OpenSSL为AES加密提供了一种流行的（但不安全 - 见下文！）命令行界面：
openssl aes-256-cbc -salt -in filename -out filename.enc
Python以PyCrypto包的形式支持AES，但它只提供工具。如何使用Python / PyCrypto解密使用OpenSSL加密的文件？
此问题过去也涉及使用相同方案的Python加密。我已经删除了那部分以阻止任何人使用它。不要以这种方式加密任何数据，因为它不符合今天的标准。你应该只使用解密，除了后向兼容性之外没有其他原因，即当你别无选择时。想要加密？如果可能的话，使用NaCl / libsodium。
答案 0 :(得分：87)
鉴于Python的普及，起初我很失望，没有完整的答案可以找到这个问题。我花了相当多的时间在这个板上阅读不同的答案，以及其他资源，以使其正确。我想我可能会分享结果以供将来参考，或许可以复习;我绝不是加密专家！但是，下面的代码似乎无缝地工作：
from hashlib import md5
from Crypto.Cipher import AES
from Crypto import Random
def derive_key_and_iv(password, salt, key_length, iv_length):
d = d_i = ''
while len(d) < key_length + iv_length:
d_i = md5(d_i + password + salt).digest()
d += d_i
return d[:key_length], d[key_length:key_length+iv_length]
def decrypt(in_file, out_file, password, key_length=32):
bs = AES.block_size
salt = in_file.read(bs)[len('Salted__'):]
key, iv = derive_key_and_iv(password, salt, key_length, bs)
cipher = AES.new(key, AES.MODE_CBC, iv)
next_chunk = ''
finished = False
while not finished:
chunk, next_chunk = next_chunk, cipher.decrypt(in_file.read(1024 * bs))
if len(next_chunk) == 0:
padding_length = ord(chunk[-1])
chunk = chunk[:-padding_length]
finished = True
out_file.write(chunk)
用法：
with open(in_filename, 'rb') as in_file, open(out_filename, 'wb') as out_file:
decrypt(in_file, out_file, password)
如果您认为有机会对此进行改进或将其扩展为更灵活（例如，使其无盐工作或提供Python 3兼容性），请随时这样做。
此答案过去也涉及使用相同方案的Python加密。我已经删除了那部分以阻止任何人使用它。不要以这种方式加密任何数据，因为它不符合今天的标准。你应该只使用解密，除了后向兼容性之外没有其他原因，即当你别无选择时。想要加密？如果可能的话，使用NaCl / libsodium。
答案 1 :(得分：20)
我正在重新发布您的代码并进行了一些更正（我不想模糊您的版本）。当您的代码有效时，它不会检测填充周围的一些错误。特别是，如果提供的解密密钥不正确，您的填充逻辑可能会做一些奇怪的事情。如果您同意我的更改，您可以更新您的解决方案。
from hashlib import md5
from Crypto.Cipher import AES
from Crypto import Random
def derive_key_and_iv(password, salt, key_length, iv_length):
d = d_i = ''
while len(d) < key_length + iv_length:
d_i = md5(d_i + password + salt).digest()
d += d_i
return d[:key_length], d[key_length:key_length+iv_length]
