cloud Archives - thefluffy007

boot2root, hacking, web application security

@Vulnhub – Solving #Cofveve

July 7, 2023July 14, 2023 thefluffy007

Another day, another challenge.

In this post we’re going to solve the Cofveve virtual machine.

Let’s get started

After downloading the virtual machine and starting it we see the following:

Using the netdiscover command we can find the IP that the virtual machine is using

Going back to our attacking machine (I’m using the parrot virtual machine).

Doing an nmap scan with the command – nmap -sV <IP address from netdiscover command>, we see the following

There are three ports that are open – ssh, and two http ports.

When http ports are opened the next step is to try to do a recursive brute force attacks to find hidden directories.

We’re going to use two programs – dirb and gobuster.

Doing the dirb of port 80, we get:

Using gobuster we see:

We see that both of the programs did not return anything.

Let’s try to the following programs with the 31337 port, let’s start

We see five files – a robots.txt, and four hidden files (start with .)

The robots.txt gives information about sites that should not be crawled by the internet. This is useful as it will give us files or directories (folders) to review to get more information.

Going to the robots.txt we see:

Hmm, there are three directories that are disallowed. Let’s try to go to them.

Let’s start with taxes. When navigating there we see:

We found the first flag!

Going to the second directory .bashrc we see

Hmm, it’s a file, let’s save it.

Going to the last directory of .profile we see

Another file to save.

Going back to either dirb or gobuster results we have one more directory, .ssh.

This is strange. The .ssh which is hidden should not be exposed on the internet. This folder has information about how to login into the system. Such information is a private key (which only the user should know) public key (which everyone knows), as well as other things.

Going to this folder we see:

We see three items, which look like additional files.

Let’s see if we can navigate to these files.

Navigating to the first item, id_rsa we see:

Another file, so let’s save it.

Let’s try the second file – authorized_keys. Going there we see:

Another file, let’s save it.

Finally, let’s go to the third item – id_rsa.pub

Now that we saved everything let’s go back to our virtual machine and open the files.

First, let’s open the id_rsa.pub file. This file is the public key that is part of a pair. The other pair is the private key. As the name suggests only the user should know their private key and not expose it to anyone else. The public key can be exposed to everyone.

Opening the file we see:

We see a ssh-rsa. RSA is a cryptographic algorithm. Looking at the end of the file we see a valid user – simon@covfefe. We need to save this user as we will need it later.

Opening the id_rsa file we see:

This file is the private key to log into the SSH server as Simon! This is bad, as I said before this file should NOT be exposed as Simon private key should only be known by him only.

Opening the authorized_keys file it has the same content as the id_pub

How can we use this information?

We know a user (simon@covfefe) and a private key.

Well we can log into the ssh server.

Let’s start.

We can use the command ssh -i id_rsa simon@covfefe

Let’s break this down.

We’re logging into the SSH server, specifying a file id_rsa (Simon’s private key) and the user of Simon

We get the following output:

Uh oh – we received an error:

The private key is too open. We need to change the make it less accessible. To do this we’re going to change the permissions from 644 to 600. What this will do is give access to the owner in this case parrot (which is us).

To do this we will use the command chmod 600 id_rsa

Trying to log in again we have the following:

We need a passphrase… which we don’t have. How do we get this?

Doing a good ol’ Google search of cracking passphrase for id_rsa we see the following link

We see there’s a command ssh2john which can be used to crack the passphrase. Let’s do it!

First, we located where the ssh2john command lives in the file system. Next, we’re going to create a hash using ssh2john. We’re going to use this for the next step.

Next, we’re going to send our hash to john which will be used to crack it. We’re going to use the wordlist rockyou.txt which is a common wordlist used to crack passwords. Doing so, we found our passphrase – starwars.

Running our SSH command again, we enter the passphrase of starwars and we’re in the system!

Doing a long listing and showing all of the files (ls -la) we see the files we found before when brute forcing the directories. Let’s look at the .bash_history. This file is good to review as it will let you know commands that were executed on the machine.

