143

The command id can be used to look up a user's uid, for example:

$ id -u ubuntu
1000

Is there a command to lookup up a username from a uid? I realize this can be done by looking at the /etc/passwd file but I'm asking if there is an existing command to to this, especially if the user executing it is not root.

I'm not looking for the current user's username, i.e. I am not looking for whoami or logname.

This also made me wonder if on shared web hosting this is a security feature, or am I just not understanding something correctly?

For examination, the /etc/passwd file from a shared web host:

root:x:0:0:root:/root:/bin/bash
bin:x:1:1:bin:/bin:/sbin/nologin
daemon:x:2:2:daemon:/sbin:/sbin/nologin
adm:x:3:4:adm:/var/adm:/sbin/nologin
lp:x:4:7:lp:/var/spool/lpd:/sbin/nologin
sync:x:5:0:sync:/sbin:/bin/sync
shutdown:x:6:0:shutdown:/sbin:/sbin/shutdown
halt:x:7:0:halt:/sbin:/sbin/halt
mail:x:8:12:mail:/var/spool/mail:/sbin/nologin
news:x:9:13:news:/etc/news:
uucp:x:10:14:uucp:/var/spool/uucp:/sbin/nologin
operator:x:11:0:operator:/root:/sbin/nologin
games:x:12:100:games:/usr/games:/sbin/nologin
gopher:x:13:30:gopher:/var/gopher:/sbin/nologin
ftp:x:14:50:FTP User:/var/ftp:/sbin/nologin
nobody:x:99:99:Nobody:/:/sbin/nologin
nscd:x:28:28:NSCD Daemon:/:/sbin/nologin
vcsa:x:69:69:virtual console memory owner:/dev:/sbin/nologin
pcap:x:77:77::/var/arpwatch:/sbin/nologin
rpc:x:32:32:Portmapper RPC user:/:/sbin/nologin
mailnull:x:47:47::/var/spool/mqueue:/sbin/nologin
smmsp:x:51:51::/var/spool/mqueue:/sbin/nologin
oprofile:x:16:16:Special user account to be used by OProfile:/home/oprofile:/sbin/nologin
sshd:x:74:74:Privilege-separated SSH:/var/empty/sshd:/sbin/nologin
dbus:x:81:81:System message bus:/:/sbin/nologin
avahi:x:70:70:Avahi daemon:/:/sbin/nologin
rpcuser:x:29:29:RPC Service User:/var/lib/nfs:/sbin/nologin
haldaemon:x:68:68:HAL daemon:/:/sbin/nologin
xfs:x:43:43:X Font Server:/etc/X11/fs:/sbin/nologin
avahi-autoipd:x:100:104:avahi-autoipd:/var/lib/avahi-autoipd:/sbin/nologin
named:x:25:25:Named:/var/named:/sbin/nologin
mailman:x:32006:32006::/usr/local/cpanel/3rdparty/mailman/mailman:/usr/local/cpanel/bin/noshell
dovecot:x:97:97:dovecot:/usr/libexec/dovecot:/sbin/nologin
mysql:x:101:105:MySQL server:/var/lib/mysql:/bin/bash
cpaneleximfilter:x:32007:32009::/var/cpanel/userhomes/cpaneleximfilter:/usr/local/cpanel/bin/noshell
nagios:x:102:106:nagios:/var/log/nagios:/bin/sh
ntp:x:38:38::/etc/ntp:/sbin/nologin
myuser:x:1747:1744::/home/myuser:/usr/local/cpanel/bin/jailshell

And here is a sample directory listing of /tmp/

drwx------  3 root     root        1024 Apr 16 02:09 spamd-22217-init/
drwxr-xr-x  2      665      664    1024 Apr  4 00:05 update-cache-44068ab4/
drwxr-xr-x  4      665      664    1024 Apr 17 15:17 update-extraction-44068ab4/
-rw-rw-r--  1      665      664   43801 Apr 17 15:17 variable.zip
-rw-r--r--  1      684      683    4396 Apr 17 07:01 wsdl-13fb96428c0685474db6b425a1d9baec

We can see root is the owner of some files, and root is also showing up in /etc/passwd , however the other users/groups all show up as numbers.

Improve this question
1
  • 8
    Keep in mind that more than one user may have the same UID. It's rare, but happens occasionally. Commented Apr 17, 2012 at 2:47

8 Answers 8

Reset to default
140

You might enjoy this little ditty. 

$ id -nu [number]

3.17.3-1-ARCH #1 SMP PREEMPT Fri Nov 14 22:56:01 CET 2014 i686 GNU/Linux

I can confirm that it returns a corresponding user name, if one exists, on Arch Linux. I can also confirm that it does not work on Ubuntu when run as a normal user, although I have not tested this as the superuser. It also does not work on Alpine Linux. Maybe a security feature prevents this from working on some systems.