# This encryption mode is no longer secure by today's standards.
# See note in original question above.
def obsolete_encrypt(in_file, out_file, password, key_length=32):
bs = AES.block_size
salt = Random.new().read(bs - len('Salted__'))
key, iv = derive_key_and_iv(password, salt, key_length, bs)
cipher = AES.new(key, AES.MODE_CBC, iv)
out_file.write('Salted__' + salt)
finished = False
while not finished:
chunk = in_file.read(1024 * bs)
if len(chunk) == 0 or len(chunk) % bs != 0:
padding_length = bs - (len(chunk) % bs)
chunk += padding_length * chr(padding_length)
finished = True
out_file.write(cipher.encrypt(chunk))
def decrypt(in_file, out_file, password, key_length=32):
bs = AES.block_size
salt = in_file.read(bs)[len('Salted__'):]
key, iv = derive_key_and_iv(password, salt, key_length, bs)
cipher = AES.new(key, AES.MODE_CBC, iv)
next_chunk = ''
finished = False
while not finished:
chunk, next_chunk = next_chunk, cipher.decrypt(in_file.read(1024 * bs))
if len(next_chunk) == 0:
padding_length = ord(chunk[-1])
if padding_length < 1 or padding_length > bs:
raise ValueError("bad decrypt pad (%d)" % padding_length)
# all the pad-bytes must be the same
if chunk[-padding_length:] != (padding_length * chr(padding_length)):
# this is similar to the bad decrypt:evp_enc.c from openssl program
raise ValueError("bad decrypt")
chunk = chunk[:-padding_length]
finished = True
out_file.write(chunk)
答案 2 :(得分：12)
下面的代码应该是Python 3与代码中记录的小变化兼容。也想用os.urandom而不是Crypto.Random。 'salted__'被salt_header取代，可以根据需要定制或留空。
from os import urandom
from hashlib import md5
from Crypto.Cipher import AES
def derive_key_and_iv(password, salt, key_length, iv_length):
d = d_i = b'' # changed '' to b''
while len(d) < key_length + iv_length:
# changed password to str.encode(password)
d_i = md5(d_i + str.encode(password) + salt).digest()
d += d_i
return d[:key_length], d[key_length:key_length+iv_length]
def encrypt(in_file, out_file, password, salt_header='', key_length=32):
# added salt_header=''
bs = AES.block_size
# replaced Crypt.Random with os.urandom
salt = urandom(bs - len(salt_header))
key, iv = derive_key_and_iv(password, salt, key_length, bs)
cipher = AES.new(key, AES.MODE_CBC, iv)
# changed 'Salted__' to str.encode(salt_header)
out_file.write(str.encode(salt_header) + salt)
finished = False
while not finished:
chunk = in_file.read(1024 * bs)
if len(chunk) == 0 or len(chunk) % bs != 0:
padding_length = (bs - len(chunk) % bs) or bs
# changed right side to str.encode(...)
chunk += str.encode(
padding_length * chr(padding_length))
finished = True
out_file.write(cipher.encrypt(chunk))
def decrypt(in_file, out_file, password, salt_header='', key_length=32):
# added salt_header=''
bs = AES.block_size
# changed 'Salted__' to salt_header
salt = in_file.read(bs)[len(salt_header):]
key, iv = derive_key_and_iv(password, salt, key_length, bs)
cipher = AES.new(key, AES.MODE_CBC, iv)
next_chunk = ''
finished = False
while not finished:
chunk, next_chunk = next_chunk, cipher.decrypt(
in_file.read(1024 * bs))
if len(next_chunk) == 0:
padding_length = chunk[-1] # removed ord(...) as unnecessary
chunk = chunk[:-padding_length]
finished = True
out_file.write(bytes(x for x in chunk)) # changed chunk to bytes(...)
答案 3 :(得分：2)
此答案基于openssl v1.1.1，与早期版本的openssl相比，它支持更强的AES加密密钥派生过程。
此答案基于以下命令：
echo -n 'Hello World!' | openssl aes-256-cbc -e -a -salt -pbkdf2 -iter 10000
此命令对明文“ Hello World！”进行加密。使用aes-256-cbc。该密钥是使用pbkdf2从密码和一个随机盐中导出的，具有sha256哈希的10,000次迭代。当提示您输入密码时，我输入了密码“ p4 $$ w0rd”。该命令产生的密文输出为：
U2FsdGVkX1/Kf8Yo6JjBh+qELWhirAXr78+bbPQjlxE=
openssl产生的上述密文解密过程如下：
以下是上述过程的python3实现：
import binascii
import base64
import hashlib
from Crypto.Cipher import AES #requires pycrypto
#inputs
openssloutputb64='U2FsdGVkX1/Kf8Yo6JjBh+qELWhirAXr78+bbPQjlxE='
password='p4$$w0rd'
pbkdf2iterations=10000
#convert inputs to bytes
openssloutputbytes=base64.b64decode(openssloutputb64)
passwordbytes=password.encode('utf-8')
#salt is bytes 8 through 15 of openssloutputbytes
salt=openssloutputbytes[8:16]
#derive a 48-byte key using pbkdf2 given the password and salt with 10,000 iterations of sha256 hashing
derivedkey=hashlib.pbkdf2_hmac('sha256', passwordbytes, salt, pbkdf2iterations, 48)
#key is bytes 0-31 of derivedkey, iv is bytes 32-47 of derivedkey
key=derivedkey[0:32]
iv=derivedkey[32:48]
#ciphertext is bytes 16-end of openssloutputbytes
ciphertext=openssloutputbytes[16:]
#decrypt ciphertext using aes-cbc, given key, iv, and ciphertext
decryptor=AES.new(key, AES.MODE_CBC, iv)
plaintext=decryptor.decrypt(ciphertext)
#remove PKCS#7 padding.