Opening this file we see a read_message and then exit. What is read_message? Is this a program?

Executing the read_message we get a prompt. Let’s enter the name Simon. We received a message stating this is is a private messaging system and the source code is in the root directory.

Navigating to the root directory and doing a long listing with all the files (ls -la) we see there’s a read_message.c file, and flag.txt!

Opening the flag.txt file we see a permission denied :-(. Looking at the long listing again this file (flag.txt) can only be accessed the owner, which in this case is root (column 4). I am the user Simon, so I can’t view the file.

Opening the read_message.c file, we see source code, and… the second flag!

And it’s giving us a hint – we need to review the source code.

Looking at the source code we have a buffer (buf) that is holding 20 characters and is executing another file program which is pointing to /usr/local/sbin/message. Whenever we see buf we need to think of buffer overflows.

Buffer overflows are a vulnerability where you overflow the input to access other parts of the computer system. In this case, since we have a holding space of 20 characters – we’re going to overload the buffer with characters OVER 20 characters. Also we’re going to try to change the shell to /bin/sh <— bourne shell. We’re going to change the shell to escalate the user. Changing from Simon to root. We need to be root to open the flag.txt file.

Let’s try it.

Running the commands id and whoami we see we’re root!

Now let’s open the flag.txt and see what’s in the file.

We solved the challenge!

boot2root, hacking, web application security

@Vulnhub – Solving Hemisphere – Gemini #boot2root

June 30, 2023June 30, 2023 thefluffy007

Another day, another challenge.

In this post, we’re going to solve the Hemisphere boot2root from Vulnhub.

Let’s get started.

When we start the Hemisphere – Gemini machine we see the following screen:

OK, we’re prompted with a login screen.

For most boot2root’s we need to find user and root.txt on the box.

Next, on our attacker machine (I’m using parrot security), I am going to type the command netdiscover.

Netdiscover will ping your network with active IPs. From there you can figure out which IP is being used for the virtual machine.

After finding the IP address, let’s open a terminal and type nmap -sV <IP address from netdiscover>. In my case, it’s going to be nmap -sV 192.168.1.132. Doing this we have the following output:

We have 5 ports open.

FTP

SSH

HTTP

And ports 139 and 445.

What do the above two ports belong to?

Well these two ports are for Samba SMB. SMB allows computers to talk with each other. Most times when SMB is enabled, it’s SMB 1.0 which is extremely insecure. So how do we exploit it?

Well, there’s an application enum4linux which will enumerate all of the SMB shares. To do this we’ll use the command enum4linux -a <IP address from netdiscover>. In my case it will be enum4linux -a <192.168.1.132>

Doing this we have:

As you can see we have A LOT of output from enumerating the different shares from SMB.

Looking through the input we see there’s a username: William. We need to save this as we’ll use it later.

Now reviewing the nmap scan again – we see that port 80 or HTTP is open.

Next step, is to enumerate/brute force to see if we can find hidden directories.

Let’s use our two favorite applications – dirb and gobuster.

We have an assortment of files/directories to review.

robots.txt – a helpful file that tells web crawlers to exclude directories to be crawled/exposed on the internet.

Going to the robots.txt file we see:

There are three directories. Let’s see if we can find any goodies!

Going to the secret folder we see:

Hmm… there’s nothing there.

Let’s try admin.

Nothing there either.

Let’s try the last folder – lol

Nothing there.

So the robots.txt was a red herring. It’s a way to knock us off our game as there’s nothing there.

Going back to dirb/gobuster we see there are two folders to review.

Let’s try assets and see what we get.

Going to the assets folder we see:

Hmm – nothing really there. It’s styling for our website.

Let’s try the second folder – images.

Going to this folder we see:

Nothing of importance. This directory is a space for all of the pictures for our site.

