Login ID cleanup


 
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Operating Systems AIX Login ID cleanup
# 1  
Old 05-11-2006
Login ID cleanup

Hello

I have many old IDs on my AIX and would like to know the
simplest way of knowing the last time an ID was used.

I am familiar with the "last" command.

Thanks for any info Smilie
# 2  
Old 05-12-2006
#!/bin/ksh
#set -vx
FALSE=false
current_time=$(perl -e 'print time (),"\n";')
let expired_time=current_time-7776000
let delete_time=current_time-15552000
awk -F: '{print $1}' /etc/passwd|while read records
do
if [ "$records" = "dm" -o "$records" = "root" -o "$records" = "daemon" -o "$records" = "bin" -o "$records" = "sys" -o "$
records" = "adm" -o "$records" = "uucp" -o "$records" = "guest" -o "$records" = "nobody" -o "$records" = "lpd" -o "$reco
rds" = "nuucp" ];then
SYS=$records
else
SYS=nonsys
fi
last_login_time=$(lsuser -a time_last_login $records|awk -F= '{print $2}')
account_status=$(lsuser -a account_locked $records|awk -F= '{print $2}')
if [ "$last_login_time" = "" ];then
if [ $records = $SYS -a "$records" != "root" -a "$account_status" = "false" ];then
# chuser account_locked=true $records
printf "%-8s never logged on. The account should be locked\n" $records
fi
if [ $records != $SYS -a "$records" != "root" ];then
# chuser account_locked=true $records
printf "%-8s never logged on. The account should be locked\n" $records
fi
elif [ $records = $SYS -a $records != "root" -a "$account_status" = "false" ];then
#chuser account_locked=true $records
printf "%-8s never logged on. The account should be locked\n" $records
elif [ $records != $SYS ];then
if [ $last_login_time -lt $delete_time ];then
printf "%-8s last logged on 6 months ago. The account needs to be deleted! Check /logs/backlogs/rmusers.log\n"
$records
echo $records must be removed. `date` >> /logs/backlogs/rmusers.log
elif [ $last_login_time -lt $expired_time ];then
# chuser account_locked=true $records
printf "%-8s last logged on 3 months ago. The account should be locked\n" $records
fi
fi
done
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