#Last byte of plaintext indicates the number of padding bytes appended to end of plaintext. This is the number of bytes to be removed.
plaintext = plaintext[:-plaintext[-1]]
#output results
print('openssloutputb64:', openssloutputb64)
print('password:', password)
print('salt:', salt.hex())
print('key: ', key.hex())
print('iv: ', iv.hex())
print('ciphertext: ', ciphertext.hex())
print('plaintext: ', plaintext.decode('utf-8'))
如预期的那样，上述python3脚本会产生以下内容：
openssloutputb64: U2FsdGVkX1/Kf8Yo6JjBh+qELWhirAXr78+bbPQjlxE=
password: p4$$w0rd
salt: ca7fc628e898c187
key: 444ab886d5721fc87e58f86f3e7734659007bea7fbe790541d9e73c481d9d983
iv: 7f4597a18096715d7f9830f0125be8fd
ciphertext: ea842d6862ac05ebefcf9b6cf4239711
plaintext: Hello World!
注意：可以在web crypto api找到javascript中的等效/兼容实现（使用https://github.com/meixler/web-browser-based-file-encryption-decryption）。
答案 4 :(得分：0)
我知道这有点晚了但here是我在2013年发表的关于如何使用python pycrypto包以openssl兼容方式加密/解密的解决方案。它已在python2.7和python3.x上测试过。可以找到源代码和测试脚本here。
此解决方案与上述优秀解决方案之间的主要区别之一是它区分管道和文件I / O，这可能会在某些应用程序中引起问题。
该博客的主要功能如下所示。
# ================================================================
# get_key_and_iv
# ================================================================
def get_key_and_iv(password, salt, klen=32, ilen=16, msgdgst='md5'):
'''
Derive the key and the IV from the given password and salt.
This is a niftier implementation than my direct transliteration of
the C++ code although I modified to support different digests.
CITATION: http://stackoverflow.com/questions/13907841/implement-openssl-aes-encryption-in-python
@param password The password to use as the seed.
@param salt The salt.
@param klen The key length.
@param ilen The initialization vector length.
@param msgdgst The message digest algorithm to use.
'''
# equivalent to:
# from hashlib import <mdi> as mdf
# from hashlib import md5 as mdf
# from hashlib import sha512 as mdf
mdf = getattr(__import__('hashlib', fromlist=[msgdgst]), msgdgst)
password = password.encode('ascii', 'ignore') # convert to ASCII
try:
maxlen = klen + ilen
keyiv = mdf(password + salt).digest()
tmp = [keyiv]
while len(tmp) < maxlen:
tmp.append( mdf(tmp[-1] + password + salt).digest() )
keyiv += tmp[-1] # append the last byte
key = keyiv[:klen]
iv = keyiv[klen:klen+ilen]
return key, iv
except UnicodeDecodeError:
return None, None
# ================================================================
# encrypt
# ================================================================
def encrypt(password, plaintext, chunkit=True, msgdgst='md5'):
'''
Encrypt the plaintext using the password using an openssl
compatible encryption algorithm. It is the same as creating a file
with plaintext contents and running openssl like this:
$ cat plaintext
<plaintext>
$ openssl enc -e -aes-256-cbc -base64 -salt \\
-pass pass:<password> -n plaintext
@param password The password.
@param plaintext The plaintext to encrypt.
@param chunkit Flag that tells encrypt to split the ciphertext
into 64 character (MIME encoded) lines.
