Enumerating Users 1

[w33mhz]

Does anyone have the basic syntax for a shell script to enumerate local user accounts

 

[w33mhz]

i am trying to write shell script to change a setting on all my user accounts except for a select few i.e. root

 

[stefanwagner]

What's about:

ls /home

?

don't visit my homepage:

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[thedaver]

cat /etc/passwd | grep "/bin/bash" | cut -f1,4 -d":"

This is a sloppy way of finding accounts with a valid shell (bash) and printing out their name and the 4th field in passwd (whatever that is)...

Conversely, you could remove 'cut' and use a 'sed' substitution ...

Or, you could do
cat /etc/passwd | grep "/bin/bash" | cut -f1 -d":" | xargs usermod .............

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[zeland]

#!/bin/bash
for i in $(cat /etc/passwd | cut -d ":" -f 1,3)
do
	user=$(echo $i | cut -d ":" -f1)
	uid=$(echo $i | cut -d ":" -f2)
	
	if [[ $uid -gt 499 && $uid -lt 65534 ]]; then
		# Do something with user
		echo $user
	fi
done

--== Anything can go wrong. It's just a matter of how far wrong it will go till people think its right. ==--

 

[w33mhz]

Excellent, thanks guys I will try that out.

 

[w33mhz]

Well that seemed to work. Just for reference for anyone here is the script I ran, it changes the password age for all users with a 90 day password.

#!/bin/bash
for i in $(cat /etc/shadow | cut -d ":" -f 1,5)
do
    user=$(echo $i | cut -d ":" -f1)
    pwdage=$(echo $i | cut -d ":" -f2)
    
    if [[ $pwdage = 90 ]; then
            passwd -x 365 $user
    fi
done

Thanks Zeland for the added logic.

 

[w33mhz]

OK, whatever does this make you happy:

#!/bin/bash
for i in $(more /etc/shadow | cut -d ":" -f 1,5)
do
    user=$(echo $i | cut -d ":" -f1)
    pwdage=$(echo $i | cut -d ":" -f2)
    
    if [[ $pwdage = 90 ]; then
            passwd -x 365 $user
    fi
done

There no Useless cat usage

 

[stevexff]

My expiry seems to be somewhere else, as I only have 'x' in field 2 of my passwd file. But here is a perl solution that doesn't have to shell out to more/cut/echo all the time. Uncomment the 'system' line when you are sure it's printing the right results...

#!/usr/bin/perl

use strict;
use warnings;

open(PASSWD, "/etc/passwd") or die "Can't open /etc/passwd $!";

while (<PASSWD>) {
   my ($user, $expiry) = (split ':')[0,1];
   print "passwd -x 365 $user\n") if $expiry == 90;
#  system("passwd -x 365 $user") if $expiry == 90;
}

close(PASSWD);

Steve

[small]"Every program can be reduced by one instruction, and every program has at least one bug. Therefore, any program can be reduced to one instruction which doesn't work." (Object:erlDesignPatterns)[/small]