Yozzang의 해킹일기 💻
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First, i include “bizness.htb” in my host file along with the machine’s IP address using the following command:

<bash />
echo "10.10.11.252 bizness.htb" >> /etc/hosts

I initiate our exploration with an Nmap scan.

I have identified open port 80. Let’s access the site in a web browser for further exploration.

Following an analysis of all pages, let’s initiate a directory search using dirsearch.

Discovered the /control/login path, explored it in the browser, and identified a login page running ApacheOFBiz.

After conducting research online, it was identified that ApacheOFBiz is vulnerable to Authentication Bypass Vulnerability.

Provided with the GitHub exploit, i will utilize it to attain a reverse shell.

 

GitHub - jakabakos/Apache-OFBiz-Authentication-Bypass: This repo is a PoC with to exploit CVE-2023-51467 and CVE-2023-49070 prea

This repo is a PoC with to exploit CVE-2023-51467 and CVE-2023-49070 preauth RCE vulnerabilities found in Apache OFBiz. - GitHub - jakabakos/Apache-OFBiz-Authentication-Bypass: This repo is a PoC w...

github.com

Run the Python exploit. Prior to execution, initiate the Netcat listener.

Success! A reverse shell has been obtained!

Now, attempt to establish a stable shell. For detailed guidance, please refer to my blog “A Step-by-Step Guide to turning a basic reverse shell into a fully interactive terminal using Python”

 

A Step-by-Step Guide to turning a basic reverse shell into a fully interactive terminal using…

Challenge

medium.com

The user flag has been discovered.

Upon exploring the machine, I found a hash value in AdminUserLoginData.xml file.

While exploring the machine, I discovered Derby database data files containing $SHA1$d$uP0_QaVBpDWFeo8-dRzDqRwXQ2I= .

This format, exemplified by “$SHA1$d$uP0_QaVBpDWFeo8-dRzDqRwXQ2I=”, signifies the use of the SHA-1 hashing algorithm, where “SHA1” denotes the algorithm, “d” serves as the salt, and “uP0_QaVBpDWFeo8-dRzDqRwXQ2I=” represents the hashed password.

Upon successfully cracking the hash, the resulting information will unveil the root password.

Here is the crack code.

<python />
import hashlib import base64 import os def cryptBytes(hash_type, salt, value): if not hash_type: hash_type = "SHA" if not salt: salt = base64.urlsafe_b64encode(os.urandom(16)).decode('utf-8') hash_obj = hashlib.new(hash_type) hash_obj.update(salt.encode('utf-8')) hash_obj.update(value) hashed_bytes = hash_obj.digest() result = f"${hash_type}${salt}${base64.urlsafe_b64encode(hashed_bytes).decode('utf-8').replace('+', '.')}" return result def getCryptedBytes(hash_type, salt, value): try: hash_obj = hashlib.new(hash_type) hash_obj.update(salt.encode('utf-8')) hash_obj.update(value) hashed_bytes = hash_obj.digest() return base64.urlsafe_b64encode(hashed_bytes).decode('utf-8').replace('+', '.') except hashlib.NoSuchAlgorithmException as e: raise Exception(f"Error while computing hash of type {hash_type}: {e}") hash_type = "SHA1" salt = "d" search = "$SHA1$d$uP0_QaVBpDWFeo8-dRzDqRwXQ2I=" wordlist = '/usr/share/wordlists/rockyou.txt' with open(wordlist,'r',encoding='latin-1') as password_list: for password in password_list: value = password.strip() hashed_password = cryptBytes(hash_type, salt, value.encode('utf-8')) # print(hashed_password) if hashed_password == search: print(f'Found Password:{value}, hash:{hashed_password}')

Successfully acquired the root flag.

Indeed! Together, i successfully navigated and Pwned the machine.

 

1. Reference

https://github.com/apache/ofbiz/blob/trunk/framework/base/src/main/java/org/apache/ofbiz/base/crypto/HashCrypt.java

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