Speaking of our site. Let’s go to the index.html file

Going to this page, we see:

Looking through the page we don’t see anything of value.

So now – we’ve enumerated all of the hidden folders and really haven’t found anything. We know we have a user of William, but how do we log in?

We need to enumerate/brute force the directories again.

How do we do this? Change our wordlist.

Going back to gobuster – we’re going to use the dirbuster directory-list-2.3-medium.txt which has more directories to check.

Entering the updated command we see there’s a new folder – portal. Let’s go to that page and see what we get.

Entering this into the URL we see:

Another page. It’s written in Spanish. We see there are three links at the top. Let’s click on the second (Sobre Nosotros) and see what we get.

Nothing to render/view.

But there is something interesting. In the address bar we have a view parameter. Which is pointing to the about-us.html page. Whenever we see a parameter field we should try to do a Local File Inclusion (LFI).

LFI is a vulnerability where you can access the files in the underlying filesystem. Let’s try to access the /etc/passwd file.

In Linux, the /etc/passwd file holds all the users. Let’s try it.

Let’s break this down.

We have five ../. What is this? Well we’re using this (../) to navigate one folder up. Meaning we want to access the parent directory of our current directory. Finally, when we’re at the topmost area of the file system. We’re telling the application to go to /etc/passwd.

Doing this the webpage displays the /etc/passwd file which reveals all of the users!

When you look in the file we see that there is a user named William. That let’s us know our enumeration from the enum4linux command worked correctly.

Now we have the user William, and we can do LFI on this application. What’s our next step.

Well we know that port 22 or SSH is open. We also know there’s a public and private key for all the users to log into the server.

We have a user of William. So we know he’s a valid user. How do we find William’s public and private key?

These keys will be in his home folder.

In Linux, every user has a home folder which allows the user to save files that can only be accessed by them.

With the public and private keys for SSH, we know it’s stored in a hidden directory titled – .ssh and saved in the id_rsa file.

id_rsa is the private key file

id_rsa.pub is the public key file

We don’t want to access the public key file as it’s accessible to EVERYONE. Hence the name.

We want the private key (id_rsa).

Let’s see if we can access this folder through LFI.

Success! We’re able to access the private key. We’re going to need this to log into the SSH server.

First we need to save this key.

To do this, right-click on the file and highlight the text, copy the text.

Open your favorite word browser, and save the contents as id_rsa (the name is important!). Also make sure you know where you saved this file!

After saving the file – let’s do a ls (listing) in the terminal

We see our id_rsa file.

Now let’s try to log in the ssh server.

Enter the command ssh -i id_rsa william@<IP address from netdiscover>. In my case the command will be ssh -i id_rsa william@192.168.1.132.

Uh-oh. We can’t log in as the permissions are too permissive or too open.

Press Control-C to get out of the prompt, we’re going to change the permissions of the id_rsa file.

To do this we’re going to enter the command chmod 600 id_rsa.

After entering this command, do a listing (ls) to view the permissions.

For the id_rsa file we see there is rw in the 2nd and 3rd position. These positions belong to the owner. In this case the owner would be parrot (column 4 in the above screenshot).

Now let’s try to log in again.

Success! We’re in the SSH server.

Now let’s do a listing to see what we see

We have the user.txt file. Let’s open it.

Enter the command cat user.txt

We have the user flag. Now to find the root.txt.

How are we going to do this?

Well there’s a few things we can do.

Since this is a Linux system, let’s enter the command sudo -l

This command will see if a user can elevate (or escalate) their privileges without being the root user.

Entering the command we see this program doesn’t exist.

OK, so we can’t use that.

Remember how we had to change the permissions to gain access to the server?

Well I wonder if we can check files that are overly permissive and use one of those files to elevate our privileges.

What command would we use?

The command we enter is find / -perm -u=s -type f 2>/dev/null

Let’s break this down

We’re using the find command

the / specifies we want to start at the top of the file system

-perm u=s specifies we want to find users that have the sticky bit on.

