumerate" in your sssd.conf Please read the warning about "enumerate TRUE" in "man sssd.conf"!

On Feb 15, 2018, at 10:19:08, David T-G wrote:

Alex & Stevens, et al --

...and then Stevens said... % % On 02/15/2018 06:09 AM, Dave Howorth wrote: % >On Thu, 15 Feb 2018 11:07:55 +0100 % >Alex Naumov wrote: % > % >>can someone tell me how to get users list of some primary group. ... % >>needing to parse files. % > % >I don't think there's a simple answer. Search unix.stackexchange to see % >what's possible.

I am inclined to agree; there is no simple and easy program that tells you that. You're gonna be parsing files.

% % A stupidly simple but not terribly accurate answer is: % #> less /etc/passwd |grep <users group number> % % I say not terribly accurate because in my system, users group is 100 % and grep returns not only 100 but anything with 100 in it. You did % say without parsing files. That is as close as I can get without % more coffee.

Well, the easy fix for this is to anchor the pattern. Something like

grep ":100:" /etc/passwd

(BTW, you don't have to cat or less through grep) will exclude 1100 and 1000 and so on. Unfortunately, it will also give you the user 100. The only way to limit to groups is to look at the GID field.

To avoid hard-coding the GID, you should first check in the /etc/groups file to see what it is. Realistically this is pretty easy to dash off in a one-liner (with a placeholder var to make it easy; if it were a shell script you'd take the group name as input):

G=groupname awk -F: -vGID=`egrep "^$G:" /etc/group | cut -d: -f3` '{if ($4==GID) print $1}'

Season to taste :-) I'd be VERY interested in how this could be

Let me see if I can write simple C program using glibc system functions for iterating over /etc/password and getting primary GID for every user. Should be doable, but it's been a while since I worked with the functions used to iterate and get information from the password struct. -- To unsubscribe, e-mail: opensuse+unsubscribe@opensuse.org To contact the owner, e-mail: opensuse+owner@opensuse.org

Hello, On Thu, 15 Feb 2018, David T-G wrote:

G=groupname awk -F: -vGID=`egrep "^$G:" /etc/group | cut -d: -f3` '{if ($4==GID) print $1}'

Season to taste :-) I'd be VERY interested in how this could be condensed, by the way!

awk -F: -vG=groupname ' BEGIN { while( getline <"/etc/group" ) { if( $1 == G ) { GID=$3; } } } $4 == GID { print $1; } ' < /etc/passwd HTH, -dnh -- Listen, three eyes, don't you try to outweird me. I get stranger things than you free with my breakfast cereal. -- Zaphod Beeblebrox -- To unsubscribe, e-mail: opensuse+unsubscribe@opensuse.org To contact the owner, e-mail: opensuse+owner@opensuse.org

Hello, On Thu, 15 Feb 2018, David T-G wrote:

% On Thu, 15 Feb 2018, David T-G wrote: ... % >Season to taste :-) I'd be VERY interested in how this could be % >condensed, by the way! % % awk -F: -vG=groupname ' % BEGIN { % while( getline <"/etc/group" ) { % if( $1 == G ) { GID=$3; } % } % } % $4 == GID { print $1; } % ' < /etc/passwd

Hmmm... Nice in that it doesn't make a separate call to look through the group file, but I don't know that I could go so far as to say that that's condensed from my previous one-liner :-)

Yes. Oh, BTW, you can also call getent for both groups and passwd. ==== $ getent passwd | awk -F: -vG=groupname ' BEGIN { while( "getent group" | getline ) { if( $1 == G ) { GID=$3; } ; } } $4 == GID { print $1; }' ==== So it'll work with NIS/LDAP too. -dnh -- 135: Druckertreiber pure virtual Araber (Martin Neumann) -- To unsubscribe, e-mail: opensuse+unsubscribe@opensuse.org To contact the owner, e-mail: opensuse+owner@opensuse.org