Hi !
How can I get the list of users who are logged on more than once ?
Can someone help me with this ?
Thanks !
--
Pozdrav
Revenger
14.6.2006 09:55:51
Jednom su proizveli Chuck Norris toalet papir, ali papir nije dopustao da
itko sere po njemu.

Revenger wrote:
> Hi !
> How can I get the list of users who are logged on more than once ?
> Can someone help me with this ?
> Thanks !
> --
> Pozdrav
> Revenger
> 14.6.2006 09:55:51
> Jednom su proizveli Chuck Norris toalet papir, ali papir nije dopustao da
> itko sere po njemu.

You might check the /var/adm/wtmp file or the output of "last" command
to
see if they could help.

Your question is not specific to AIX system, so I guess if you post the

question in comp.unix.shell or comp.unix.admin, you might get
responses in quick succession.

James

Revenger wrote:
> Hi !
> How can I get the list of users who are logged on more than once ?
> Can someone help me with this ?
> Thanks !
> --
> Pozdrav

This is probably not the ideal way to do it if you have a lot of users
on the system, as you have to grep a file once for EVERY user entry,
but it works. It shows a list of any duplicate user logins at the time
of the sampling. I think this could be done using awk in a better way,
but didnt have time to sit down and figure it out.

# START OF SCRIPT
#!/bin/ksh
#set -vx
/bin/rm /tmp/user.snapshot 2>/dev/null
TIMESTAMP="$(date +"%c")"
who|sort|sed -e "s/ /!/g">/tmp/user.snapshot

TITLEBAR=Y
for LOOP in $(cat /tmp/user.snapshot)
do
USRNAM=$(echo "${LOOP}"|cut -c1-8)
if test $(cat /tmp/user.snapshot|grep "^${USRNAM}"|wc -l) -gt 1
then
if test "${TITLEBAR}" = "Y"
then
TITLEBAR=N
echo "\nList of Duplicate Users as of ${TIMESTAMP}:\n"
fi
echo "${LOOP}"|sed -e "s/!/ /g"
fi
done
/bin/rm /tmp/user.snapshot
# END OF SCRIPT