What is the sticky bit? Sticky bit in Linux specifies we can run an application as the owner of the file.

We want to access the root.txt, so we know that the superuser root is probably the only user who can access this file.

-type f = specifies we want files

2>/dev/null specifies that any errors go to the /dev/null file.

Entering this command we see:

Most of these files are always there but we have a file that shouldn’t be there /etc/passwd.

Why shouldn’t this file in this list? Well the /etc/passwd file holds all the users in the operating system.

This file is overly permissive. We see we have read, write, and execute.

So what do we do? Add another root user.

How do we do this?

First we need to create a password. We’re going to use the openssl command to create a suitable password for the /etc/passwd file. I’m going to select a password of password. See the screenshot below.

Next, we’re going to add our new user in the /etc/passwd file. How are we going to do this?

We’re going to use the echo command.

Let’s break this line down.

echo is a command in Linux. We’re going to echo what is in quotation marks.

We’re going to create a user of test

Next, we enter the password we created with openssl

Next, we have the numbers of 0 for userid and groupid. 0 represent root.

Next, we have the name of root

Next, we specify the home folder for root which is /root

Next, we have the shell for this user which is the bourne again shell (/bin/bash)

Finally, we’re appending this input to the /etc/passwd

Pressing Enter we are prompt back to the command line

Using the command su (switch user)

and enter our new user of test and password of password

We see that the login was successful.

Entering the whoami command we see that our user is now root. Which means our privilege escalation worked!

Now that we’re root let’s navigate to the /root folder and do a listing

Cool, we see the root.txt file.

Opening this file we see:

Success! We found the root flag!

*****If you’ve read to the end – I wanted to add that I am now on @buymeacoffee! If you like the above content and want more of it. Please support: https://www.buymeacoffee.com/thefluffy007

boot2root, hacking, web application security

@RealTryHackMe – RootMe

June 16, 2023April 9, 2023 thefluffy007

Another day, another challenge.

In today’s post we’re going to solve the RootMe room in TryHackMe.

Let’s get started.

Going to the room let’s deploy the machine. This will give us the IP target IP address.

Note: Make sure you’re logged into TryHackMe’s network through OVPN.

After the deployment is complete, we see the following.

Note: Your IP will be different.

Let’s answer the questions

First question:

We can press the completed button as we have successfully deployed our machine.

Second question:

We need to see how many ports are open. Let’s use the application – nmap. Nmap or Network mapper is used to find open/active services on a server.

We’re going to use the -sV (all) command to get find the version numbers of the active services.

Let’s open a terminal and enter the below command.

The complete command is nmap -sV <IP address from deployment> in my case the command will be nmap -sV 10.10.239.135

A screenshot below shows:

2 ports are open. 22 and 80. Corresponding to SSH and HTTP.

So the answer is two.

Entering this into the text box and pressing Enter, we see that is the right answer.

Moving on to question 3.

Going back to our screenshot above of our nmap results we see the version is Apache 2.4.29. Entering 2.4.29 into the text box and pressing Submit we see that’s the correct answer.

Now to question 4.

Going back to our nmap scan we see that for port 22 the service is SSH (case matters!)

Entering this into the text box and pressing Submit we see that is the correct answer.

Question 5

We need to find directories on the web server using the GoBuster Tool.

First we need to figure out – what is the web server?

Going back to our nmap results, the web server is port 80 or HTTP. HTTP stands for Hyper Text Transport Protocol. You’re using the HTTP protocol right now by viewing this blog post. When accessing the internet, and typing in http is invoking the above protocol.

Now that we know that the web server is port 80 or HTTP. How do we access the GoBuster tool?

Well if you’re using the Parrot Security or Kali virtual machine (or attack box on TryHackMe) all you need to do is open a terminal and type gobuster

To access the different directories we’re going to enter the following command gobuster dir -u http://<ip address from deployment> -w /usr/share/wordlists/dirb/common.txt. In my case the command will be gobuster dir -u http://10.10.239.135 -w /usr/share/wordlists/dirb/common.txt.

