I'd like to know basically how can I encrypt data with a generated salt key and then decrypt it using python ?
i've gone trough a lot of websites and modules, and they all look great at encrpytion part, but none can decrypt as it seems.
My main concern is to have strong salt key, that'd be probably generated over few hunderd times, then use that key to encrypt data - in particular I'm looking into encrypting JSON encoded data with the salt key, sending the encrypted data to the other side ( listening client ) and then decrypt the data there based on the algorithm that is used to generate the salt key.
I've found that mcrypt module would work best with this, but there isn't much documentation for python-mcrypt module ( that's currently being outdated and not maintained).
The short answer to your question is that you combine the password and the salt and hash them repeatedly to create your key. Then you append the salt onto the ciphertext so that you can generate the key for decryption. To ensure that I had the right answer, I made a few functions to do the work. They are given below.
In my answer, I've made use of pycrypto, so we need to import a few of those libraries.
import Crypto.Random from Crypto.Cipher import AES import hashlib
To aid readability, I've defined a few constants that I'll use later on.
# salt size in bytes SALT_SIZE = 16 # number of iterations in the key generation NUMBER_OF_ITERATIONS = 20 # the size multiple required for AES AES_MULTIPLE = 16
To use a salt, I've done a password-based encryption scheme. I've used the RSA PKCS #5 standard for password-based encryption key generation and padding, adapted for the AES encryption algorithm.
To generate the key, the password and the salt are concatenated. This combination is hashed as many times as requested.
def generate_key(password, salt, iterations): assert iterations > 0 key = password + salt for i in range(iterations): key = hashlib.sha256(key).digest() return key
To pad the text, you figure out how many extra bytes you have beyond an even multiple of 16. If it is 0, you add 16 bytes of padding, if it is 1, you add 15, etc. This way you always add padding. The character you pad with is the character with the same value as the number of padding bytes (
chr(padding_size)), to aid the removal of the padding at the end (
def pad_text(text, multiple): extra_bytes = len(text) % multiple padding_size = multiple - extra_bytes padding = chr(padding_size) * padding_size padded_text = text + padding return padded_text def unpad_text(padded_text): padding_size = ord(padded_text[-1]) text = padded_text[:-padding_size] return text
Encryption requires generating a random salt and using that along with the password to generate the encryption key. The text is padded using the above
pad_text function and then encrypted with a cipher object. The ciphertext and salt are concatenated and returned as a result. If you wanted to send this as plaintext, you would need to encode it with base64.
def encrypt(plaintext, password): salt = Crypto.Random.get_random_bytes(SALT_SIZE) key = generate_key(password, salt, NUMBER_OF_ITERATIONS) cipher = AES.new(key, AES.MODE_ECB) padded_plaintext = pad_text(plaintext, AES_MULTIPLE) ciphertext = cipher.encrypt(padded_plaintext) ciphertext_with_salt = salt + ciphertext return ciphertext_with_salt
Decryption proceeds backwards, pulling the salt off of the ciphertext and using that to decrypt the remainder of the ciphertext. Then the plaintext is unpadded using
def decrypt(ciphertext, password): salt = ciphertext[0:SALT_SIZE] ciphertext_sans_salt = ciphertext[SALT_SIZE:] key = generate_key(password, salt, NUMBER_OF_ITERATIONS) cipher = AES.new(key, AES.MODE_ECB) padded_plaintext = cipher.decrypt(ciphertext_sans_salt) plaintext = unpad_text(padded_plaintext) return plaintext
Let me know if you have any other questions/clarifications.