Trying to get list of logged on users sorted

Shell Programming and Scripting
1

I'm trying to execute a single shell command that will give me a sorted list of all the users currently logged into the system, displaying the users name as it appears in /etc/passwd.

I've tried

awk -F: '{print $1}' /etc/passwd | xargs finger -s | cut -c11-28 | uniq

This list whoever does not give me the current logged in users, since if I run who the list is much smaller.

Can anyone help me out, and give me some guidance here?

2

Kindly use the command "users" and sort it with the help of pipe.

Cheers.

3

That works to give me a list of usernames, but what I'm looking for is the users real names, sorted.

I have tried using

users | xargs grep /etc/passwd | cut -d: -f5 | sort -fd | uniq

But that does not work either since grep seems to try and execute the usernames instead of looking for them in the /etc/passwd file.

4

I know it's not pretty but

usrs=`users | sort`
for i in `echo $usrs`
do
grep $i /etc/passwd | awk -F \: '{print $1}'
done

should work.

5

That didn't really do what I wanted but I've modified it to give me the Real names at least, the problem still being that this list is not sorted, and this has to be executed from a command line.

I'm really lost here guys any help?

usrs=`who | cut -d" " -f1 | sort -df | uniq`
for i in $usrs
do
  grep $i /etc/passwd | awk -F \: '{print $5}'
done
6

it is sorted by user name not by real name. Put the sorting code in after you get the real names.

Try something like this:

if [ -f tmp.txt ]
then
  rm -f tmp.txt
fi
usrs=`users`
for i in $usrs
do
  grep $i /etc/passwd | awk -F \: '{print $5}' >> tmp.txt
done
sort tmp.txt 
rm tmp.txt

Again it's not pretty but it should work and sort by real name.

7 
who | while read username junk ; do awk -F: '($1 == "'$username'") {print $5}' /etc/passwd ; done | sort -u
8

Thanks that worked perfectly