Let’s break this command down

We enter gobuster to invoke the program

dir to specify we want to brute force or view all directories

-u http://<ip address from deployment> – specifies we want to view all directories from this website

-w /usr/share/wordlists/dirb/common.txt – specifies we want to use this file as our wordlist to find the hidden directories.

Entering the above command into a terminal and pressing enter we have the following:

Let’s explain the numbers next to the results

The numbers are HTTP codes and can tell you a lot of information

Let’s break give a brief breakdown of the different codes

HTTP Code 200 – OK – meaning the website rendered correctly

HTTP Code 301/302 – Redirection – meaning the website will redirect to another page. Pages with this number you should delve deeper into by visiting the actual page

HTTP Code 401 – Unauthorized – meaning you’re not authenticated (or logged in) to view the page

HTTP Code 403 – Forbidden – meaning you’re not authorized to view the page

From the screenshot above we see a few 301’s (redirects) that we should check out.

Question 5 is a gimme/free question as it says no answer is needed, so press Submit to collect the points.

Now, on to question 6

We need to find the hidden directory.

Looking at our gobuster results and what we know about HTTP codes, the 302 are the results we should focus on. Looking at the results and the number of characters the question is looking for – we can surmise the answer is panel.

Entering /panel/ into the text box and pressing Enter we see the assumption was correct.

We can also double check this by opening a web browser and entering the following http://<IP address from deployment>/panel

Doing this on my machine – I see the following

Again – just because we see a redirect doesn’t mean the page will not render. Always check 301 and 302 HTTP codes!

Now on to question 7.

First we’re in a new section of the challenge titled – getting a shell. This is where things get interesting…

What is a shell? Well there are two types of shells

Bind shell – need to have a listener running on the target machine. How bind shells work is the attacker connect to the listener on the target machine to gain a remote shell. This is a two step process. Also, the listener has to be on the target machine. If it’s not this type of shell will not work.

An visual example of a bind shell:

Note: The -e /bin/sh specifies to send a Bourne shell to the attacker’s box

Reverse shell – listener is on the attacker machine, and the target connects to the listener on the attacker machine with a shell. This is the best option as it removes having a listener on the target machine. Also reverse shells allows to be done on popular ports such as HTTP

In our case we’re going to use a reverse shell.

How are we going to do this?

We know that we have a secret directory named – panel. When we went to this page we also noticed that we can upload files.

Let’s try to upload a php file that has a reverse shell attached to it.

Going to this site, we see a file of PHP code.

Copy the code and open a text editor and paste the results.

Scrolling down we see a few lines that need to be changed.

I’m going to explain this below.

The two lines we need to change are our IP address and port.

The IP address is going to point to our IP from our TryHackMe account. You can find the address at the top of the page in green. My address is 10.13.2.231. This will be considered the Attacker’s box.

The port we can make anything we want. In this case let’s make it 1234

Save the file and exit the editor.

Now doing a listing (ls) we see the following

We need to change the permission to have the file execute. To do this we enter the command chmod +x test.php

Now let’s go to the panel directory.

Opening a browser and entering http://<IP address from deployment>/panel, we see the following. Let’s try to update our test.php file.

After pressing upload we see there’s an error. PHP files are not permitted.

How can we fix this? Well, let’s try changing the extension from PHP to PHP5.

Going back to the terminal let’s enter the command: cp test.php test.php5. This will create a new file named test.php5.

Doing a listing (ls -la) we see that the file is created.

Going back to the panel directory let’s browse and select test.php5

After pressing Upload we see that the file was uploaded successfully.

We see the file was uploaded, but how do we get to the file? Going back to to our gobuster results, we see there’s another 301 named uploads. Let’s try to navigate to this directory.

Opening another tab and entering http://<IP address from deployment>/uploads we see the following:

Our test file is here!

