Loop that prints twice in bash - bash

Question

I am writing this bash script that is supposed to print out all the users that have never logged in with an option to sort them. I have managed to get all the input working, however, I am encountering issues when it comes to printing the output. The loop goes as follows:
for user in $(lastlog | grep -i 'never' | awk '{print $1}'); do
grep $user /etc/passwd | awk -F ':' '{print $1, $3}'
done
of course, this loop doesn't sort the output, however, from my limited understanding of shells and shell scripting it should only be a matter of putting a ' | sort' after the first "awk '{print $1}'". my problem is that the output of this loop prints every user at least twice, and in some instances, four times. Why is that and how can I fix it?

Answer 1

Well, let's try to debug it:
for user in $(lastlog | grep -i 'never' | awk '{print $1}'); do
echo "The user '$user' matches these lines:"
grep $user /etc/passwd | awk -F ':' '{print $1, $3}'
echo
done
This outputs:
The user 'daemon' matches these lines:
daemon 1
colord 112
The user 'bin' matches these lines:
root 0
daemon 1
bin 2
sys 3
sync 4
games 5
man 6
(...)
And indeed, the entry for colord does contain daemon:
colord:x:112:120:colord colour management daemon,,,:/var/lib/colord:/bin/false
^-- Here
And the games entry does match bin:
games:x:5:60:games:/usr/games:/usr/sbin/nologin
^-- Here
So instead of matching the username string anywhere, we just want to match it from the start of the line until the first colon:
for user in $(lastlog | grep -i 'never' | awk '{print $1}'); do
echo "The user '$user' matches these lines:"
grep "^$user:" /etc/passwd | awk -F ':' '{print $1, $3}'
echo
done
And now each entry only shows the singly entry it was supposed to, so you can remove the echos and keep going.
If you're interested in finesse and polish, here's an alternative solution that works efficiently across language settings, weird usernames, network auth, large lists, etc:
LC_ALL=C lastlog |
awk -F ' ' '/Never logged in/ {printf "%s\0", $1}' |
xargs -0 getent passwd |
awk -F : '{print $1,$3}' |
sort

Answer 2

Just think what happens with a user named sh. How many users would grep sh match? Probably all of them, since each is using some shell in the shell field.
You should think about
awk -F ':' '$1 == "'"$user"'" {print $1, $3}' /etc/passwd
or with an awk variable for user:
awk -F ':' -vuser="$user" '$1 == user {print $1, $3}' /etc/passwd

Answer 3

Your grep will match multiple lines, (man will match manuel and norman etc.) anchor it to the beginning of the line and add a trail :.
grep "^${user}:" /etc/passwd | awk -F ':' '{print $1, $3}'
A better option might be to forget about grepping /etc/passwd completely and use the id command to get the user id:
id=$(id -u "${user}" 2>/dev/null) && printf "%s %d\n" "${user}" "${id}"
If the id command fails nothing is printed, or it could be modified to be:
id=$(id -u "${user}" 2>/dev/null)
printf "%s %s\n" "${user}" "${id:-(User not found)}"
In gnu linux I'm pretty sure that the found users id not existing isn't possible as lastlog will only report existing users so the second example may be pointless.