For loops with multiple variables

sparcguy · March 24, 2011, 3:15am

Hi script gurus.

I have need to know how to use for loop with multiple variable.

Basically lets take for example /etc/passwd file has following entries

The above cat command will basically first greps the real users that have email addresses then converts ':' to '+' then using cut to extract the userid and email addresses using the '+' as a delimiter to produce the below output

sparcguy sparcguy@unix.com
aixguy aixguy@unix.com
hpuxguy hpuxguy@unix.com

Assuming I pipe the above output to a file called /tmp/list

I know I can just cat the file to display the above but how to I use for loop to display it instead?

the above does not seems to work, does anybody knows how to use multiple variables in a for i in `cat file` loop?

thx in advance

Franklin52 · March 24, 2011, 3:40am

Is something like this what you're looking for?

sed 's/\([^:]*\).*+\([^:]*\).*/\1 \2/' /etc/passwd | while read name email
do
  echo "$name"
  echo "$email"
  ...
  ...
done

sparcguy · March 24, 2011, 4:16am

Hi Franklin52,

Thanks for your replies. Well your method works but I need to do some more further manipulation. I'm trying to rewrite the solaris password aging script "pwage" for hpux Trusted systems. I already figured out the calculation part I just need to find a way for the script to be able to reply to the user that's why I'm trying to use a for loop.

http://www.unix.com/shell-programming-scripting/33854-check-password-age.html?bcsi_scan_9DF8B7809DCC25ED=x3TVLmO5mutY4fcVp4y3XB4AAAAwTGsP&bcsi_scan_filename=33854-check-password-age.html

for i in `cat /tmp/list | awk '{ print $1 }'`
do

for j in `find /tcb -name $i -exec ls -1 {} \;`
do
LASTPWCHG=`cat $j | grep u_succhg | sed 's/:/#/g' | cut -d# -f3`
DAYSEC=`echo "60*60*24" | bc`
DAWNOFTIME=`/usr/contrib/bin/perl -e 'print int(time)'`
SECSAGO=`echo "$DAWNOFTIME - $LASTPWCHG" | bc`
echo $DAYSEC
echo $DAWNOFTIME
echo $LASTPWCHG
echo $SECSAGO
DAYSAGO=`echo $SECSAGO/$DAYSEC | bc`
echo $DAYSAGO
done

MAXAGE=90
LEFTDAYS=`echo "$MAXAGE - $DAYSAGO" | bc`
echo $LEFTDAYS
if [ "$LEFTDAYS = 7 ]
then
show a reminder and send to user email address.... etc

.
.
.

done

Maybe a little explaination for benefit of others

DAYSEC= number of seconds in a day = 60*60*24

DAWNOFTIME = total number of seconds from 1970 to the present time

LASTPWCHG = on trusted hpux servers encrypted passwords are stored in directories and has a field called u_succhg with a huge number like example :u_succhg#1300355419: this number is from 1970 to the date where the last password was successfully changed in seconds.

so basically find the time from 1970 minus the time of the last password changed you get the number of days ago password was changed compares that to 90 days password expiry you will get remainder in days when the password going to expire

Franklin52 · March 24, 2011, 5:55am

Not sure if this is what you're looking for, but it's unnecessary to use a temporary file for the userid.
You could do something like this:

MAXAGE=90

while IFS=: read uid dummy
do
  for j in `find /tcb -name "$uid" -exec ls -1 {} \;`
  do
    LASTPWCHG=`awk -F: '/u_succhg/{print $3}' "$j"`
    DAYSEC=86400
    DAWNOFTIME=`/usr/contrib/bin/perl -e 'print int(time)'`
    SECSAGO=`echo "$DAWNOFTIME - $LASTPWCHG" | bc`
    DAYSAGO=`echo "$SECSAGO/$DAYSEC" | bc`
    LEFTDAYS=`echo "$MAXAGE - $DAYSAGO" | bc`
    if [ "$LEFTDAYS" -le 7 ]
    then
       # show a reminder and send to user email address.... etc
    fi
  done
done < /etc/passwd

sparcguy · March 24, 2011, 6:00am

thanks for your reply again dude.

I found a less painful way.

no need to loop

Will share the script for password aging for hpux trusted servers in another thread.