Create a file to automatically check user list for users specified

Question

I am realitively new to bash scripting, but have created this from my intense googling

I am attempting to create a bash script to run on Ubuntu that will check the user list and find any users that aren’t in a specified list. I have written this so far:

Cat /etc/passwd | grep -o -P ‘.{0,40}:1[0-9][0-9][0-9].{0,0}.’ | cut -d: -f1

views passwd file, then finds only the lines with a user ID of 1000 or above, and everything before the userID of each line is piped into cut which removes all except the username of each user.

I then want to have the script check a file for usernames I specify (probably copied from a list) and compare each to the output of the above. Removing all usernames specified.

So for instance:

I have users John, Ben and Tom on my computer If I put John and Ben in the file the script is accessing, it should output Tom since he is not specified

How would I go about doing this?

Answer 1

This code should work for you.

for username in $( awk -F':' '$3 >= 1000 {print $1}' /etc/passwd )
do
        if ! grep -q "$username" list.txt; then
                echo "$username"
        fi
done

Keep the usernames to check in list.txt as one user per line

Answer 2

grep allows specifying patterns from file. Thus you can use grep -f option for that purpose and -v for inverse pattern matching, i.e. print lines not matching the pattern. Thus,

 printf "^%s\n" 'John' 'Ben' > /tmp/list.txt
 grep -v -f /tmp/list.txt  /etc/passwd | cut -d ':' -f 1
 rm l/tmp/ist.txt

We could also take into account UID greater than 999, since by default that refers to human users unless changed by your sysadmin.

awk -F ':' '$1 !~ /John|Ben|nobody/  && $3 > 999' /etc/passwd