如何在Linux(可能还有其他unix)中列出一个组的所有成员?
当前回答
下面的命令将列出属于<your_group_name>的所有用户,但只列出由/etc/group数据库管理的用户,不包括LDAP、NIS等。它也只适用于次要组,它不会列出将该组设置为主要组的用户,因为主要组存储为/etc/passwd.文件中的GID(数字组ID)
grep <your_group_name> /etc/group
其他回答
下面是一个非常简单的awk脚本,它考虑了其他答案中列出的所有常见陷阱:
getent passwd | awk -F: -v group_name="wheel" '
BEGIN {
"getent group " group_name | getline groupline;
if (!groupline) exit 1;
split(groupline, groupdef, ":");
guid = groupdef[3];
split(groupdef[4], users, ",");
for (k in users) print users[k]
}
$4 == guid {print $1}'
我将它用于支持ldap的设置,运行在任何具有符合标准的getent和awk的设备上,包括solaris 8+和hpux。
下面的命令将列出属于<your_group_name>的所有用户,但只列出由/etc/group数据库管理的用户,不包括LDAP、NIS等。它也只适用于次要组,它不会列出将该组设置为主要组的用户,因为主要组存储为/etc/passwd.文件中的GID(数字组ID)
grep <your_group_name> /etc/group
Zed的实现应该扩展到其他一些主要UNIX上。
有人可以使用Solaris或HP-UX硬件吗?没有测试这些案例。
#!/usr/bin/perl
#
# Lists members of all groups, or optionally just the group
# specified on the command line
#
# Date: 12/30/2013
# Author: William H. McCloskey, Jr.
# Changes: Added logic to detect host type & tailor subset of getent (OSX)
# Attribution:
# The logic for this script was directly lifted from Zed Pobre's work.
# See below for Copyright notice.
# The idea to use dscl to emulate a subset of the now defunct getent on OSX
# came from
# http://zzamboni.org/\
# brt/2008/01/21/how-to-emulate-unix-getent-with-macosxs-dscl/
# with an example implementation lifted from
# https://github.com/petere/getent-osx/blob/master/getent
#
# Copyright © 2010-2013 by Zed Pobre (zed@debian.org or zed@resonant.org)
#
# Permission to use, copy, modify, and/or distribute this software for any
# purpose with or without fee is hereby granted, provided that the above
# copyright notice and this permission notice appear in all copies.
#
use strict; use warnings;
$ENV{"PATH"} = "/usr/bin:/bin";
# Only run on supported $os:
my $os;
($os)=(`uname -a` =~ /^([\w-]+)/);
unless ($os =~ /(HU-UX|SunOS|Linux|Darwin)/)
{die "\$getent or equiv. does not exist: Cannot run on $os\n";}
my $wantedgroup = shift;
my %groupmembers;
my @users;
# Acquire the list of @users based on what is available on this OS:
if ($os =~ /(SunOS|Linux|HP-UX)/) {
#HP-UX & Solaris assumed to be like Linux; they have not been tested.
my $usertext = `getent passwd`;
@users = $usertext =~ /^([a-zA-Z0-9_-]+):/gm;
};
if ($os =~ /Darwin/) {
@users = `dscl . -ls /Users`;
chop @users;
}
# Now just do what Zed did - thanks Zed.
foreach my $userid (@users)
{
my $usergrouptext = `id -Gn $userid`;
my @grouplist = split(' ',$usergrouptext);
foreach my $group (@grouplist)
{
$groupmembers{$group}->{$userid} = 1;
}
}
if($wantedgroup)
{
print_group_members($wantedgroup);
}
else
{
foreach my $group (sort keys %groupmembers)
{
print "Group ",$group," has the following members:\n";
print_group_members($group);
print "\n";
}
}
sub print_group_members
{
my ($group) = @_;
return unless $group;
foreach my $member (sort keys %{$groupmembers{$group}})
{
print $member,"\n";
}
}
如果有更好的方法来分享这个建议,请告诉我;我考虑了很多方法,这就是我想到的。
再加上grep和tr:
$ grep ^$GROUP /etc/group | grep -o '[^:]*$' | tr ',' '\n'
user1
user2
user3
下面的命令将列出属于<your_group_name>的所有用户,但只列出由/etc/group数据库管理的用户,不包括LDAP、NIS等。它也只适用于次要组,它不会列出将该组设置为主要组的用户,因为主要组存储为/etc/passwd.文件中的GID(数字组ID)
awk -F: '/^groupname/ {print $4;}' /etc/group
推荐文章
- 在Bash中检查变量是否存在于列表中
- 查看PS命令的全部输出
- Linux命令将域名转换为IP
- 如何从命令行通过mysql运行一个查询?
- 在创建守护进程时执行双fork的原因是什么?
- 如何从命令行在windows中找到mysql数据目录
- 有效地测试Linux上的端口是否打开?
- 如何从另一个文件A中删除文件B中出现的行?
- 对以制表符分隔的文件进行排序
- 使用sudo时未找到命令
- 当有命令行参数时,如何使用GDB分析程序的核心转储文件?
- 如何使用查找命令从列表中查找所有具有扩展名的文件?
- 如何强制从另一个SSH会话分离屏幕?
- 如何将文件指针(file * fp)转换为文件描述符(int fd)?
- Linux Bash中双&和分号有什么区别?