1

I'm trying to remove all default user passwords across a bunch of servers with ansible. Firstly, I'd like to output the name of every user whose current password is foobar. How can I achieve this?

My first intent was to get the hash from /etc/shadow and grep for it, but this won't work because of salting.

Do I need to calculate my own hashes for this and compare them? Or is there a faster and easier approach?

  • 1
    The reason why password hashes are saluted is to prevent this exact search. This is supposed to be difficult. Any solution will require re-hashing guesses (like foobar) for every users. – Philip Couling Jun 3 at 10:59
4

There's a specialized tool for password weakness check: John the Ripper available and probably packaged in all common Unix & Linux flavours.

Here's an example of usage on Debian GNU/Linux 9 (unshadow comes along john). Some care should be taken when manipulating password files, this is just a PoC. Note that the john command could be run remotely (and thus not installed anywhere else than a dedicated system) as long as it's provided suitable password files.

Setup (including setting password foobar to account test):

# echo test:foobar | chpasswd
# grep ^test: /etc/shadow
test:$6$84SIejUB$qM5UulJEIiwjOc4PWXYupWoyU/jMP0rKA8cM1g8CEOgxMlC.x4ndbbdRq438rjKb.6UwCoTqzvgxoi0h51Kpm1:18050:0:99999:7:::
# unshadow /etc/passwd /etc/shadow > /root/workpasswd
# echo foobar > /tmp/wordlist

Test for forbidden/default passwords:

# john -wordlist:/tmp/wordlist /root/workpasswd
Created directory: /root/.john
Loaded 3 password hashes with 3 different salts (crypt, generic crypt(3) [?/64])
Press 'q' or Ctrl-C to abort, almost any other key for status
foobar           (test)
1g 0:00:00:00 100% 5.882g/s 5.882p/s 17.64c/s 17.64C/s foobar
Use the "--show" option to display all of the cracked passwords reliably
Session completed

Result:

# john -show /root/workpasswd 
test:foobar:1001:1001:,,,:/home/test:/bin/bash

1 password hash cracked, 2 left

Cleanup:

# rm -r /root/workpasswd /root/.john /tmp/wordlist
3

Can you try logging in as each user? e.g.

echo "foobar" | su username

You would need to temporarily disable the TTY check.

2

So here is a little C snippet that checks for the existing users password:

Save the following snippet in a file called: checkpass.c

#include <pwd.h>
#include <shadow.h>
#include <stdio.h>
#include <unistd.h>
#include <string.h>
#include <crypt.h>

static int pwcheck (char *user, char *passwd)
{
    struct passwd *pw;
    struct spwd *spwd;
    char *epasswd, *cpasswd;
    char *tty;

    if ((pw = getpwnam(user)) == NULL) {
        return 1;
    }
     /*
     * XXX If no passwd, let them login without one.
     */
    if (pw->pw_passwd[0] == '\0') {
        return 0;
    }
    spwd = getspnam(user);
    cpasswd = (spwd ? spwd->sp_pwdp : pw->pw_passwd);

    epasswd = crypt(passwd, cpasswd);
    if (epasswd == NULL) {
        return 2;
    }
    if (strcmp (epasswd, cpasswd)) {
        return 1;
    }

    return 0;
}

int main (int argc, char *argv[])
{
    if (argc < 3) return 4;
    return pwcheck (argv[1], argv[2]);
}

Compile the above code using:

gcc -o checkpass checkpass.c -lcrypt

Now from the command line just run the following:

while IFS=: read -r user _; do
  if ./checkpass "$user" foobar; then
    printf 'The ollowing user %s has the password set to foobar\n' "$user";
  fi;
done </etc/passwd

It's maybe a long shot but should work!

0

Because I'm not a fan of installing extra software and don't want to mess around with sudoers, what I ended up doing was

sshpass -p foobar ssh -o PreferredAuthentications=keyboard-interactive,password -o PubkeyAuthentication=no user@host

and afterwards check the exit code in Ansible. If the password was correct the exit code will be 0, otherwise 5.

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