Develop Reference

Error when decrypt message from blowfish/base64

I'm trying to decrypt some message encrypted by FiSH plugin from mirc. The plugin use a blowfish encryption as mode ECB based on a key but first it encrypted the messages as code64 and then with blowfish ECB. The problem is when I try decrypt the message with blowfish after decode it from base64 get the same error always. "Data must be aligned to block boundary in ECB mode"
The des encrypted message is: "Probando un mensaje cifrado"
from Crypto.Cipher import Blowfish
from Crypto.Util.Padding import pad, unpad
from os import urandom
import base64
key = b"passw0rd"
text =b"+OK Tnkrh0sIoWb1oS1FT.RQop/.JPXNc.lclFO/gueZ4/ZwN1H0"
decode64 = base64.b64decode(text)
decrypt = Blowfish.new(key,Blowfish.MODE_ECB)
msg = decrypt.decrypt(decode64)
print(msg)
site-packages\Crypto\Cipher\_mode_ecb.py", line 163, in decrypt
raise ValueError("Data must be aligned to block boundary in ECB mode")
ValueError: Data must be aligned to block boundary in ECB mode
WITH NON-STANDARD TABLE:
custom_a = "./0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
standard_a = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
DECODE_TRANS = str.maketrans(custom_a,standard_a)
key = b"passw0rd"
text ="Tnkrh0sIoWb1oS1FT.RQop/.JPXNc.lclFO/gueZ4/ZwN1H0"
decode64 = base64.b64decode(text.translate(DECODE_TRANS))
decrypt = Blowfish.new(key,Blowfish.MODE_ECB)
msg = decrypt.decrypt(decode64)
print(msg)

There are two obvious problems I see here. There may be more, but these will need to be dealt with first.
You need to remove the prefix "+OK " from the message before decrypting it. It's not part of the encrypted data.
The data transmitted by FiSH is not standard Base64. It uses a nonstandard encoding table which you'll have to account for.

How to decrypt RSA encrypted file (via PHP and OpenSSL) with pyopenssl?

Simple n00b question: I am trying to replicate the behavior of the openssl_private_decrypt function in PHP to decrypt a file that a vendor is sending me which was encrypted via the openssl_public_encrypt function. I am using python 3.4 and thus the only library I can see available is pyopenssl, but it's sufficiently low-level that I cannot easily find out how to do what I want to do. It's probably very simple, but does anyone have an exemple of what I want to do?

With the Cryptography module, which you can install with:
$ pip install cryptography
Assuming you have the private key stored in a file called "path/to/key.pem", first you load the private key:
from cryptography.hazmat.primitives import serialization
with open("path/to/key.pem", "rb") as key_file:
private_key = serialization.load_pem_private_key(
key_file.read(),
password=None,
backend=default_backend()
)
And then you decrypt with:
plaintext = private_key.decrypt(
ciphertext,
padding.OAEP(
mgf=padding.MGF1(algorithm=hashes.SHA1()),
algorithm=hashes.SHA1(),
label=None
)
)

Thanks to #mnistic it got to work, with a couple of modifications though. Here is the final working code (you have to keep in mind the defaults of openssl_private_decrypt):
from cryptography.hazmat.primitives import serialization
from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives.asymmetric import padding
# It's critical that the file be opened in mode "rb"!
with open("private.key", 'rb') as key_file:
private_key = serialization.load_pem_private_key(key_file.read(), password=None, backend=default_backend())
with open('encrypted_file', 'rb') as encrypted_file:
ciphertext = encrypted_file.read()
plaintext = private_key.decrypt(ciphertext, padding.PKCS1v15())
Please note that ciphertext needs to be shorter than the maximum chunk size of the key (which is the number of bits in the key divided by 8 for RSA). Hope that helps future Googlers!

How to check Python's AES decrypt error?

I'm using python to encrypt and decrypt files. When file encrypted, then try to decrypt like this:
from Crypto.Cipher import AES
from Crypto import Random
def decrypt(in_file, out_file, pwd, key_len=32):
bs = AES.block_size
salt = in_file.read(bs)[len('Salted__'):]
key, iv = derive_keyiv(pwd, salt, key_len, bs)
cipher = AES.new(key, AES.MODE_CBC, iv)
next_chunk = ''
finished = False
try:
while not finished:
chunk, next_chunk = next_chunk, cipher.decrypt(in_file.read(1024*bs))
if len(next_chunk) == 0:
padding_len = ord(chunk[-1])
chunk = chunk[:-padding_len]
finished = True
out_file.write(chunk)
return True, None
except Exception as e:
return False, e
But if the password input error, the decrypt still decrypt in_file and write to out_file and no exception throw.
How to check the password error during decrypt?

