Answer
To list all users on a Unix system, even the ones who are not logged in, look at the /etc/password file.
$ cat /etc/passwd
Use the ‘cut’ command to only see one field from the password
file.
For example, to just see the Unix user names, use the command “$ cat /etc/passwd | cut -d: -f1.”
$ cat /etc/passwd | cut -d: -f1
Or to only see the GECOS field (i.e. the account holder’s real name), try this:
$ cat /etc/passwd | cut -d: -f5
Note that users will also see Unix system accounts such
as “root,” “bin,” and “daemon” in the /etc/passwd file. These system accounts
are not Unix users.
List ALL users in a Unix Group (Primary and Secondary)
Is there a command or better combination of cmds that will give me the list of Unix users in a particular Unix group?
Answer
If you have Python, an alternative
Code:
#!/usr/bin/env python
d={}
for line in open("/etc/group"):
line=line.strip().split(":")
users=line[3]
if users:
for u in users.split(","):
d.setdefault(u,[])
d[u].append(line[0])
for i,j in d.iteritems():
print "%s is in groups: %s" %(i,j)
output
Code:
# ./test.py
daemon is in groups: ['bin']
user5 is in groups: ['dialout', 'video']
user2 is in groups: ['dialout', 'video']
user3 is in groups: ['dialout', 'video']
user1 is in groups: ['dialout', 'video']
nobody is in groups: ['nogroup']
Or
Code:
groupName="$1"
# Save some processing, no need to call getent so much.
#
groupEntry=$(getent group | grep "^${groupName}:")
if [ ! "$groupEntry" ]; then
echo "Group $groupName does not exist." >&2
exit 1
fi
# Note it IS possible that the same group is found in a local /etc/group
# as well as another source, you'll get two results. We'll assume the
# first one wins.
#
groupId=$(echo "$groupEntry" | cut -d: -f3 | head -1)
# A username could be in a primary group and have that SAME primary
# group redundantly set as a secondary group.
#
passwdUser=$(getent passwd | cut -d: -f1,4 | grep ":${groupdId}$" | cut -d: -f1)
echo "${groupEntry},${passwdUser}" | cut -d: -f4 | tr ',' '\012' | sed '/^$/d' | sort -u
