Hello World and welcome to haxez, today we’re going to be looking at Hack This Site Extended Basic 6. Solving this challenge requires some basic knowledge of PHP or any other language for that matter. It’s a simple challenge that shows how poorly coded web application authentication mechanisms can be bypassed.

The Extended Basic 6 Challenge

The image below shows the PHP that makes up the authentication mechanism. Furthermore, This is the code that we need to exploit in order to bypass the authentication mechanism. The introduction text at the top explains that the sysadmin is a noob and that the script is located at http://moo.com/moo.php. Therefore, to solve this challenge we need to append the correct syntax to the end of the URL and submit it to the submission form.

The PHP Code

The snippet below is the exact code we’re going to be exploiting. Furthermore, there is no backend database to worry about, just some simple PHP logic that we can leverage for our own nefarious purposes.

<?php
        $user = $_GET['user'];
        $pass = $_GET['pass'];
        if (isAuthed($user,$pass))
        {
                $passed=TRUE;
        }
        if ($passed==TRUE)
        {
                echo 'you win';
        }
?>
        <form action="me.php" method="get">
        <input type="text" name="user" />
        <input type="password" name="pass" />
        </form>
<?php
        function isAuthed($a,$b)
        {
                return FALSE;
        }
?>

Breaking Down The Code

We’re going to break the code down line by line in order to solve this challenge. Once we understand what the code is doing, we will be able to use its own logic against it and bypass authentication.

First, the code starts with it a tag telling us what language it is.

<?php

Next, we have two variables being set from user input. The $user and $pass variables are populated by the values of ‘user’ and ‘pass’. This is more evident later on when looking at the HTML form.

$user = $_GET['user'];
$pass = $_GET['pass'];

This is where the logic beings. When the user submits their username and password, the data is passed to an if statement. If the values of the variables ‘$user’ and ‘$pass’ are correct then the variable ‘$passed’ is set to true.

if (isAuthed($user,$pass))
{
        $passed=TRUE;
}

After that, the application uses another if statement to check whether the value of the variable ‘$passed’ is set to true. If ‘$passed’ is set to true then the application echos out “you win”. This also ends the first section of PHP.

if ($passed==TRUE)
{
        echo 'you win';
}
?>

Now, we move on to the HTML. This is the login form that the user will send when loading the page in their browser. It is fairly standard and just performs a get request to me.php. The contents of the get request or the values of the input for ‘user’ and ‘password’. The user input type is text and the password input type is password.

<form action="me.php" method="get">
<input type="text" name="user" />
<input type="password" name="pass" />
</form>

Finally, we have the last section of PHP. This is the function that checks to see whether the username and password are correct. This function compares the values of variables ‘$user’ and ‘$pass’ with the variables ‘$a’ and ‘$b’. If they match then it is the first if statement is executed.

<?php
        function isAuthed($a,$b)
        {
                return FALSE;
        }
?>

The Extended Basic 6 Solution

With that rather long explanation out the way, we can now move on to solving the challenge. It’s pretty simple if you know how to PHP site URL’s work. We know that in order for us to authenticate, the variable ‘$passed’ needs to be set to ‘TRUE’

PHP allows you to specify variables and their contents in the URL. For example, a dynamic PHP website using a CMS like WordPress may have a URL like HTTP://site.com/index.php?page=1. The question mark indicates that what follows is a variable and in this case, the variable is ‘$page’. Furthermore, we can specify that we want the value of the page variable to be 1.

With this in mind, we can take the provided URL of HTTP://moo.com/moo.php and append a question mark followed by the variable passed equals TRUE, as seen below.

http://moo.com/moo.php?passed=TRUE

Submitting this to the input box will solve the challenge.

Hello world, welcome to HaXeZ where today we will be looking at the Hack This Site Extended Basic mission 5. This mission is another programming mission that requires you to review the contents of a PHP file and then subsequently a shell script that is used to edit the PHP file. There is an error in the shell script that prevents it from doing what it’s supposed to do. We need to fix it.

