Debian: get the login name from UID

Question

What I need is something like:

$ who-has-uid 1000
cyrus

I know the file /etc/passwd contains such informations, I'm not asking a script that parses it.

Answer 1

If you have root access, it's as simple as:

sudo -u \#${pid} whoami

Answer 2

You could try getent (getent man page):

$ getent passwd 1005
shufler:x:1005:119:shufler,,,:/home/local/shufler:/bin/bash

You could parse out just the username with sed.

Answer 3

$ who-has-uid() { perl -e 'print +(getpwuid('$1'))[0], "\n"'; }
$ who-has-uid 0
root
$ who-has-uid 1
daemon

Note that this will work (assuming Perl is configured properly) even if the information comes from somewhere other than the /etc/passwd file.

There's no real error checking; who-has-uid 999 prints an empty line if there's no such UID on the system.

If you don't insist on a one-liner, you can put this somewhere in your $PATH:

#!/usr/bin/perl

use strict;
use warnings;

my $ok = 1;
foreach my $uid (@ARGV) {
    my @pw = getpwuid $uid;
    if (@pw) {
        print "$pw[0]\n";
    }
    else {
        warn "$uid: No such user\n";
        $ok = 0;
    }
}

exit 1 if not $ok;

Answer 4

Use awk, (usually installed GNU awk, gawk) Tell it to use colon as a delimiter, see if field 3 (the UID) is 1000, then print the first field (username)

gawk -F: '$3 == 1000{print $1}' < /etc/passwd

This is if the info is in /etc/passwd, if you have network login info (e.g. on NIS or LDAP) getent works better

getent passwd | gawk -F: '$3 == 1000{print $1}'

You do need single quotes: if you use double quotes, bash will try to figure out what $1 means to the shell, and not pass it to gawk.