This does not affect the decrypt operation.
@param msgdgst The message digest algorithm.
'''
salt = os.urandom(8)
key, iv = get_key_and_iv(password, salt, msgdgst=msgdgst)
if key is None:
return None
# PKCS#7 padding
padding_len = 16 - (len(plaintext) % 16)
if isinstance(plaintext, str):
padded_plaintext = plaintext + (chr(padding_len) * padding_len)
else: # assume bytes
padded_plaintext = plaintext + (bytearray([padding_len] * padding_len))
# Encrypt
cipher = AES.new(key, AES.MODE_CBC, iv)
ciphertext = cipher.encrypt(padded_plaintext)
# Make openssl compatible.
# I first discovered this when I wrote the C++ Cipher class.
# CITATION: http://projects.joelinoff.com/cipher-1.1/doxydocs/html/
openssl_ciphertext = b'Salted__' + salt + ciphertext
b64 = base64.b64encode(openssl_ciphertext)
if not chunkit:
return b64
LINELEN = 64
chunk = lambda s: b'\n'.join(s[i:min(i+LINELEN, len(s))]
for i in range(0, len(s), LINELEN))
return chunk(b64)
# ================================================================
# decrypt
# ================================================================
def decrypt(password, ciphertext, msgdgst='md5'):
'''
Decrypt the ciphertext using the password using an openssl
compatible decryption algorithm. It is the same as creating a file
with ciphertext contents and running openssl like this:
$ cat ciphertext
# ENCRYPTED
<ciphertext>
$ egrep -v '^#|^$' | \\
openssl enc -d -aes-256-cbc -base64 -salt -pass pass:<password> -in ciphertext
@param password The password.
@param ciphertext The ciphertext to decrypt.
@param msgdgst The message digest algorithm.
@returns the decrypted data.
'''
# unfilter -- ignore blank lines and comments
if isinstance(ciphertext, str):
filtered = ''
nl = '\n'
re1 = r'^\s*$'
re2 = r'^\s*#'
else:
filtered = b''
nl = b'\n'
re1 = b'^\\s*$'
re2 = b'^\\s*#'
for line in ciphertext.split(nl):
line = line.strip()
if re.search(re1,line) or re.search(re2, line):
continue
filtered += line + nl
# Base64 decode
raw = base64.b64decode(filtered)
assert(raw[:8] == b'Salted__' )
salt = raw[8:16] # get the salt
# Now create the key and iv.
key, iv = get_key_and_iv(password, salt, msgdgst=msgdgst)
if key is None:
return None
# The original ciphertext
ciphertext = raw[16:]
# Decrypt
cipher = AES.new(key, AES.MODE_CBC, iv)
padded_plaintext = cipher.decrypt(ciphertext)
if isinstance(padded_plaintext, str):
padding_len = ord(padded_plaintext[-1])
else:
padding_len = padded_plaintext[-1]
plaintext = padded_plaintext[:-padding_len]
return plaintext
答案 5 :(得分：-2)
但如果你想要做的只是加密和解密文件，这是合适的。
我从here复制的自我答案。我认为这可能是一个更简单，更安全的选择。虽然我对一些关于它有多安全的专家意见感兴趣。
我使用Python 3.6和SimpleCrypt来加密文件然后上传它。
我认为这是我用来加密文件的代码：
from simplecrypt import encrypt, decrypt
f = open('file.csv','r').read()
ciphertext = encrypt('USERPASSWORD',f.encode('utf8')) # I am not certain of whether I used the .encode('utf8')
e = open('file.enc','wb') # file.enc doesn't need to exist, python will create it
e.write(ciphertext)
e.close
这是我在运行时解密的代码，我运行getpass("password: ")作为参数，所以我不必在内存中存储password变量
from simplecrypt import encrypt, decrypt
from getpass import getpass
# opens the file
f = open('file.enc','rb').read()
print('Please enter the password and press the enter key \n Decryption may take some time')
# Decrypts the data, requires a user-input password
plaintext = decrypt(getpass("password: "), f).decode('utf8')
print('Data have been Decrypted')
注意，在python 2.7中UTF-8编码行为是不同的，因此代码会略有不同。