Command to list all users with their UID?

List all users with a /home folder:

awk -F: '/\/home/ {printf "%s:%s\n",$1,$3}' /etc/passwd

or all users with a UID >= 1000:

awk -F: '($3 >= 1000) {printf "%s:%s\n",$1,$3}' /etc/passwd

a combination

awk -F: '/\/home/ && ($3 >= 1000) {printf "%s:%s\n",$1,$3}' /etc/passwd

or for all entries

awk -F: '{printf "%s:%s\n",$1,$3}' /etc/passwd

More information here

You can find it easily by just using cut :

cut -d: -f1,3 /etc/passwd

-d: sets the delimiter as : for cut
-f1,3 extracts the field 1 and 3 only delimited by : from the /etc/passwd file

Check man cut to get more idea.

Example :

$ cut -d: -f1,3 /etc/passwd
root:0
daemon:1
bin:2
sys:3
sync:4
games:5
......

If you have ldap configured, to include the ldap users in the output :

getent passwd | cut -d: -f1,3

Alternatively to list all users including UID and GID information.

for user in $(cat /etc/passwd | cut -f1 -d":"); do id $user; done

Cheers,

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