Now how do we connect to the target box?

First, we need to open a new terminal and enter nc -nvlp 1234

Let’s explain what’s happening:

nc – is a program named netcat. You can think of netcat as a swiss-army program that can do a lot of information. In our case, we’re going to use netcat to set up our listener on our machine.

nvlp – this is a series of parameters that do the following not resolve names, verbose printing, listen, on a specific port

1234 – is the port we’re going to listen on. **Make sure this matches the port inside your test.php5 file, otherwise the next steps will not work**

Going back to the tab with the uploads folder, click on the test.php5. You will notice the application is running and seems to hang. This is what we want.

Going back to our terminal where we set up the netcat listener, we have input! Our reverse shell worked successfully! We’re officially on the target machine!

We can prove this by entering the command whoami which will give us our current user. The current user is www-data which signifies the web user.

OK, we’re on the machine, but how do we find and open the user.txt file (we need this to answer the question). We need to find it on the file system.

Entering the command find / -name=user.txt 2>/dev/null

Let’s explain this command:

find – is the program we’re using

/ – specifies we want to start at the root

-name user.txt – specifies the file we want to find on the file system

2 >/dev/null – specifies if there are any errors – such as we access denied, send that output to /dev/null. In other words do not output it to the screen.

Entering this in the terminal we see the following:

The user.txt is at /var/www/user.txt

Navigating to the /var/www directory using the cd (change directory) command, we need to view the user.txt file. We’re going to do that with the cat (concatenate) command.

Doing this we see:

We found the answer!

Entering THM{y0u_g0t_a_sh3ll} into the text field we see it’s the correct answer.

Now on to question 8.

OK, we need to search SUID permissions to find a weird file.

First, we need to describe what is a SUID. SUID or a user sticky bit is a permission inside of Linux that allows an application to run as it’s root owner. This is good for system administrators when they want to run commands without switching users. However, there are times where these files are overly permissive or too open for anyone to run. With these files we can do something called privilege escalation which means we can upgrade our user from a regular user to an admin/super user.

How do we find files that have the sticky bit turned on in Linux?

Well we enter the command find / -perm -u=s -type f 2>/dev/null.

Let’s break this down

find – the program we’re using

/ – start at the root of the file system

-perm – specifies we’re looking at permissions

-u=s – specifies we’re looking for the SUID. u is user, and s specifies the sticky bit with execution turned on. If we didn’t want execution turned on we would use S (this wouldn’t help us as we need to execute the program)

2>/dev/null – specifies if there are any errors (such as permission denied) send it to /dev/null

Looking at the output, some of these files are standard. There is one file that looks suspect. The file is /usr/bin/python.

Entering this in the text box we see this is the correct answer.

On to question 9

We need to escalate our privileges to change from www-data to root.

From question 8, we see that /usr/bin/python is the weird file in question that shouldn’t have the SUID bit on. We’re going to use this file to escalate our privileges from www-data to root.

Going to this site and scrolling down we see a section to escalate our privileges using python. Let’s go back to the terminal and enter the command

Let’s break this down

/usr/bin/python – specifies we want to execute the python program

-c = execute the following command

‘import os; os.execl(“/bin/sh”, “sh”, “-p”)’ – specifies we want to import the os library. Then we’re going to execute another command execute command line (execl) with the following parameters:

/bin/sh – specifies we want ot use the Bourne shell

sh – we’re specifying the file or the shell

-p – specifies the command we want to use

This question is a gimme so we can click the completed button.

Now, on to the last question – question 10.

We need to find the root.txt file

After the above command and doing a whoami command we see that we’re root!

Now we need to read the root.txt file

Navigating to the /root folder and doing a listing (ls -la) we see the root.txt file!

Opening the root.txt file with the cat command, we see:

Entering the above THM{pr1v1l3g3_3sc4l4t1on} into the text box we see that’s the correct answer and we have solved all the challenges.

