The line the God throws down to you and me makes a pleasing geometry.

[reddragdiva]

The fun part of shell scripting is writing apparently perfectly Wirth-friendly code of immaculate Algol-like structure, peppered with inline bits of line noise that do the actual work. This is management-friendly because then they think they understand the immaculate structure, and because they don't understand the line noise it must therefore be simple.

#!/usr/bin/ksh
# Script XXXX_passwd_over_57_days_old
#
# D. Gerard   | 10-Jan-06   | Adapted from XXXXXXXXX version (oracle password)
#             |             | for XXXX (Unix password). Is there a Solaris
#             |             | command which just gives us this info? Should be.


# function days_since_epoch()
# Number of days since 1st January, 1970 (= day 0)
# adapted from from SysAdmin Magazine, July 2003:
# http://www.samag.com/documents/s=8284/sam0307b/0307b_l3.htm
# arguments: $1 = day, $2 = month, $3 = year in format YYYY
days_since_epoch()
{
typeset -i DSE

DSE=$((($1-32075+1461*($3+4800+($2-14)/12)/4+367*($2-2-($2-14)/12*12)/12-3*(($3+4900+($2-14)/12)/100)/4)-2440588))
echo $DSE
}

# Output file of names for deletion
LISTFILE=/tmp/oldlist
echo > $LISTFILE

# Today in days since 01 01 1970
TODAY= `days_since_epoch \`date +"%d %m 20%y"\``

# Get list of usernames from /etc/passwd. We only want those in the form aa99999
for i in `/usr/bin/awk -F:  '{print $1}' /etc/passwd | /usr/bin/grep '^[a-z]\{2\}[0-9]\{5\}$' `
do

# Get date of last password change (mmddyy, which we shuffle to yymmdd)
# NOTE: If never changed, logins gives the date as "010170"
# PS means password set; NP means no password set.
# NP will typically never have been used and have change date "010170"
DATE=`/usr/bin/logins -x -l $i | /usr/bin/awk '/PS|NP/  {print  substr($2,5,2) substr($2,1,4)}'`

# Solaris two year dates 69-99 are 1969-1999, dates 00-68 are 2000-2068
if [[ "690000" > $DATE ]]
        then DATE=`printf "20%s" $DATE`
        else DATE=`printf "19%s" $DATE`
fi

# Is $DATE more than 57 days before today? Add to list
if  [[ $(( $TODAY - `days_since_epoch \`(echo $DATE|/usr/bin/awk '{print substr($1,7,2) " " substr ($1,5,2) " " substr ($1,1,4)}')\``)) -gt 57 ]]
        then echo $i >> $LISTFILE
fi

done

# In future, we may take action on the list automatically.
exit 0

Note I've missed my chance to inline even more of the magic. The YY to YYYY conversion can be inlined with nawk and ?: and the script is less than polished in general. But that can be fixed next version. In fact, I'll see if I can get everything from logins on inside the final if.