[SOLVED] Finding out connected users in shell

Question

Hi there!
This may be a simple question at first glance: who is connected and how many sessions per user are opened?  I thought a 'who' or 'w' would do it.
Sadly, it will only work with TTY sessions. Any session started from a login manager for X will not be taken into account by any of the aforementioned commands.
'loginctl list-... ' does the trick for systemd-based systems. But I'm looking for a more portable solution (BSD, other GNU, etc.).
Any thoughts?
Last edited by Ambrevar (2014-01-11 08:51:00)

Answer 1

Thanks, Trilby, this seems to be a good lead!
I've been playing with 'ps' a little to see what I could get. First I think the following command is equivalent and more portable:
$ ps -ax -o user='' -o tty='' | sort -u
'-e' is GNU-specific, and --no-header can be avoided by setting all header entries to the null string.
This will display all 'users', even 'root', 'polkitd', 'dbus', etc. Every X terminal will add another superfluous entry, but we can get rid of it easily:
$ ps -ax -o user='' -o tty='' | grep -v ' pts/' | sort -u
To display a list of all users with the number of started sessions:
$ ps -ax -o user='' -o tty=''| sort -u | awk '$0 !~ / pts\// {x[$1]++} END{for(i in x) print i, x[i]}'
doe 2
polkitd 1
dbus 1
root 3
To display the number of started session for the specified user:
$ ps -o user='' -o tty='' -u $USER | sort -u | grep -cv ' pts/'
Do not use '-U $USER', since it will display one root entry per X session (X euid is $USER but X ruid is root).
I've not tested the above commands on BSD systems, but according to the 'ps' man page from FreeBSD, this should work well. But it may fail if the TTY are not reported the same way. It also relies on the pts names.
Don't know if anyone can come up with a better suggestion, but for now I'm quite satisfied with the result, so I'll mark it as a 'solved'. Thanks again!
Last edited by Ambrevar (2014-01-11 08:53:36)