Allowing awk to use shell variables

[SiberianKhatru]

I am trying to run a script which pulls out all users from the /etc/passwd file, subjects that list of users to exclusion criteria (removing system users from the list such as root, lp, sys etc) and produces a new list of users. I need to pass the new list of users to awk to determine the users home directories in the passwd file.

I then need to 'find' files in the users home directories which they own. This all sounds complicated because there are in-consistencies on the file system which deems this necessary.

So far I have set up variables:
############
CRITERIA='(^root|^bin|^sys|^lp|^uucp|^nuucp|^listen|^noaccess|^daemon)' # this sets out the list of not-needed users

USERS=`nawk -F: '$1 !~ /${CRITERIA}/ {print $1}' /etc/passwd` # this produces the new list variable
############

#The following attempts to pull out the home directories and search in those directories for files

############
for people in $USERS
do

HOME=`grep "^$people" /etc/passwd |nawk -F: ' {print $6}'`

find $HOME -user $people -exec ls -l {} \; | awk '$3 !~ /^root/ {print}'
done
############

My 'excluding root,lp etc users from my list is not recognised in awk - the original list of users is passed unaltered to the find command and all users have their home directories interogated.

What is my syntactical error?

 

[SiberianKhatru]

passing variables to awk has been looked at before with the -v, but I have a list of variables to set up.......do I need to enter -v ... -v ... -v ... etc etc a mile long?

 

[marsd]

This has saved me much time.

for xx in $(var_list)
do

awk -v inX=$xx ' {
script
}'
done

 

[aigles]

nawk -F: '$1 !~ /${CRITERIA}/ {print $1}' /etc/passwd

The shell doesn't subtitute the variable because ${CRITERIA} is in a single quote delimited string.
So nawk selects lines with first word ($1) begining with end of line ($), in other words nawk selects all not empty lines.


Consult faq271-1281 "Shell variables in awk"

If you don't want to use the -v option, you can access to exported shell variables with the ENVIRON array :

export VAR1='the value...'
awk ' { print "The value of VAR1 is " ENVIRON["VAR1"] ; ... } ' input_file


A possible solution to your problem :

export CRITERIA=^root|^bin|^sys|^lp|^uucp|^nuucp|^listen|^noaccess|^daemon)'
awk -F: -f UsersFiles.awk /etc/passwd

with UsersFiles.awk :


$1 !~ ENVIRON["CRITERIA"] {
User = $1 ;
Home = $6 ;
FindCmde = "find \"" Home "\" -user \"" User "\" -exec ls -ld {} \\;" ;
while ( FindCmde | getline ) {
if ( $3 != "root" )
print $0 ;
}
close(FindCmde) ;
}

Jean Pierre.