AES by itself cannot check if the key is "correct". It is simply a pure function that transforms some bytes to some other bytes.
To achieve what you want, you need to implement it yourself. One way to do is to add a fixed header (like 16 bytes of zero) to the plaintext before encryption. Upon decryption, check and discard the said header if it matches or raise an error if the header mismatches.
A side note: you are doing encryption without any authentication, which is probably insecure.
Edit
First of all, you should add authentication. Encryption without authentication easily leads to many security flaws, many not obvious to the untrained. Especially since you are using AES in CBC mode, you may open yourself to padding oracle attacks without authentication.
When you do authenticated encryption the right way (encrypt-then-mac), you will get an authentication error if the user input the wrong password. If you want to further distinguish a wrong password from tampered data, you have to devise your own method, like prepending a ciphertext of magic number.

Django encryption key integrity

I am implementing a Django website in which uploaded files are encrypted with a user provided key before they are saved on the server (/media). When users wish to view them, they are prompted for the key, the encrypted file is decrypted, and then displayed for them. Here's my encrypt/decrypt file code:
from Crypto import Random
from Crypto.Cipher import AES
from Crypto.Hash import SHA256
def encryption_pad(string):
pad = b"\0" * (AES.block_size - len(string) % AES.block_size)
padded_string = string + pad
return padded_string
def encrypt_file(key, file):
with open(file, 'rb') as out:
byte_output = out.read()
hash = SHA256.new()
hash.update(key)
byte_output = encryption_pad(byte_output)
initialization_vector = Random.new().read(AES.block_size)
cipher = AES.new(hash.digest(), AES.MODE_CBC, initialization_vector)
encrypted_output = initialization_vector + cipher.encrypt(byte_output)
with open(file + ".enc", 'wb') as out:
out.write(encrypted_output)
def decrypt_file(file, key):
with open(file, 'rb') as input:
ciphertext = input.read()
hash = SHA256.new()
hash.update(key)
initialization_vector = ciphertext[:AES.block_size]
cipher = AES.new(hash.digest(), AES.MODE_CBC, initialization_vector)
decrypted_output = cipher.decrypt(ciphertext[AES.block_size:])
decrypted_output = decrypted_output.rstrip(b"\0")
with open(file[:-4], 'wb') as output:
output.write(decrypted_output)
I am relatively new to security, so my question is: For this setup the keys must exist in the server's memory for some duration of time, so what is the proper way for my views.py function to pass them to this module and then properly dispose of them after?

There are some existing questions on how to securely handle (or not) in-memory objects in Python: see here and here.
If security is that important, though, you might want to consider an even more secure option: doing the encryption and decryption on the client, in Javascript. That way the key never gets sent over the wire, and never exists on the server. That's how LastPass works, for example.

AES: how to detect that a bad password has been entered?

A text s has been encrypted with:
s2 = iv + Crypto.Cipher.AES.new(Crypto.Hash.SHA256.new(pwd).digest(),
Crypto.Cipher.AES.MODE_CFB,
iv).encrypt(s.encode())
Then, later, a user inputs the password pwd2 and we decrypt it with:
iv, cipher = s2[:Crypto.Cipher.AES.block_size], s2[Crypto.Cipher.AES.block_size:]
s3 = Crypto.Cipher.AES.new(Crypto.Hash.SHA256.new(pwd2).digest(),
Crypto.Cipher.AES.MODE_CFB,
iv).decrypt(cipher)
Problem: the last line works even if the entered password pw2 is wrong. Of course the decrypted text will be random chars, but no error is triggered.
Question: how to make Crypto.Cipher.AES.new(...).decrypt(cipher) fail if the password pw2 is incorrect? Or at least how to detect a wrong password?
Here is a linked question: Making AES decryption fail if invalid password
and here a discussion about the cryptographic part (less programming) of the question: AES, is this method to say “The password you entered is wrong” secure?
.

AES provides confidentiality but not integrity out of the box - to get integrity too, you have a few options. The easiest and arguably least prone to "shooting yourself in the foot" is to just use AES-GCM - see this Python example or this one.
You could also use an HMAC, but this generally requires managing two distinct keys and has a few more moving parts. I would recommend the first option if it is available to you.
A side note, SHA-256 isn't a very good KDF to use when converting a user created password to an encryption key. Popular password hashing algorithms are better at this - have a look at Argon2, bcrypt or PBKDF2.
Edit: The reason SHA-256 is a bad KDF is the same reason it makes a bad password hash function - it's just too fast. A user created password of, say, 128 bits will usually contain far less entropy than a random sequence of 128 bits - people like to pick words, meaningful sequences etc. Hashing this once with SHA-256 doesn't really alleviate this issue. But hashing it with a construct like Argon2 that is designed to be slow makes a brute-force attack far less viable.

