I'm trying to get a ruby implementation of an encryption lib that's apparently popular in the Java world -- PBEWithMD5AndDES
Does anyone know how to use openssl or another open source gem to perform encryption/decryption that's compatible with this format?
Updated:
I used a gem chilkat to implement it but it is paid, i need an opensource solution.
You don't need to actually implement PBEWithMD5andDES assuming ruby has a DES implementation. What you need to implement is the key derivation function ( who you get a key out of a password) and then feed that derived key to DES with the appropriate mode and padding.
Thankfully, the key derivation function is not particularly security critical in implementation, so you can do it yourself safely enough. According to the rfc, PBEwithMD5AndDES is actually the PBKDF1 ( a ker derivation function) used with DES in CBC mode .
PBKDF1 does not look that hard to implement . Looks like you can do it with a for loop and an md5 call.
Note that you may still get some odd results because of the possibility of a different padding scheme being used in Java and Ruby. I assume that the spec one is pkcs 1.5 padding, but at a quick glance, I can't confirm this
PBKDF1 applies a hash function, which shall be MD2 [6], MD5 [19] or SHA-1 [18], to derive keys. The length of the derived key is bounded
by the length of the hash function output, which is 16 octets for MD2 and MD5 and 20 octets for SHA-1. PBKDF1 is compatible with the key
derivation process in PKCS #5 v1.5.PBKDF1 is recommended only for compatibility with existing
applications since the keys it produces may not be large enough for
some applications.PBKDF1 (P, S, c, dkLen)
Options: Hash underlying hash function
Input: P password, an octet string S salt, an eight-octet string c iteration count, a positive integer dkLen intended length in octets of derived key, a positive integer, at most 16 for MD2 or MD5 and 20 for SHA-1
Output: DK derived key, a dkLen-octet string
Steps:
1. If dkLen > 16 for MD2 and MD5, or dkLen > 20 for SHA-1, output "derived key too long" and stop. 2. Apply the underlying hash function Hash for c iterations to the concatenation of the password P and the salt S, then extract the first dkLen octets to produce a derived key DK: T_1 = Hash (P || S) , T_2 = Hash (T_1) , ... T_c = Hash (T_{c-1}) , DK = Tc<0..dkLen-1> 3. Output the derived key DK.
I know it is super old but I had the same problem and just solved it so here it goes to encrypt, where salt is your salt sting, passkey is your password key string and iterations is number of iterations you want to use
def encrypt_account_number cipher = OpenSSL::Cipher::Cipher.new("DES") cipher.encrypt cipher.pkcs5_keyivgen passkey, salt,iterations,digest encrypted_account_number = cipher.update(account_number) encrypted_account_number << cipher.final Base64.encode64(encrypted_account_number ) end def decrypt_account_number cipher = OpenSSL::Cipher::Cipher.new("DES") base_64_code = Base64.decode64(account_number) cipher.decrypt cipher.pkcs5_keyivgen passkey, salt,iterations,digest decrypted_account_number = cipher.update base_64_code decrypted_account_number << cipher.final decrypted_account_number end For what its' worth, I'm posting my python code, which actually works (I have tons of encrypted values which were done using org.jasypt.util.text.BasicTextEncryptor, I needed to decrypt them.)
import base64 import hashlib from Crypto.Cipher import DES """ Note about PBEWithMD5AndDES in java crypto library: Encrypt: Generate a salt (random): 8 bytes <start derived key generation> Append salt to the password MD5 Hash it, and hash the result, hash the result ... 1000 times MD5 always gives us a 16 byte hash Final result: first 8 bytes is the "key" and the next is the "initialization vector" (there is something about the first 8 bytes needing to be of odd paraity, therefore the least significant bit needs to be changed to 1 if required. We don't do it, maybe the python crypto library does it for us) <end derived key generation> Pad the input string with 1-8 bytes (note: not 0-7, so we always have padding) so that the result is a multiple of 8 bytes. Padding byte value is same as number of bytes being padded, eg, \x07 if 7 bytes need to be padded. Use the key and iv to encrypt the input string, using DES with CBC mode. Prepend the encrypted value with the salt (needed for decrypting since it is random) Base64 encode it -> this is your result Decrypt: Base64 decode the input message Extract the salt (first 8 bytes). The rest is the encoded text. Use derived key generation as in Encrypt above to get the key and iv Decrypt the encoded text using key and iv Remove padding -> this is your result (I only have implemented decrypt here since that's all I needed, but encrypt should be straighforward as well) """ def get_derived_key(password, salt, count): key = password + salt for i in range(count): m = hashlib.md5(key) key = m.digest() return (key[:8], key[8:]) def decrypt(msg, password): msg_bytes = base64.b64decode(msg) salt = msg_bytes[:8] enc_text = msg_bytes[8:] (dk, iv) = get_derived_key(password, salt, 1000) crypter = DES.new(dk, DES.MODE_CBC, iv) text = crypter.decrypt(enc_text) # remove the padding at the end, if any return re.sub(r'[\x01-\x08]','',text) I've updated python script from user3392439, with encrypt support. Wish it helpful.
