Hello. I have an RS/6000 running AIX 4 and I need to be able to see if there are any users that are logged on more than once from the same terminal so I can kick them off to make room for other terminals. 64 connections is the limit. Currently I am doing this:

who | more
and then manually looking for duplicate IP's and then doing
ps -t pts/whatever or tty
to find the PID then killing the PID. Is there an easier and faster way of doing this? Thanks

-paul

Instead of manually looking for duplicate IDs you can pipe the output to sort for getting unique records...

who | sort -uk6.2,6

Thanks - but that is not showing me duplicates, instead it looks like there are no duplicates. I just tested by opening up two more connections from my machine and ran that command.

By removing the U, so it is just
who | sort -k6.2,6

it works. Thanks again

It can sometimes be difficult to "imagine" what your output looks like. Thus, examples are always a good way to convey the requirments.

Does this code:

who | tr -s " " | cut -d" " -f1,6 | sort | uniq -c

Here is a snippet of who | sort -uk6.2,6 :

# who | sort -uk6.2,6
bhb tty11 Jun 16 11:06
rf1 pts/67 Jun 16 10:21 (10.203.32.105)
rf1 pts/20 Jun 16 10:58 (10.203.32.109)
bhb pts/4 Jun 16 06:24 (10.203.32.110)
rf1 pts/28 Jun 16 09:54 (10.203.32.117)
bhb pts/63 Jun 16 09:00 (10.203.32.123)
rf1 pts/9 Jun 16 09:42 (10.203.32.124)
bhb pts/50 Jun 16 09:26 (10.203.32.126)
root pts/54 Jun 16 08:30 (10.203.32.132)
bhb pts/25 Jun 16 07:21 (10.203.32.151)
bhb pts/10 Jun 16 10:49 (10.203.32.152)
bhb pts/35 Jun 16 07:40 (10.203.32.157)
bhb pts/2 Jun 16 10:47 (10.203.32.159)
bhb pts/23 Jun 16 07:20 (10.203.32.163)
bhb pts/26 Jun 16 07:22 (10.203.32.171)
bhb pts/1 Jun 16 06:26 (10.203.32.172)
bhb pts/41 Jun 16 07:53 (10.203.32.173)
bhb pts/27 Jun 16 11:13 (10.203.32.183)
bhb pts/21 Jun 16 10:28 (10.236.1.131)
bhb pts/24 Jun 16 11:54 (10.236.32.93)
bhb pts/42 Jun 16 09:24 (10.238.0.148)
bhb pts/39 Jun 16 07:52 (10.238.1.29)
bhb pts/31 Jun 16 10:34 (10.238.1.42)

here is a snippet of who | sort -k6.2,6 :

# who | sort -k6.2,6
bhb tty11 Jun 16 11:06
bhb tty14 Jun 16 09:44
bhb tty17 Jun 16 09:47
bhb tty37 Jun 16 08:25
bhb tty49 Jun 16 09:21
bhb tty8 Jun 16 07:54
root tty1 Jun 16 08:18
rf1 pts/67 Jun 16 10:21 (10.203.32.105)
rf1 pts/20 Jun 16 10:58 (10.203.32.109)
bhb pts/4 Jun 16 06:24 (10.203.32.110)
rf1 pts/28 Jun 16 09:54 (10.203.32.117)
bhb pts/63 Jun 16 09:00 (10.203.32.123)
rf1 pts/9 Jun 16 09:42 (10.203.32.124)
bhb pts/50 Jun 16 09:26 (10.203.32.126)
root pts/54 Jun 16 08:30 (10.203.32.132)
bhb pts/25 Jun 16 07:21 (10.203.32.151)
bhb pts/10 Jun 16 10:49 (10.203.32.152)
bhb pts/35 Jun 16 07:40 (10.203.32.157)
bhb pts/2 Jun 16 10:47 (10.203.32.159)
bhb pts/23 Jun 16 07:20 (10.203.32.163)
bhb pts/26 Jun 16 07:22 (10.203.32.171)
bhb pts/1 Jun 16 06:26 (10.203.32.172)
bhb pts/41 Jun 16 07:53 (10.203.32.173)

(un)fortunately, I have killed all the duplicate sessions, so none are showing up in my examples. To get really lazy, is there a way to ONLY show the duplicate IPs?

who | \
sort -k6.2 | \
awk '{
  if (x[$6]++) {
    if (prev)
       printf("%-8s %s\n", prev, $6)
    printf("%-8s %s\n", $2, $6)
    prev=""
  } else
    prev = $2
}'