Like this content and want more? Well support me by Buying Me A Coffee. Link –> https://www.buymeacoffee.com/thefluffy007

boot2root, hacking, OSCP, web application security

OSCP Series: Kioptrix Level 3

March 10, 2019March 3, 2021 thefluffy007

Another day, another challenge.

In today’s blog post we’re going to solve level 3 of the Kioptrix series.

For the level 2 walk through, click HERE.

Let’s begin.

Starting the VM, we’re presented with the following login screen.

Going back to the login screen the disclaimer states that we need to modify the /etc/hosts to include our IP address that our kioptrix3 vm is running.

Let’s see which services are running on the virtual machine

Hmm… we have two services open – ssh and http. Let’s go to the web application and do a manual crawl of the site.

Navigating to kioptrix3.com we see the following:

Clicking the Login button we’re presented with a login screen. At the bottom we notice the page is using LotusCMS. Let’s keep this in mind, just in case we it to exploit the application.

Let’s run nikto to see if we can find more vulnerabilities in the web application.

We see that the application found a phpmyadmin folder. Let’s go to the folder and see what we can find.

Going to the php page, we realize we don’t have login credentials, so at this point we can’t log into the phpmyadmin page. We know the application is using LotusCMS let’s see if there’s a LotusCMS module that we can use in Metasploit.

After opening and searching Metasploit for LotusCMS we see there’s a perfect remote code execution we can use.

After selecting the exploit let’s review the options to see what we need to add to the module to make the exploit work. We need a remote host (RHOST), remote port (RPORT) and the URI. Let’s add the information below.

After entering the information, and typing run, we notice that we’re presented with a meterpreter shell! Let’s go to the home directory and see how many users are on the box.

We have three users, dreg, loneferret, and www (web). Let’s see what’s in the dreg folder.

Searching the dreg folder we don’t see much. We see a bash_logout, bashrc, and profile scripts. Let’s move on to loneferret.

Hmm… Loneferret has a lot of information. We see there’s a file titled, “.sudo_as_admin_successful”, a company police readme, and a checksec script. Let’s review the company policy readme.

Well… it seem if we want to edit, create, or view files we need to use the command sudo ht. Let’s keep this in our toolbox because I am sure we will use it later. Remember we found a phpmyadmin page, let’s see if we can find a config file with login credentials.

Doing a search on the kioptrix3 folder, we see there are three config files. Let’s review the last file, gconfig.php.

Opening the gconfig.php file we notice there are login credentials for the phpmyadmin page! Let’s keep this in our toolbox because we might have to use it later.

At this point, we have php login credentials, but that’s not enough to gain root privileges. Let’s see if we can review the web application again for more clues.

We notice there’s a gallery section of the application. Maybe we can be lucky and the application is susceptible to sql injection.

Adding an apostrophe to the end of the number there’s no change in the application. We notice at the bottom of the page there’s a sorting option. Let’s see if changing the sorting option will invoke a SQL injection.

Changing the sorting option to Photo Id, and adding the apostrophe to the end of the number, we get the following…

Success! The application is susceptible to SQL injection. Let’s fire up sqlmap and see what goodies we can find.

Executing a preliminary SQLMap run we see that SQLMap verified our manual testing results that the application is susceptible to SQL injection. SQLMap also found the technologies the application is using. Let’s do a more extensive probe. Firing SQLMap again we use the dump all to see what data we can find.

We found the dev accounts database, and notice that we have two users dreg, and loneferret, along with their passwords. Remember these are the same users we found when we used Metasploit with the LotusCMS remote code execution.

Now that we have login credentials, and we know that loneferret has more promising information than dreg, let’s connect to loneferret’s account through ssh.

Connecting to loneferret’s account, we try to access the root folder, but receive a permission denied (we’re not root… yet). We also try to execute the sudo ht command that was listed in the company readme file, and we get the error, “error opening terminal: xterm-256color”.