import base64 import hashlib import re import os from Crypto.Cipher import DES """ Note about PBEWithMD5AndDES in java crypto library: Encrypt: Generate a salt (random): 8 bytes <start derived key generation> Append salt to the password MD5 Hash it, and hash the result, hash the result ... 1000 times MD5 always gives us a 16 byte hash Final result: first 8 bytes is the "key" and the next is the "initialization vector" (there is something about the first 8 bytes needing to be of odd paraity, therefore the least significant bit needs to be changed to 1 if required. We don't do it, maybe the python crypto library does it for us) <end derived key generation> Pad the input string with 1-8 bytes (note: not 0-7, so we always have padding) so that the result is a multiple of 8 bytes. Padding byte value is same as number of bytes being padded, eg, \x07 if 7 bytes need to be padded. Use the key and iv to encrypt the input string, using DES with CBC mode. Prepend the encrypted value with the salt (needed for decrypting since it is random) Base64 encode it -> this is your result Decrypt: Base64 decode the input message Extract the salt (first 8 bytes). The rest is the encoded text. Use derived key generation as in Encrypt above to get the key and iv Decrypt the encoded text using key and iv Remove padding -> this is your result (I only have implemented decrypt here since that's all I needed, but encrypt should be straighforward as well) """ def get_derived_key(password, salt, count): key = password + salt for i in range(count): m = hashlib.md5(key) key = m.digest() return (key[:8], key[8:]) def decrypt(msg, password): msg_bytes = base64.b64decode(msg) salt = msg_bytes[:8] enc_text = msg_bytes[8:] (dk, iv) = get_derived_key(password, salt, 1000) crypter = DES.new(dk, DES.MODE_CBC, iv) text = crypter.decrypt(enc_text) # remove the padding at the end, if any return re.sub(r'[\x01-\x08]','',text) def encrypt(msg, password): salt = os.urandom(8) pad_num = 8 - (len(msg) % 8) for i in range(pad_num): msg += chr(pad_num) (dk, iv) = get_derived_key(password, salt, 1000) crypter = DES.new(dk, DES.MODE_CBC, iv) enc_text = crypter.encrypt(msg) return base64.b64encode(salt + enc_text) def main(): msg = "hello, world" passwd = "mypassword" s = encrypt(msg, passwd) print s print decrypt(s, passwd) if __name__ == "__main__": main() For @cooljohny
I do not really recall how this code works, but I'm 99% sure it does. I wrote it by carefully digging through the spec from the Java implementation of pbewithmd5anddes, and testing this python against that. I delivered exactly this code to a client, and it worked fine for them. I changed the constants before pasting it here, but that's all. You should be able to confirm that it produces the same encrypted output as the Java lib, and then replicate it in ruby. Good luck!
import base64 from Crypto.Cipher import DES from passlib.utils.pbkdf2 import pbkdf1 password = 'xxxxxxx' iterations = 22 salt_bytes = [19,15,78,45,34,90,12,11] # convert saltBytes to a string salt_string = ''.join([chr(a) for a in salt_bytes]) # a sample request raw_data = '''{"something":"to","encrypt":"here"}''' # from the standard... padding_value = (8 - (raw_data.__len__() % 8)) padding_data = chr(padding_value) * padding_value padded_data = raw_data + padding_data # 22 iterations, 16 is the # of bytes in an md5 digest pbkres = pbkdf1(password, salt_string, iterations, 16, 'md5') # split the digest into two 8-byte halves # this gives the DES secret key and initializing vector des_key, iv = pbkres[0:8], pbkres[8:16] # encrypt with DES cipher = DES.new(des_key, DES.MODE_CBC, iv) cmsg = cipher.encrypt(padded_data) # and base64 encode base64.b64encode(cmsg)