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Full Discussion: login audit bash script
Top Forums Shell Programming and Scripting login audit bash script Post 302599945 by agama on Sunday 19th of February 2012 01:29:38 PM
Old 02-19-2012
Yes, sorting the output would have been disaster!

You can try adding a sort after the grep in the pipeline. This will sort based on the start login time and write the most recent events first. In addition, add an exit after you print the first bit of data. This will allow any 'still logged in' records to be discarded, and will print the last complete session for the user.

Code:
grep $user /tmp/$USER.wtmp| sort  -k 7rn,7 -k 4rM,4 -k 5rn,5 -k 6rn,6  | while read  uname tty st_dow st_mon st_day st_time st_yr junk end_dow end_mon end_day end_time end_yr ttime host 
do
    case $end_dow in
        crash|gone|down|still) ;;       # ignore those with missing info
        *)
            printf "\n"
            printf "       User:\t%s\n" $uname
            printf "      Start:\t%s %s %s %s\n" $st_dow $st_mon $st_day $st_time
            printf "        End:\t%s %s %s %s\n" $end_dow $end_mon $end_day $end_time
            printf "    Time On:\t%s\n" $ttime
            printf "Remote Host:\t%s\n" ${host:-localhost}
            exit 0
        ;;
    esac
done

---------- Post updated at 13:29 ---------- Previous update was at 13:26 ----------

I forgot to mention that you might want to tighten up the sort a bit. It sorts based on year, month, day and hour, but doesn't take into account minutes, so if a user logged in and out twice in the same hour, you might not get the most recent one.
 

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