如何在Python中解密OpenSSL aes加密文件?

OpenSSL provides a popular (but insecure – see below!) command line interface for AES encryption:

OpenSSL为AES加密提供了一个流行(但不安全)的命令行接口:

openssl aes-256-cbc -salt -in filename -out filename.enc

Python has support for AES in the shape of the PyCrypto package, but it only provides the tools. How to use Python/PyCrypto to decrypt files that have been encrypted using OpenSSL?

Python以PyCrypto包的形式支持AES，但它只提供工具。如何使用Python/PyCrypto来解密使用OpenSSL加密的文件?

Notice

This question used to also concern encryption in Python using the same scheme. I have since removed that part to discourage anyone from using it. Do NOT encrypt any more data in this way, because it is NOT secure by today's standards. You should ONLY use decryption, for no other reasons than BACKWARD COMPATIBILITY, i.e. when you have no other choice. Want to encrypt? Use NaCl/libsodium if you possibly can.

这个问题也用于使用相同方案的Python中的加密。从那以后，我删除了这个部分，以阻止任何人使用它。不要用这种方式加密更多的数据，因为按照今天的标准它是不安全的。您应该只使用解密，而不是出于向后兼容性，也就是说，当您别无选择时。要加密?如果可能的话，使用NaCl/ lib钠。

5 个解决方案

#1

Given the popularity of Python, at first I was disappointed that there was no complete answer to this question to be found. It took me a fair amount of reading different answers on this board, as well as other resources, to get it right. I thought I might share the result for future reference and perhaps review; I'm by no means a cryptography expert! However, the code below appears to work seamlessly:

考虑到Python的流行，一开始我对没有找到这个问题的完整答案感到失望。我花了相当多的时间阅读这个板上不同的答案，以及其他资源，才使它正确。我想我可以分享一下结果，以供以后参考，也许可以复习一下;我绝不是密码学专家!然而，下面的代码似乎可以无缝地工作:

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)

Usage:

用法:

with open(in_filename, 'rb') as in_file, open(out_filename, 'wb') as out_file:
    decrypt(in_file, out_file, password)

If you see a chance to improve on this or extend it to be more flexible (e.g. make it work without salt, or provide Python 3 compatibility), please feel free to do so.

如果您看到有机会改进它或将它扩展为更灵活的(例如，使它在没有盐的情况下工作，或提供Python 3兼容性)，请放心这样做。

Notice

This answer used to also concern encryption in Python using the same scheme. I have since removed that part to discourage anyone from using it. Do NOT encrypt any more data in this way, because it is NOT secure by today's standards. You should ONLY use decryption, for no other reasons than BACKWARD COMPATIBILITY, i.e. when you have no other choice. Want to encrypt? Use NaCl/libsodium if you possibly can.

这个答案在Python中也使用了相同的方案。从那以后，我删除了这个部分，以阻止任何人使用它。不要用这种方式加密更多的数据，因为按照今天的标准它是不安全的。您应该只使用解密，而不是出于向后兼容性，也就是说，当您别无选择时。要加密?如果可能的话，使用NaCl/ lib钠。

#2

I am re-posting your code with a couple of corrections (I didn't want to obscure your version). While your code works, it does not detect some errors around padding. In particular, if the decryption key provided is incorrect, your padding logic may do something odd. If you agree with my change, you may update your solution.

我正在重新发布您的代码和一些更正(我不想让您的版本模糊)。当代码工作时，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。如果你同意我的改变，你可以更新你的解决方案。

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)

#3

The code below should be Python 3 compatible with the small changes documented in the code. Also wanted to use os.urandom instead of Crypto.Random. 'Salted__' is replaced with salt_header that can be tailored or left empty if needed.

下面的代码应该与代码中记录的小更改兼容。也想使用操作系统。urandom代替Crypto.Random。“Salted__”被替换为salt_header，如果需要，可以对其进行裁剪或保持为空。

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(...)

#4

I know this is a bit late but here is a solution that I blogged in 2013 about how to use the python pycrypto package to encrypt/decrypt in an openssl compatible way. It has been tested on python2.7 and python3.x. The source code and a test script can be found here.

我知道这有点晚了，但这是我在2013年写的一个解决方案，关于如何使用python pycrypto包以openssl兼容的方式加密/解密。它已经在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。

One of the key differences between this solution and the excellent solutions presented above is that it differentiates between pipe and file I/O which can cause problems in some applications.

该解决方案与上面介绍的优秀解决方案之间的一个关键区别是，它区分管道和文件I/O，这可能会在某些应用程序中导致问题。

The key functions from that blog are shown below.

这个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://*.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

#5

Note: this method is not OpenSSL compatible

But it is suitable if all you want to do is encrypt and decrypt files.

但是，如果您想要做的只是加密和解密文件，那么它是合适的。

A self-answer I copied from here. I think this is, perhaps, a simpler and more secure option. Although I would be interested in some expert opinion on how secure it is.

我从这里抄的一个自述。我认为这可能是一个更简单、更安全的选择。尽管我对一些专家的意见很感兴趣关于它有多安全。

I used Python 3.6 and SimpleCrypt to encrypt the file and then uploaded it.

我使用Python 3.6和SimpleCrypt对文件进行加密，然后上传。

I think this is the code I used to encrypt the file:

我想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

This is the code I use to decrypt at runtime, I run getpass("password: ") as an argument so I don't have to store a password variable in memory

这是我在运行时用来解密的代码，我运行getpass(“password:”)作为参数，这样我就不必在内存中存储密码变量。

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')

Note, the UTF-8 encoding behaviour is different in python 2.7 so the code will be slightly different.

注意，在python 2.7中，UTF-8编码行为不同，因此代码略有不同。