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
我认为最简单的方法是以下步骤,你不需要安装任何软件包或软件:
首先,你找出你想知道的用户组的GID,有很多方法: cat /etc/group(最后一列是GID) Id用户(用户是属于组的人) 现在,您将在文件/etc/passwd中列出所有用户,但是您将使用以下后续命令应用一些过滤器,以获得前一个组的成员。
cut -d: -f1,4 /etc/passwd |grep GID (GID是你从步骤1中得到的数字)
切命令将选择一些“列”的文件,参数d设置分隔符”:“在这种情况下,参数- f选择“字段”(或列)在案例1和4所示(在/ etc / passwd文件,1º列是用户的名称和4º是用户所属的组的GID),完成| grep GID将滤波器组(4º列),你选择了。
其他回答
lid -g groupname | cut -f1 -d'('
使用Python列出组成员:
import grp;打印grp.getgrnam (GROUP_NAME)[3]”
参见https://docs.python.org/2/library/grp.html
下面是一个脚本,返回/etc/passwd和/etc/group中的用户列表 它不检查NIS或LDAP,但它显示了将该组作为默认组的用户 在Debian 4.7和solaris 9上测试
#!/bin/bash
MYGROUP="user"
# get the group ID
MYGID=`grep $MYGROUP /etc/group | cut -d ":" -f3`
if [[ $MYGID != "" ]]
then
# get a newline-separated list of users from /etc/group
MYUSERS=`grep $MYGROUP /etc/group | cut -d ":" -f4| tr "," "\n"`
# add a newline
MYUSERS=$MYUSERS$'\n'
# add the users whose default group is MYGROUP from /etc/passwod
MYUSERS=$MYUSERS`cat /etc/passwd |grep $MYGID | cut -d ":" -f1`
#print the result as a newline-separated list with no duplicates (ready to pass into a bash FOR loop)
printf '%s\n' $MYUSERS | sort | uniq
fi
或者作为一行程序,您可以直接从这里剪切和粘贴(在第一个变量中更改组名)
MYGROUP="user";MYGID=`grep $MYGROUP /etc/group | cut -d ":" -f3`;printf '%s\n' `grep $MYGROUP /etc/group | cut -d ":" -f4| tr "," "\n"`$'\n'`cat /etc/passwd |grep $MYGID | cut -d ":" -f1` | sort | uniq
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。
我已经尝试了grep 'sample-group-name' /etc/group,它将根据这里的示例列出您指定的组的所有成员
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