The best way is to use authenticated encryption, and a modern memory-hard entropy-stretching key derivation function such a scrypt to turn the password into a key. The cipher's nounce can be used as salt for the key derivation. With PyCryptodome that could be:
from Crypto.Random import get_random_bytes
from Crypto.Cipher import AES
from Crypto.Protocol.KDF import scrypt
# initialize an AES-128-GCM cipher from password (derived using scrypt) and nonce
def cipherAES(pwd, nonce):
# note: the p parameter should allow use of several processors, but did not for me
# note: changing 16 to 24 or 32 should select AES-192 or AES-256 (not tested)
return AES.new(scrypt(pwd, nonce, 16, N=2**21, r=8, p=1), AES.MODE_GCM, nonce=nonce)
# encryption
nonce = get_random_bytes(16)
print("deriving key from password and nonce, then encrypting..")
ciphertext, tag = cipherAES(b'pwdHklot2',nonce).encrypt_and_digest(b'bonjour')
print("done")
# decryption of nonce, ciphertext, tag
print("deriving key from password and nonce, then decrypting..")
try:
plaintext = cipherAES(b'pwdHklot2', nonce).decrypt_and_verify(ciphertext, tag)
print("The message was: " + plaintext.decode())
except ValueError:
print("Wrong password or altered nonce, ciphertext, tag")
print("done")
Note: Code is here to illustrate the principle. In particular, the scrypt parameters should not be fixed, but rather be included in a header before nonce, ciphertext, and tag; and that must be somewhat grouped for sending, and parsed for decryption.
Caveat: nothing in this post should be construed as an endorsement of PyCryptodome's security.
Addition (per request):
We need scrypt or some other form of entropy stretching only because we use a password. We could use a random 128-bit key directly.
PBKDF2-HMAC-SHAn with 100000 iterations (as in the OP's second code fragment there) is only barely passable to resist Hashcat with a few GPUs. It would would be almost negligible compared to other hurdles for an ASIC-assisted attack: a state of the art Bitcoin mining ASIC does more than 2*1010 SHA-256 per Joule, 1 kWh of electricity costing less than $0.15 is 36*105 J. Crunching these numbers, testing the (62(8+1)-1)/(62-1) = 221919451578091 passwords of up to 8 characters restricted to letters and digits cost less than $47 for energy dedicated to the hashing part.
scrypt is much more secure for equal time spent by legitimate users because it requires a lot of memory and accesses thereof, slowing down the attacker, and most importantly making the investment cost for massively parallel attack skyrocket.

Doesn't use the Crypto package, but this should suit your needs:
import base64
import os
from cryptography.fernet import Fernet
from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives.kdf.scrypt import Scrypt
def derive_password(password: bytes, salt: bytes):
"""
Adjust the N parameter depending on how long you want the derivation to take.
The scrypt paper suggests a minimum value of n=2**14 for interactive logins (t < 100ms),
or n=2**20 for more sensitive files (t < 5s).
"""
kdf = Scrypt(salt=salt, length=32, n=2**16, r=8, p=1, backend=default_backend())
key = kdf.derive(password)
return base64.urlsafe_b64encode(key)
salt = os.urandom(16)
password = b'legorooj'
bad_password = b'legorooj2'
# Derive the password
key = derive_password(password, salt)
key2 = derive_password(bad_password, salt) # Shouldn't re-use salt but this is only for example purposes
# Create the Fernet Object
f = Fernet(key)
msg = b'This is a test message'
ciphertext = f.encrypt(msg)
print(msg, flush=True) # Flushing pushes it strait to stdout, so the error that will come
print(ciphertext, flush=True)
# Fernet can only be used once, so we need to reinitialize
f = Fernet(key)
plaintext = f.decrypt(ciphertext)
print(plaintext, flush=True)
# Bad Key
f = Fernet(key2)
f.decrypt(ciphertext)
"""
This will raise InvalidToken and InvalidSignature, which means it wasn't decrypted properly.
"""
See my comment for links to the documentation.

For future reference, here is a working solution following the AES GCM mode (recommended by #LukeJoshuaPark in his answer):
from Crypto.Cipher import AES
from Crypto.Random import get_random_bytes
# Encryption
data = b"secret"
key = get_random_bytes(16)
cipher = AES.new(key, AES.MODE_GCM)
ciphertext, tag = cipher.encrypt_and_digest(data)
nonce = cipher.nonce
# Decryption
key2 = get_random_bytes(16) # wrong key
#key2 = key # correct key
try:
cipher = AES.new(key2, AES.MODE_GCM, nonce=nonce)
plaintext = cipher.decrypt_and_verify(ciphertext, tag)
print("The message was: " + plaintext.decode())
except ValueError:
print("Wrong key")
It does fail with an exception when the password is wrong indeed, as desired.
The following code uses a real password derivation function:
import Crypto.Random, Crypto.Protocol.KDF, Crypto.Cipher.AES
def cipherAES(pwd, nonce):
return Crypto.Cipher.AES.new(Crypto.Protocol.KDF.PBKDF2(pwd, nonce, count=100000), Crypto.Cipher.AES.MODE_GCM, nonce=nonce)
# encryption
nonce = Crypto.Random.new().read(16)
cipher = cipherAES(b'pwd1', nonce)
ciphertext, tag = cipher.encrypt_and_digest(b'bonjour')
# decryption
try:
cipher = cipherAES(b'pwd1', nonce=nonce)
plaintext = cipher.decrypt_and_verify(ciphertext, tag)
print("The message was: " + plaintext.decode())
except ValueError:
print("Wrong password")
#fgrieu's answer is probably better because it uses scrypt as KDF.