Improve this answer
8
  • 6
    id -u jimmij => 1000. id -nu 1000 => id: 1000: no such user. Commented Nov 21, 2014 at 12:39
  • 3
    works for me, with id --version = id (GNU coreutils) 8.23 Commented Jan 5, 2016 at 16:49
  • 2
    note: this works on Red Hat/CentOS >=7 (not on Red Hat/CentOS <=6) Commented Dec 29, 2016 at 10:38
  • 5
    FYI this does not work on busybox/alpine linux/alpine docker images Commented Sep 6, 2017 at 19:10
  • 3
    works on Ubuntu 20.04 (probably with older versions as well) Commented Apr 2, 2021 at 20:32
101

Try

getent passwd "$uid" | cut -d: -f1
Improve this answer
9
  • 1
    If this returns nothing does that mean I don't have access to translate the id to a username? Commented Apr 17, 2012 at 4:07
  • 2
    More likely you didnt set "$uid"` or that uid doesn't exist. Does grep ":$uid:" /etc/passwd find it? Does getent passwd produce any output? Commented Apr 17, 2012 at 4:14
  • @cwd: You should always have access to translate an id to a username. For instance, ls -l is always doing this. Commented Apr 17, 2012 at 7:00
  • Just curious because looking at an ls listing on a shared host was showing numbers in the user / group name columns with ls. Perhaps it is a security precaution or jailshell thing? Commented Apr 17, 2012 at 19:50
  • @cwd More likely the shared host is using an /etc/passwd mounted / shared from elsewhere which wasn't mounted at the time. Commented Apr 17, 2012 at 19:56
53

ls already performs that lookup. You can perform a user information lookup from the command line with getent passwd.

If ls shows a user ID instead of a user name, it's because there's no user by that name. Filesystems store user IDs, not user names. If you mount a filesystem from another system, or if a file belongs to a now-deleted user, or if you passed a numerical user ID to chown, you can have a file that belongs to a user ID that doesn't have a name.

On a shared host, you may have access to some files that are shared between several virtual machines, each with their user database. This is a bit weird (why share files but not the users that own them?), but it's technically possible.

Improve this answer
3
  • 2
    I'm seeing the number instead of the name on ls -l if the name is over eight characters in length. Commented Apr 20, 2017 at 14:44
  • 1
    @kgrittn It might depend on the Unix variant. GNU ls shows the whole username. BusyBox ls truncates it to 8 characters (at least on the build I have here). Commented Apr 20, 2017 at 16:32
  • Try also id -u <username> as per this under-appreciated answer. Commented Sep 16, 2023 at 12:42
15

The command id can be used to both to look up UID and/or USER name

  1. look up a UID by USER, for example:
    $ id -u ubuntu
    1000
  1. look up a USER by UID, for example:
    $ id -un 1000
    ubuntu

If the UID or USER is not found on current machine it prints no such user message to stderr, for example:

$ id -un 1234
id: 1234: no such user

From man id:

id - print real and effective user and group IDs
   -n, --name
          print a name instead of a number, for -ugG
   -u, --user
          print only the effective user ID
   -g, --group
          print only the effective group ID
   -G, --groups
          print all group IDs
Improve this answer
2
  • 2
    Note, that the no such user message is printed on stderr. You get an empty string if you capture the name like User=$(id -un 1000 2> /dev/null). Commented May 24, 2021 at 16:00
  • @RaoulSteffen good point. I have updated my original post with stderr output. Commented May 25, 2021 at 7:59
1

I realize this is an old question, but here's another answer

awk -F: '{print $1,$3}' /etc/passwd | grep <UID>
Improve this answer
2
  • 1
    Pipe to grep what? (Also if you add four spaces at the start of the line, it will render it as fixed width text.) Commented Mar 31, 2016 at 21:42
  • 1
    I prefer the answer with getent passwd. But if parsing /etc/passwd, it would be nicer to use awk only ID=0 ; awk -F: '{if ( $3 == '"$ID"' ) print $1}' /etc/passwd. Commented Dec 29, 2016 at 10:47
0

Parse /etc/passwd:

% awk -F: "/:$(id -u ubuntu):/{print \$1}" /etc/passwd
ubuntu
Improve this answer
4
  • 8
    Unlike getent, this does not work if the system uses LDAP. Commented Apr 17, 2012 at 8:51
  • 3
    this doesn't work either if NIS is used or whatever other distributed authentication protocol. Commented Apr 18, 2012 at 1:25
  • @choroba : if he had a specific requirement for LDAP or NIS, he probably would have mentioned it. My solution works on typical linux setups. Commented Apr 18, 2012 at 2:55
  • 2
    downvote beacause it's not reliable (it could match either the user or group). Commented Dec 29, 2016 at 10:36
0 
id | awk '{print $1}' | sed 's/.*(//;s/)$//'

This oneliner does the following:

  • get user information
  • extract the first column ("uid=ID(NAME)")
  • remove anything before brackets, and the brackets themselves
Improve this answer
0
0

Just to add another method for use on systems without a suitably recent id utility:

perl -le 'print scalar getpwuid $ARGV[0]' "$uid"

Unlike grepping the passwd file, this works fine on systems that use other methods of username/uid mapping (e.g. LDAP or NIS), and it has high probability of working even on really weird or ancient unix-ish systems thanks to perl's extreme portability and its long legacy of being used as a swiss army knife for system admins.

Improve this answer

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.