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Simple Shell Script to remind user of upcoming pw change

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  • 24-09-2015 11:32am
    #1
    Stuxnet
    Registered Users Posts: 3,735 ✭✭✭


    Lads,

    There is a password expiration policy as you would expect on all of our Linux servers globally at work, in the domains I'm part of, users are both sent reminder emails and the shell lets them know when passwords are about to expire, thus reducing tickets etc....but we have one domain I do not have any admin access on, I am a regular user...and the admins of this domain do not provide any for-warning of impending password expiration, pain in the arse, especially when it can can takes days for your ticket for pw reset to be serviced on that domain...

    I cant use the Linux command chage on this domain, to edit passwd file to give notification.

    Short of setting up a goggle calender reminder to myself...does anybody have an idea for a bash script that I could load into my .profile on that domain, echoing a backwards countdown from a future date to remind me to change pre-emptively !

    pseudo code: $expiry_date; $current_date; $date_diff ; echo 'Reminder: Change password in $date_diff days. '

    `date` is complicated when you haven't fooled around with it :-p

    bash or ksh ...Im not good at scripts...any help appreciated cheers.


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    #2
    Skrynesaver
    Registered Users Posts: 1,109 ✭✭✭Skrynesaver


    Something like this would work (could be tidied up mind you... 
    chage -l ${USER} | perl -MTime::Local -ne ' # capture expiry date from chage
if (($mon,$day,$year)=$_=~/Password expires\s+:\s+(\S+)\s+(\d+),\s+(\d+)/){  # If this is the expiry date line
%mon=(Jan=>0,Feb=>1,Mar=>2,Apr=>3,May=>4,Jun=>5,Jul=>6,Aug=>7,Sep=>8,Oct=>9,Nov=>10,Dec=>11); # create a month=>localtime hash
$exp_time=timelocal(0,0,0,$day,$mon{$mon},$year - 1900); #generate epoch for midnight on $date
$remaining=int(($exp_time - time())/(24 * 60 * 60)); #get remaining days
print "You have $remaining days to password expiry\n"}' #Display remaining days


  • Options
    #3
    Stuxnet
    Registered Users Posts: 3,735 ✭✭✭Stuxnet


    Cheers for that, I managed to do it !!

    Should of tried harder before posting here :)
    #!/bin/sh

expiry=`date -d "Dec 22" +%j`
today=`date +%j`
difference=$(($expiry - $today))

if [[ ! $difference =~ ^[0-9+$ ]]; then
echo Change password in $difference days.

else
        echo Password has expired.
fi


  • Options
    #4
    PrzemoF
    Registered Users Posts: 1,931 ✭✭✭PrzemoF 


    #!/bin/sh
expiry=`date -d "Sep 19 2016" +%s`
today=`date +%s`
diff_days=$(echo "($expiry - $today) / 86400" | bc)

if [ $diff_days -gt 0 ]; then
        echo Change password in $diff_days days.
else
        echo Password has expired $(echo "-1*$diff_days" | bc) days ago
fi

    I could not understand your code, so I tweaked it a bit :)


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