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Top Forums UNIX for Advanced & Expert Users Finding user accounts not accessed for a specific number of days Post 302556157 by admin_xor on Saturday 17th of September 2011 04:13:08 PM
Old 09-17-2011
Java

Thank you all for your reply. This weekend I'm planning to sit tight and write a script for this. Once done, I'll post that here and would love to get your reviews. Smilie

---------- Post updated 09-17-11 at 04:13 PM ---------- Previous update was 09-16-11 at 07:45 PM ----------

Well thanks for your input guys. Here is the bash script I have written for our Linux and Solaris servers. It's almost similar to the ksh script I wrote for the same purpose on AIX servers. The difference is with AIX's lsuser command, it was easy for me to get user details.

Review the code and let me know your thoughts Smilie
Code:
#!/bin/bash

# this script searches for the user ids which have not been used for a specific
# number of days


# the number of days to be monitored
expdays=30
expiry=$(( 86400*$expdays ))

LOG1=/tmp/inactive_usr_$expdays_days.$( date '+%m-%d-%Y' ).log
LOG2=/tmp/never_logged_usrs.$( date '+%m-%d-%Y' ).log
tmp1=/tmp/tmp1.log.$$

# list out the accounts which can be used to log in to the server

cat /etc/passwd | grep -Ev ^"root|daemon|nobody|adm|sys" | cut -d ":" -f1 >>$tmp1

# date and time calculation with respect to unix epoch

year=$( date +%Y )

today=$( date +%s )

# now, for each of the user listed above, run a check if the last logged in time exceedes the specified number of days.

cat $tmp1 | while read usr; do
        umnt=$( last -1 $usr | head -1 | awk '{ print $5 }' )
        uday=$( last -1 $usr | head -1 | awk '{ print $6 }' )
        uhour=$( last -1 $usr | head -1 | awk '{ print $7 }' | cut -d ":" -f1 )
        umin=$( last -1 $usr | head -1 | awk '{ print $7 }' | cut -d ":" -f2 )

        # the below line checks and stores the IDs which were never used
        if [ ! $uday ] || [ ! $uhour ] || [ ! $umin ]; then
            echo "$usr:__NeveR__" >>$LOG2
        else
            llogin=$( date +%s -d"$umnt $uday, $year $uhour:$umin:01" )
            mini=$(( $today-$llogin ))
            if [[ $mini -gt $expiry ]]; then
                login=$(( ($today-$llogin)/86400 ))
                echo "$usr:$login" >>$LOG1
            fi
        fi
done

rm -f $tmp1

 

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lppasswd(1)							    Apple Inc.							       lppasswd(1)

NAME
lppasswd - add, change, or delete digest passwords. SYNOPSIS
lppasswd [ username ] lppasswd -a [ -g groupname ] username lppasswd -x username DESCRIPTION
lppasswd adds, changes, or deletes passwords in the CUPS digest password file, passwd.md5. When run by a normal user, lppasswd will prompt for the old and new passwords. When run by the super-user, lppasswd can add new accounts (-a username), change existing accounts (user- name), or delete accounts (-x username) in the digest password file. Digest usernames do not have to match local UNIX usernames. OPTIONS
lppasswd supports the following options: -g groupname Specifies a group other than the default system group. SECURITY ISSUES
By default, the lppasswd program is not installed to allow ordinary users to change their passwords. To enable this, the lppasswd command must be made setuid to root with the command: chmod u+s lppasswd While every attempt has been made to make lppasswd secure against exploits that could grant super-user privileges to unprivileged users, paranoid system administrators may wish to use Basic authentication with accounts managed by PAM instead. SEE ALSO
lp(1), lpr(1), http://localhost:631/help COPYRIGHT
Copyright 2007-2013 by Apple Inc. 22 February 2008 CUPS lppasswd(1)
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