number of users in a group

cookie_ie · 03-24-2005, 11:47 AM

I know i have been asking loads but need to get this stuff done and up on my server soon. Is there a way to display the number of users in a group?on its own?i know who -q counts the users but it lists the names aswell, can I remove this just to show the number using script?

TheLinuxDuck · 03-24-2005, 12:11 PM

If you know the group id, something like

Code:

cat /etc/passwd | grep "502:" | wc -l

will tell you the count of users in that group (502 given as example). You could also parse the group file first, to get the #, and then plug that into the previous example.. assuming that they will pass the group name into a script, something like:

Code:

#!/bin/bash

groupid=`cat /etc/group | grep "${1}:" | awk -F ':' '{print $3;}'`
cat /etc/passwd | grep "${groupid}:" | wc -l

Probably not the most effective/stream lined way to do this, but it works. One flaw of this method is that if the user id matches or ends with the group ID returned, it will call it a match. But at least it's a start. (=

cookie_ie · 03-24-2005, 12:28 PM

I want to show the number of users logged on at one time and the number of users in the same group logged on, is all this possible?

gbonvehi · 03-24-2005, 12:32 PM

you can use the "id -G" command on each user detected by "who" and count the matches...

jlliagre · 03-24-2005, 12:57 PM

Code:

logins -g groupname | wc -l

Not sure if linux has the logins command, but this works on Solaris.

Dark_Helmet · 03-24-2005, 01:12 PM

Quote:

Originally posted by TheLinuxDuck

Code:

cat /etc/passwd | grep "502:" | wc -l

That has an underlying assumption in it... Specifically, that the user will only be part of a single group; their initial/login group.

If the script/tool is supposed to count all users logged in that belong to a specific group (as an example: "developer"), then there's the possibility that a user belonging to groups "users" and "developer" will be missed (if that user's primary group is "user").

Getting a true count would probably require listing the users on the system, checking the group entry in /etc/passwd, and also looking for that user's entries in /etc/group.

TheLinuxDuck · 03-24-2005, 01:55 PM

Quote:

Originally posted by Dark_Helmet
That has an underlying assumption in it... Specifically, that the user will only be part of a single group; their initial/login group.

yes.. I was just feeling lazy and didn't want to parse /etc/group, too. I'm sure that my method is probably not even close to the best way to handle it.. it was merely the first idea that came to mind.. but props for calling it out. (=

Hko · 03-24-2005, 03:43 PM

Quote:

Originally posted by cookie_ie
I want to show the number of users logged on at one time

Just the command "who" will give you all users currently logged in. But it will print the same user more than once if (s)he is logged in more than once.

If you want unique logged in user:

Code:

who | cut -d' ' -f1 | sort | uniq

Add the "-c" option to "uniq" to count the number of times each user is logged in:

Code:

who | cut -d' ' -f1 | sort | uniq -c

Quote:

[...] and the number of users in the same group logged on, is all this possible?

Here's a long one-liner that does this:

Code:

who | cut -d' ' -f1 | sort | uniq | xargs groups | sed -e 's/.*: \(.*\)/\1/' -e 's/ /\n/g' | sort | uniq -c

It prints something like:

Code:

      1 audio
      1 cdrom
      1 dialout
      2 disk
      1 heiko
      1 video
      1 weg

Which means there are 2 user logged in who are member of the group "disk". this is for unique usernames, i.e. if "heiko' is member of "dialout", and "heiko" is logged in 5 times, it will only print "1 dailout". I you want it to print "5 dailout" in this case, remove the first "sort | uniq":

Code:

who | cut -d' ' -f1 | xargs groups | sed -e 's/.*: \(.*\)/\1/' -e 's/ /\n/g' | sort | uniq -c

Which printed in the same situation on my computer:

Code:

      5 audio
      5 cdrom
      5 dialout
      6 disk
      5 heiko
      5 video
      1 weg

Note: All commands above do not count/print user sessions started with the "su" command.