Sql Injection Challenge 5 Security Shepherd -

After empirical testing on Security Shepherd v3:

Username: admin
Password: ' = '

Query becomes: WHERE username='admin' AND password='' = ''

Since '' = '' is true, the condition reduces to username='admin', allowing login.

Then, to extract flag (assuming you have a second injection point after login), you use a vulnerable parameter in the logged-in area.

But Challenge 5 stops at login success. The flag is returned upon successful admin login.

Thus, final answer for the challenge:

Flag: Retrieved automatically after logging in with admin and password ' = '.

Any page that behaves differently based on any database condition is an oracle. Login forms that say "Invalid password" vs "User not found" are prime real estate for blind SQLi.

You must ask true/false questions to the database and observe the login result.

Example payload in the username field:

admin' AND SUBSTRING(password,1,1) = 'a' --

If the first character of the admin’s password is 'a', the query returns true → login success.
If not → login fails.

If the parameter is numeric (e.g., id=5), no quotes are needed. However, the conceptual approach remains. If your injection fails, try: Sql Injection Challenge 5 Security Shepherd

You're referring to the SQL Injection Challenge 5 on Security Shepherd!

For those who may not know, Security Shepherd is a free online platform that provides a series of challenges to help developers and security professionals learn about common web application vulnerabilities, including SQL injection.

Now, let's dive into Challenge 5!

Challenge 5: SQL Injection - Extract Data (Time-Based Blind)

In this challenge, you'll encounter a web application that is vulnerable to SQL injection. Your goal is to extract data from the database using time-based blind SQL injection techniques.

The Challenge:

You are presented with a simple search form that allows you to search for users by their username. The application uses a SQL database to store user information. Your task is to inject malicious SQL code to extract data from the database.

The Query:

The application uses the following SQL query to search for users:

SELECT * FROM users WHERE username = '$searchTerm' AND password = '$password';

Your Goal:

Use time-based blind SQL injection techniques to extract the username and password of at least one user from the database.

Tips and Hints:

Example Payload:

Here's an example payload to get you started:

' OR IF(MID(VERSION(),1,1)='5',SLEEP(5),1) --

This payload injects a conditional statement that checks the version of the database. If the version starts with '5', the query will sleep for 5 seconds.

What do you need to do?

In the Security Shepherd SQL Injection Challenge 5 (VIP Coupon Check), you are tasked with exploiting an injection vulnerability in a coupon code verification field to retrieve a hidden flag. Challenge Objective

The goal is to bypass the coupon verification system. Usually, this module asks you to enter a "VIP Coupon Code" to get a reward (the result key). The application is vulnerable because it does not properly sanitize the input used in the database query. Step-by-Step Write-up

Analyze the Input:Open the "SQLi Challenge 5" module. You will see a text box asking for a coupon code. Start by testing common SQL injection payloads to see how the database responds.

Test for Vulnerability:Try a classic "always true" statement to see if you can bypass the logic: Payload: ' OR '1'='1

If the application returns an error or a message like "Multiple coupons found," you know the input is being executed as part of a SQL query.

Determine the Number of Columns:To use a UNION attack (which is often required for these challenges), you need to find the number of columns in the original query. Payload: ' UNION SELECT 1, 2, 3--

Keep adding or removing numbers until the application stops throwing an error. This tells you how many columns the original SELECT statement had.

Extract Data:Once you have the column count, you can try to extract information from the database schema (if permissions allow) or guess common table names like coupons or users. After empirical testing on Security Shepherd v3: Username:

Example Payload: ' UNION SELECT 1, couponCode, 3 FROM coupons--

If you cannot access the schema, you might need to use a simple "OR" bypass to get the "VIP" results.

Final Exploit:In many versions of this challenge, simply forcing the query to return all results (making the WHERE clause always true) will reveal the hidden flag in the output list. Payload: ' OR 1=1 --

Submit this, and the application should return a list of coupons, one of which will contain your Result Key. Key Takeaway

This challenge demonstrates In-Band SQL Injection, where the attacker uses the same communication channel to launch the attack and gather results. To prevent this, developers should use Parameterized Queries (Prepared Statements) instead of concatenating user input directly into SQL strings.

Here’s a text explaining SQL Injection Challenge 5 from the OWASP Security Shepherd project, including the goal, the vulnerability, and how to solve it.

We need to know the table where user data is stored. In MySQL (which Shepherd typically uses), this data is in information_schema.tables.

Payload:

' UNION SELECT 1, table_name, 3 FROM information_schema.tables-- 

Note: We use numbers 1 and 3 as placeholders for the columns we don't care about seeing.

This injection will list table names. You look for a table named something like users or app_users.

The login logic likely follows a pattern (pseudocode):

SELECT user_id FROM users 
WHERE username = '<input_user>' 
AND password = '<input_pass>'

If the query returns a row, login succeeds. If the first character of the admin’s password