Simple CTF [TRYHACKME - EASY]

Challenge link: https://tryhackme.com/room/easyctf

Challenge Difficulty: Easy

Challenge Tags: security, enumeration, privesc

Reconnaissance & Scanning

From the THM platform, I got 10.10.56.199 as the target machine.

Question 1: How many services are running under port 1000?

• Using the command nmap -sV -O -p1-1000 10.10.56.199 I was able to identify two (2) ports between under 1000.

• The -p1-1000 switch specifies the range of ports to consider during a nmap scan. If not explicitly specified during the scan, nmap scan 1000 of the most common ports.

Question 2: What is running on the higher port?

• Using the command nmap -sV -O 10.10.56.199 I identified ssh as being the service running on the highest port.

Question 3: What's the CVE you're using against the application?

• I began further enumeration starting with port 80 hosting the HTTP server.

• Pasting the IP address onto a browser did not yield results, as I was directed to an Apache landing page.

• I then decided to do subdomain enumeration to find hidden directories on the HTTP port using gobuster dir -u http://10.10.56.199 -w /usr/share/dirbuster/wordlists/directory-list-2.3-medium.txt

• I found a directory /simple and instantly tried accessing it from the browser.

• It was another landing page for a content management system (CMS) which I assumed was going to be our entry point or possible angle for an initial foothold.

• After some scrutinization of the webpage, I found the CMS name and version at the bottom of the page, which I then did a Google search for to find the Common Vulnerabilities and Exposures (CVEs) it can succumb to.

• After visiting the CVE details website and going through the possible exploits I managed to identify CVE-2019-9053 as being the correct exploit for this challenge.

Question 4: To what kind of vulnerability is the application vulnerable?

• From reading the CVE description I gathered that it was an SQL Injection vulnerability, also known as 'SQLi'.

Exploitation

Question 5: What's the password?

• I moved on to download the exploit script of the CVE from exploit-db.

• The script was saved as 46635.py so running nano 46635.py displayed the contents of the script.

• The script was made in Python but because it was compiled in python2 I had to change the syntax to python3 because it was giving errors. I downloaded 2to3 to change file compatibility with the code below:

pip install 2to3
pip3 install 2to3

python -m pip install 2to3
python3 -m pip install 2to3
py -m pip install 2to3

• After running 2to3 -w 46635.py the file was successfully converted and I could finally run the exploit.

• To run the exploit, I had to add the target URL and a wordlist path as the parameters of the command: python3 46635.py -u http://10.10.56.199/simple/ -w /usr/share/wordlists/rockyou.txt

• From the results I got, I found the username mitch and a password hash. I later discovered that I had to add another switch --crack within the command for the password hash to be cracked.

• I tried running the exploit again python3 46635.py -u http://10.10.56.199/simple/ --crack -w /usr/share/wordlists/rockyou.txt with the switch but it showed null results.

NB: I assume that part of the code got broken when I converted it from python2 to python3, so I left this angle as a dead-end to try out a different attack vector.

• I returned to the FTP port ftp 10.10.56.199 and tried the popular 'anonymous' login - this gave me access to the FTP server!

• I used ls to view the contents of the server and found one folder labelled 'pub', I navigated to the folder with cd pub and ran ls again to view the contents of this folder, and I discovered a text file labelled 'ForMitch.txt'

• I downloaded the file onto my local machine using mget ForMitch.txt and I viewed the file with cat ForMitch.txt.

• From the message context, I figured that there was a user called mitch considering that we found the same username from the exploit script earlier. And I also assumed that there was a way to brute force the password since that message also suggested that it cracked in a short time.

• Using hydra -l mitch -P /usr/share/wordlists/rockyou.txt ssh://10.10.56.199:2222 I tried brute forcing the SSH port using mitch as the username.

NB: It was necessary to specify the port 2222 since the tools are accustomed to port 22 as the default SSH port.

• Results from Hydra gave us the password 's*****'.

Initial Foothold

Question 6: Where can you log in with the details obtained?

• Attempting to gain access to the server via SSH with mitch as the username and s***** as the password gave us an initial foothold.

• 'SSH' was the answer to question 6.

Question 7: What's the user flag?

• ls showed that there was a user.txt file in the home directory, running cat user.txt displayed the first flag.

Question 8: Is there any other user in the home directory? What's its name?

• I ran pwd to see the name of my current directory, then I ran cd .. to navigate back to the home directory. ls gave me the name of the other folder present in the home directory 'sunbath' which was the answer to this question.

Privilege Escalation

Question 9: What can you leverage to spawn a privileged shell?

• For privilege escalation, I first used sudo -l to find binaries I could run as the current user and found usr/bin/vim.

• I went to GTFOBins and looked for a SUDO shell escape script for Vim.

• The script I found was as follows: sudo vim -c ':!/bin/sh'

• The answer to this question is 'vim'.

Question 10: What's the root flag?

• I ran cd ../.. to navigate into the root directory.

• By navigating to the root folder using cd root and ls to view the contents of the directory I found the root.txt file, cat root.txt to reveal the final flag!