The Extended Basic 5 Code

The image below shows the code that Sam has written. The introduction message advises that Sam created a function called safeeval to run commands. However, on the page, he neglected to use safeeval and use eval() instead. Furthermore, it explains that he wrote a shell script to go through the PHP file and replace all values of eval() with safeeval. Unfortunately, there is an error in the shell script that prevents the script from working.

The PHP Code

<?php
        include ('safe.inc.php');
        if ($access=="allowed") {
                eval($_GET['cmd']);
                if (!empty($_GET['cmd2'])) {
                        eval($_GET['cmd2']);
                }
        }
?>

The Shell Script

#!/bin/sh
rm OK
sed -E "s/eval/safeeval/" <exec.php >tmp && touch OK
if [ -f OK ]; then
        rm exec.php && mv tmp exec.php
fi

Hack This Site Extbasic 5 Code Breakdown

We don’t need to break down the PHP code as that is the code we’re trying to amend with the shell script. So let’s break the shell script down line by line.

First, as with all shell scripts, we get a comment to explain that it is in fact a shell script. Nothing out of the ordinary here.

!/bin/sh

Next, we have the ‘rm’ command which on Linux means to remove something and after a bit of research, it appears that it is the same on FreeBSD too. So this line is saying remove ok.

rm OK

The next line is using the sed (stream editor) command which appears to be what is substituting eval for safeeval. It is then passing the exec.php file to the sed command using a less than sign. After that, it appears to be taking the results of the sed command and appending them to ‘tmp’ and creating a file.

sed -E "s/eval/safeeval/" <exec.php >tmp && touch OK

Next, we appear to have some logic that checks if the command executed ok, and if it did it moves on to the next line.

if [ -f OK ]; then

Finally, the script removes exec.php and moves tmp to exec.php.

rm exec.php && mv tmp exec.php

Hack This Site Extbasic 5 Solution

The final line is the end of the if statement so we don’t need to explain that any further. Upon closer inspection and reading the Wikipedia page for sed it appears substitutions with sed requires 2 characters. First, it explains that in some cases you need to start the argument with -E. This is true on MacOS which is a variant of free BSD. We know Sam is using free BSD so the -E at the start of his sed statement is correct. However, we also need the characters s and g. The s character tells sed to substitute one word for the other (eval with safeeval). The g character tells it to do it globally. In short, Sam needs a g at the end of his sed statement to replace all instances of eval with safeeval. The correct syntax should be as follows.

sed -E "s/eval/safeeval/g" <exec.php >tmp && touch OK

Without the g, the script only replaces one instance of eval. The script has multiple instances of eval so the script fails to complete its purpose. Fun challenge.

Hello world, welcome to haxez where today we will be looking at the Hack This Site Extended Basic mission 4. This mission is another programming mission that requires you to examine the source code of an application to determine its output. Again, please be advised that I’m terrible at programming so my explanation might be terrible.

The Extended Basic 4 Code

The image below shows the introduction message and explains that sometimes we may need to decipher a language. Furthermore, it explains that sometimes the language may not be on google or encrypted in some way.

Below that we can see there appears to be a user input of the numbers 6 and 7.

We then have a number of lines of code that appear to perform operations on the user-submitted values.

The Solution

As with Extended Basic 3, I’m going to attempt to break this down line by line and explain what is happening.

BEGIN F.ake

This appears to be the start of the program. I don’t think there is much more to it than that other than indicating the start of the program.

var int as in

What this appears to be doing is assigning whatever value the user has submitted to the variable var. The ‘in’ is the user input and var is the variable name. In this case, the value will be 6.

int var as in

This is similar to the line above and is assigning whatever the user submits to a variable called ‘int’. Again the ‘in’ part of the statement appears to be the user input prompt. In this case, the value will be 7.

out var int

Finally, the script is printing or echoing both the variables ‘var’ and ‘init’ to the screen meaning it should output 67.

Extended Basic – Mission 4 Conclusion

I can’t think of any other way to solve this or what any of the other parts of the code would be doing, other than what I have explained. I hope this has helped you solve the challenge. Feel free to check out parts 1 to 3 and drop by my youtube channel and subscribe.

