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
我所做的与上面的perl代码类似,但是用本地perl函数替换了getent和id。它要快得多,应该可以跨不同的*nix口味工作。
#!/usr/bin/env perl
use strict;
my $arg=shift;
my %groupMembers; # defining outside of function so that hash is only built once for multiple function calls
sub expandGroupMembers{
my $groupQuery=shift;
unless (%groupMembers){
while (my($name,$pass,$uid,$gid,$quota,$comment,$gcos,$dir,$shell,$expire)=getpwent()) {
my $primaryGroup=getgrgid($gid);
$groupMembers{$primaryGroup}->{$name}=1;
}
while (my($gname,$gpasswd,$gid,$members)=getgrent()) {
foreach my $member (split / /, $members){
$groupMembers{$gname}->{$member}=1;
}
}
}
my $membersConcat=join(",",sort keys %{$groupMembers{$groupQuery}});
return "$membersConcat" || "$groupQuery Does have any members";
}
print &expandGroupMembers($arg)."\n";
其他回答
lid -g groupname | cut -f1 -d'('
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
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
我所做的与上面的perl代码类似,但是用本地perl函数替换了getent和id。它要快得多,应该可以跨不同的*nix口味工作。
#!/usr/bin/env perl
use strict;
my $arg=shift;
my %groupMembers; # defining outside of function so that hash is only built once for multiple function calls
sub expandGroupMembers{
my $groupQuery=shift;
unless (%groupMembers){
while (my($name,$pass,$uid,$gid,$quota,$comment,$gcos,$dir,$shell,$expire)=getpwent()) {
my $primaryGroup=getgrgid($gid);
$groupMembers{$primaryGroup}->{$name}=1;
}
while (my($gname,$gpasswd,$gid,$members)=getgrent()) {
foreach my $member (split / /, $members){
$groupMembers{$gname}->{$member}=1;
}
}
}
my $membersConcat=join(",",sort keys %{$groupMembers{$groupQuery}});
return "$membersConcat" || "$groupQuery Does have any members";
}
print &expandGroupMembers($arg)."\n";
再加上grep和tr:
$ grep ^$GROUP /etc/group | grep -o '[^:]*$' | tr ',' '\n'
user1
user2
user3