Shell Programming and Scripting

What would be the best way to extract a users home from /etc/passwd. I intended to use cut but have been semi advised that a can do it using and eval statement and the ~ operator. Unfortunately this was just a casual conversation so made little sense at the time. Any help much appreciated.

cheers

Look at the password file on your machine.

Type:
cat /etc/passwd at the prompt. Entries in /etc/passwd look something like this:

arushkin:Igljf78DS:132:20:Amy Rushkin:/usr/people/arushkin:/bin/csh
trsmith:*:543:20:Trent Smith, sys adm:/usr/people/trsmith/:/bin/tcsh

To extract a field, where NAME is set to the UserID of the user you are trying to query.

awk -v NAME=${NAME} 'BEGIN{FS=":"} /NAME/ {print $7}'

or

cat /etc/passwd | awk -F ":" '{print $7}'

Would that be a UUOC?

Code:

awk -F\: '{print $7}' /etc/passwd 

or 

awk -F\: '/username/{print $7}'  /etc/passwd 

or

awk -F\: '/'${USERNAME}'/{print $7}'  /etc/passwd

where username in the second example is replaced by the user you want to search for.

A user's home directory can be found by using tide-expansion,e.g...

Code:

echo ~username

dir=~username
echo $dir

its ok i have solved it

cat /etc/passwd | awk -F ":" '{print $6}' #this displays 6th field ie home directories

awk -F\: '{print $6}' /etc/passwd |grep pauline #this displays pauline home directory

or echo ~pauline.fowler

thanks