Hello world and welcome back to haxez, thank you for surfing by. This post is a walkthrough of the Hack This Site Extended Basic Mission 3. The purpose of this challenge is to deduce the function of a bespoke programming language’s application. A basic understanding of programming and assigning variables is required for this challenge. However, I’m terrible at programming and was still able to solve the challenge.

The Extended Basic 3 Function

As mentioned above, the image below informs the user that the challenge creator has created a bespoke programming language. In order to solve the challenge, we need to walk through the application step by step and determine the output.

Therefore, I believe the best method of solving this challenge is to analyse each line individually and identify what it is doing.

The Solution

BEGIN notr.eal

Firstly, the application starts with ‘BEGIN notr.eal’. Granted, this appears fairly self-explanatory and denotes the start of the application.

CREATE int AS 2

Secondly, it appears as though the application is creating an integer with the value of 2. However, as with other programming languages, the position of ‘CREATE’ suggests it is more likely that the integer value of 2 is being assigned to the variable ‘CREATE’.

DESTROY int AS 0

Thirdly, the same can be said about the ‘DESTROY’ variable. This could easily be mistaken for a function of the program. However, since this function isn’t previously described in the program, I’m going to assume that an integer value of 0 is being added to the variable ‘DESTROY’.

ANS var AS Create + TO

Fourthly, it would appear that the value of the ‘create’ variable (currently 2) or ‘CREATE’ as previously written is being add to the ‘TO’ variable. I’m not too sure about this one but it resulted in the correct answer so my logic (however flawed) seems correct.

out TO

Finally, the value of the ‘TO’ variable is printed out to the screen. So in this instance the answer should be 2. If you submit that to the submission box then it should solve the challenge.

BEGIN notr.eal /* Starts the program
CREATE int AS 2 /* Adds the integer 2 to variable 'CREATE'
DESTROY int AS 0 /* Adds the integer 0 to variable 'DESTROY'
ANS var AS Create + TO /* Appears to take the value of variable 'CREATE' and adds to varable 'TO'
out TO /* Prints the value of 'TO'

Extended Basic Mission 3 — Conclusion

While my explanation might be incorrect, it resulted in the correct answer. Furthermore, I tried to solve the challenge in other ways but wasn’t able to. If we break the program down again and look at lines 2 and 3 we could infer that the program is simply creating an integer of 2 and then destroying it. If it destroys the integer then the value of the variable ‘TO’ would be 0 which is the wrong answer. I’ve also looked at whether ‘AS’ could be a variable but we end up with the same result of the value of the variable being destroyed and ending up with 0. There could be something else I’m missing and if you spot it then please let me know. Anyway this was a fun challenge, please check out my other posts in this series ExtBasic1 and ExtBasic2.

Hello world, welcome to haxez where we will be covering Hack This Site Extended Basic Mission 2. This challenge is fairly simple provided you have an understanding of application structures. It requires us to slightly modify the provided script in order to access the index.php page at the root of the web application. In order to do that we need to perform a directory traversal up two directories to grab the index.html page.

The Extended Basic 2 Function

As you can see from the screenshot below, we have some fairly basic PHP code that is attempting to get the contents of the filename specified by the value ‘filename‘. Furthermore, it specifies the type of extension for the filename which in this case is ‘.php‘. Underneath the code, we have a submission box where we need to submit the solution to the challenge.

The Solution

Given these points, all we need to do to solve this mission is to tell the script to navigate up two directories. We are currently in the ‘extbasic‘ directory looking at the file named 2 ‘/missions/extbasic/2‘. So by traversing up two directories we should be in the root directory. Once there, we need to specify the ‘index.php‘, however the file extension ‘.php‘ has already been appended for us so we only need to specify the word index. The correct solution should be ‘../../index‘. Paste that into the check form and you should complete the mission and be able to proceed on to the next one.

Extended Basic Mission 2 — Conclusion

This is a simple but fun challenge that tests your knowledge of web application directory structures and code reading ability. While I wouldn’t have a clue how to write this off the top of my head, I easily worked out what the code is attempting to do. Once you understand what the code is doing, and you understand the rules of the mission then it’s fairly simple. This type of attack is known as a directory traversal attack and can be prevented by validating user input and by having strict permissions policies on directories. Anyway, I hope this helped you solve the mission.