OpenSSL 为 AES 加密提供了一个流行的（但不安全的 - 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？
这个问题过去也涉及使用相同方案的 Python 中的加密。此后，我删除了该部分以阻止任何人使用它。不要以这种方式加密任何更多数据，因为按照今天的标准它并不安全。您应该只使用解密，除了向后兼容性之外没有其他原因，即当您别无选择时。想要加密？如果可能，请使用 NaCl/libsodium。
鉴于 Python 的流行，起初我很失望，因为找不到这个问题的完整答案。我花了相当多的时间阅读这个板上的不同答案以及其他资源，才能把它弄好。我想我可能会分享结果以供将来参考和审查；我绝不是密码学专家！但是，下面的代码似乎可以无缝运行：
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)
如果您发现有机会对此进行改进或将其扩展为更灵活（例如，使其无需盐即可工作，或提供 Python 3 兼容性），请随时这样做。
这个答案过去也涉及使用相同方案的 Python 中的加密。此后，我删除了该部分以阻止任何人使用它。不要以这种方式加密任何更多数据，因为按照今天的标准它并不安全。您应该只使用解密，除了向后兼容性之外没有其他原因，即当您别无选择时。想要加密？如果可能，请使用 NaCl/libsodium。
我正在重新发布您的代码并进行一些更正（我不想掩盖您的版本）。虽然您的代码有效，但它不会检测到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)
下面的代码应该与代码中记录的小更改兼容 Python 3。还想使用 os.urandom 而不是 Crypto.Random。'Salted__' 被替换为 salt_header，可以根据需要对其进行定制或留空。
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(...)
这个答案基于 openssl v1.1.1，它支持比以前版本的 openssl 更强大的 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
此命令加密明文“Hello World！” 使用 aes-256-cbc。密钥是使用 pbkdf2 从密码和随机盐中派生的，具有 10,000 次 sha256 散列迭代。当提示输入密码时，我输入了密码“p4$$w0rd”。该命令产生的密文输出为：
U2FsdGVkX1/Kf8Yo6JjBh+qELWhirAXr78+bbPQjlxE=
上面由openssl产生的密文解密过程如下：
下面是上述过程的python3实现：
import binascii
import base64
import hashlib
from Crypto.Cipher import AES #requires pycrypto
#inputs
openssloutputb64='U2FsdGVkX1/Kf8Yo6JjBh+qELWhirAXr78+bbPQjlxE='
password='p4$$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'))
正如预期的那样，上面的 python3 脚本产生以下内容：
openssloutputb64: U2FsdGVkX1/Kf8Yo6JjBh+qELWhirAXr78+bbPQjlxE=
password: p4$$w0rd
salt: ca7fc628e898c187
key: 444ab886d5721fc87e58f86f3e7734659007bea7fbe790541d9e73c481d9d983
iv: 7f4597a18096715d7f9830f0125be8fd
ciphertext: ea842d6862ac05ebefcf9b6cf4239711
plaintext: Hello World!
注意：可以在https://github.com/meixler/web-browser-based-file-encryption-decryption找到javascript 中的等效/兼容实现（使用web crypto api ） 。
我知道这有点晚了，但这是我在 2013 年发表的关于如何使用 python pycrypto 包以兼容 openssl 的方式加密/解密的解决方案。已经在python2.7和python3.x上测试过了。源代码和测试脚本可以在这里找到。
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了管道和文件 I/O，这可能会在某些应用程序中导致问题。
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
尝试了上面的所有内容以及其他线程的更多内容，这对我有用，相当于 openssl 中的内容：
不是最好的 encrpython 但那些是要求
解密：openssl enc -d -aes256 -md md5 -in {->path_in} -out {->path_out} -pass pass:{->pass}
加密：openssl enc -e -aes256 -md md5 -in {->path_in} -out {->path_out} -pass pass:{->pass}
Python：
from os import urandom
from hashlib import md5
from Crypto.Cipher import AES
import typer
def filecrypto(in_file, out_file, password, decrypt: bool = True):
salt_header = 'Salted__'
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_f(in_file, out_file, password, salt_header=salt_header, key_length=32):
bs = AES.block_size
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)
with open(out_file, 'wb') as f_out:
# write the first line or the salted header
f_out.write(str.encode(salt_header) + salt)
with open(in_file, 'rb') as f_in:
f_out.write(cipher.encrypt(f_in.read()))
def decrypt_f(in_file, out_file, password, salt_header=salt_header, key_length=32):
bs = AES.block_size
with open(in_file, 'rb') as f_in:
# retrieve the salted header
salt = f_in.read(bs)[len(salt_header):]
key, iv = derive_key_and_iv(password, salt, key_length, bs)
cipher = AES.new(key, AES.MODE_CBC, iv)
with open(out_file, 'wb') as f_out:
f_out.write(cipher.decrypt(f_in.read()))
return decrypt_f(in_file, out_file, password) if decrypt else encrypt_f(in_file, out_file, password)
if __name__ == "__filecrypto__":
typer.run(filecrypto)
但是，如果您只想加密和解密文件，则它是合适的。
我从这里复制的自我回答。我认为这也许是一个更简单、更安全的选择。尽管我会对有关它的安全性的一些专家意见感兴趣。
我使用 Python 3.6 和SimpleCrypt加密文件，然后上传。
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
这是我在运行时用来解密的代码，我getpass("password: ")作为参数运行，所以我不必password在内存中存储变量
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')
请注意，UTF-8 编码行为在 python 2.7 中有所不同，因此代码会略有不同。