Doing a quick google search we find that we need to use the following command “export TERM=xterm”. Doing this and running the sudo ht command again we’re presented with the following screenshot. Let’s see if we can view and modify the /etc/sudoers file and see if we can escalate our privileges from loneferret to root.

Opening the /etc/sudoers file we notice that the loneferret has a user privilege escalation where a password is not required. Right now it’s for the commands: su, and sh. Let’s add our ht command to the list. After adding the command, saving, and exiting we’re presented back to the command prompt.

Executing the command sudo /bin/sh, and running a whoami we see that we’ve been escalated to root! Running the ls command we see that we’re presented with the same information we had in our meterpreter shell. Changing our directory to /root we notice there’s a Congrats.txt file.

Opening the Congrats.txt file we have found the flag and completed the challenge!

hacking, web application security

Looking to pass the OSCP, CEH, eCPPT, or LPT? Keep reading… @j0emccray

July 9, 2018March 3, 2021

Greetings!

This blog post will be short.

I received the following email (from Strategic Security owned by Joe McCray) in my inbox:

As many know, I’m embarking on the OSCP challenge (full disclosure: my job is paying for the cert), and I was happy to see this email.

$50 for a self-paced course that will give you a primer before starting the real (in my case OSCP) course seems like a deal too good to be true…

In this case, it’s not. The email is 100% factual. I signed up for the course about an hour ago, and was presented with the coursework. I also received an email stating a customer service rep will reach out to me to make sure everything is going well. This customer service rep will be an accountability buddy to make sure that all assignments and quizzes are completed.

So, if you want to study the concepts of the OSCP, eCPPT, LPT, or CEH course(s) before actually taking the class, then look no further than this course. At $50 what do you REALLY have to lose? At the worst you’re out of a week’s worth of lunch, at best you acquire knowledge that will last a lifetime!

If you want more information about the course, check out Joe McCray on twitter at @j0emccray.

NOTE: I will do a review of the course when I have completed it – so stay tuned!

capture the flag, hacking, web application security

PicoCTF 2017 – WorldChat

June 19, 2017March 3, 2021 thefluffy007

Another day, another challenge…

In today’s blog post we will be solving the “WorldChat” challenge from the PicoCTF.

Let’s get started!

Going to the challenge we see:

OK so we need to find the flag inside of the WorldChat app. According to the description when connecting to this app there will be many people on the app besides us.

Let’s look at the hints to see if it will help us.

We need to us the nc command (we’ve used this in another challenge) and use the grep command to filter output.

Let’s try it.

Connecting to the server we see a bunch of chats from different people. I pressed Ctrl + C to stop it.

Let’s use the hints and use the grep command with the “|” (pipe) command.

Press Ctrl + C to end the chat.

I have only captured screenshots of output that has the flag. Entering the flag into the input box we acquired 30 points.

capture the flag, hacking, web application security

InfoSec Institute CTF Challenge #14

June 10, 2017March 3, 2021 thefluffy007

Another day, another challenge.

Today’s challenge comes from the InfoSec Institute CTF program.

Going to the following link we see the following:

Doing a right click, view page source we see the following:

Hmm… there’s a file, titled level14 inside the misc folder. Let’s go that file and see what’s there…

Going to the file we see the following:

Hmm… it looks like we have a SQL dump that’s showing us all the tables and values inside of a php application.

Scrolling down we see something that looks interesting, and strange…

Could this be some type of encoding? Possibly hexadecimal encoding?

First, we don’t need the double forward slash, we just need one. Removing the extra slashes we get the following:

Using a Hex to ASCII converter here, we get:

We found the flag – infosec_flagis_whatsorceryisthis

Lessons learned:

Our trick still works! We were able to find valuable information when looking at the page source. Going to the file listed we noticed it was a dump of SQL tables. Looking through the tables we noticed suspicious output, which we guessed was some type of encoding. Using information we learned from a previous challenge we were able to deduce that the encoding was hexadecimal encoding. From there we were able to find the flag.