如果它不是立即显而易见的，那么首先我要说的是我不是一个加密人。
我的任务是在Python 2.7中复制Java的PBEWithMD5AndDES（使用DES加密的MD5摘要）的行为。
我可以访问Python的加密工具包PyCrypto。
这是我试图复制其行为的Java代码：
import java.security.spec.KeySpec;
import javax.crypto.spec.PBEKeySpec;
import javax.crypto.SecretKey;
import javax.crypto.SecretKeyFactory;
import java.security.spec.AlgorithmParameterSpec;
import javax.crypto.spec.PBEParameterSpec;
import javax.crypto.Cipher;
import javax.xml.bind.DatatypeConverter;
public class EncryptInJava
{
public static void main(String[] args)
{
String encryptionPassword = "q1w2e3r4t5y6";
byte[] salt = { -128, 64, -32, 16, -8, 4, -2, 1 };
int iterations = 50;
try
{
KeySpec keySpec = new PBEKeySpec(encryptionPassword.toCharArray(), salt, iterations);
SecretKey key = SecretKeyFactory.getInstance("PBEWithMD5AndDES").generateSecret(keySpec);
AlgorithmParameterSpec paramSpec = new PBEParameterSpec(salt, iterations);
Cipher encoder = Cipher.getInstance(key.getAlgorithm());
encoder.init(Cipher.ENCRYPT_MODE, key, paramSpec);
String str_to_encrypt = "MyP455w0rd";
byte[] enc = encoder.doFinal(str_to_encrypt.getBytes("UTF8"));
System.out.println("encrypted = " + DatatypeConverter.printBase64Binary(enc));
}
catch (Exception e)
{
e.printStackTrace();
}
}
}
对于给定的值，它输出以下内容：
encrypted = Icy6sAP7adLgRoXNYe9N8A==
这是我的火腿尝试将上面的内容移植到Python，encrypt_in_python.py：
from Crypto.Hash import MD5
from Crypto.Cipher import DES
_password = 'q1w2e3r4t5y6'
_salt = '\x80\x40\xe0\x10\xf8\x04\xfe\x01'
_iterations = 50
plaintext_to_encrypt = 'MyP455w0rd'
if "__main__" == __name__:
"""Mimic Java's PBEWithMD5AndDES algorithm to produce a DES key"""
hasher = MD5.new()
hasher.update(_password)
hasher.update(_salt)
result = hasher.digest()
for i in range(1, _iterations):
hasher = MD5.new()
hasher.update(result)
result = hasher.digest()
key = result[:8]
encoder = DES.new(key)
encrypted = encoder.encrypt(plaintext_to_encrypt + ' ' * (8 - (len(plaintext_to_encrypt) % 8)))
print encrypted.encode('base64')
它输出一个完全不同的字符串。
是否可以将Java实现移植到使用标准Python库的Python实现？
显然，Python实现要求我加密的明文是八个字符的倍数，我甚至不确定如何填充我的明文输入以满足该条件。
谢谢你的帮助。
感谢GregS的评论，我能够对这个转换进行排序！
为了将来参考，这个Python代码模仿上面Java代码的行为：
from Crypto.Hash import MD5
from Crypto.Cipher import DES
_password = 'q1w2e3r4t5y6'
_salt = '\x80\x40\xe0\x10\xf8\x04\xfe\x01'
_iterations = 50
plaintext_to_encrypt = 'MyP455w0rd'
# Pad plaintext per RFC 2898 Section 6.1
padding = 8 - len(plaintext_to_encrypt) % 8
plaintext_to_encrypt += chr(padding) * padding
if "__main__" == __name__:
"""Mimic Java's PBEWithMD5AndDES algorithm to produce a DES key"""
hasher = MD5.new()
hasher.update(_password)
hasher.update(_salt)
result = hasher.digest()
for i in range(1, _iterations):
hasher = MD5.new()
hasher.update(result)
result = hasher.digest()
encoder = DES.new(result[:8], DES.MODE_CBC, result[8:16])
encrypted = encoder.encrypt(plaintext_to_encrypt)
print encrypted.encode('base64')
这个Python代码在Python 2.7中输出以下内容：
Icy6sAP7adLgRoXNYe9N8A==
再次感谢GregS指出我正确的方向！
对于Python 3.6，我使用以下代码进行了测试，它可以从上面稍作改动：
from Crypto.Hash import MD5
from Crypto.Cipher import DES
import base64
import re
_password = b'q1w2e3r4t5y6'
_salt = b'\x80\x40\xe0\x10\xf8\x04\xfe\x01'
_iterations = 50
plaintext_to_encrypt = 'MyP455w0rd'
# Pad plaintext per RFC 2898 Section 6.1
padding = 8 - len(plaintext_to_encrypt) % 8
plaintext_to_encrypt += chr(padding) * padding
if "__main__" == __name__:
"""Mimic Java's PBEWithMD5AndDES algorithm to produce a DES key"""
hasher = MD5.new()
hasher.update(_password)
hasher.update(_salt)
result = hasher.digest()
for i in range(1, _iterations):
hasher = MD5.new()
hasher.update(result)
result = hasher.digest()
encoder = DES.new(result[:8], DES.MODE_CBC, result[8:16])
encrypted = encoder.encrypt(plaintext_to_encrypt)
print (str(base64.b64encode(encrypted),'utf-8'))
decoder = DES.new(result[:8], DES.MODE_CBC, result[8:])
d = str(decoder.decrypt(encrypted),'utf-8')
print (re.sub(r'[\x01-\x08]','',d))
输出：
Icy6sAP7adLgRoXNYe9N8A ==
MyP455w0rd
我从here找到了一个
import base64
import hashlib
import re
import os
from Crypto.Cipher import DES
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 = '\xA9\x9B\xC8\x32\x56\x35\xE3\x03'
enc_text = msg_bytes
(dk, iv) = get_derived_key(password, salt, 2)
crypter = DES.new(dk, DES.MODE_CBC, iv)
text = crypter.decrypt(enc_text)
return re.sub(r'[\x01-\x08]','',text)
def encrypt(msg, password):
salt = '\xA9\x9B\xC8\x32\x56\x35\xE3\x03'
pad_num = 8 - (len(msg) % 8)
for i in range(pad_num):
msg += chr(pad_num)
(dk, iv) = get_derived_key(password, salt, 2)
crypter = DES.new(dk, DES.MODE_CBC, iv)
enc_text = crypter.encrypt(msg)
return base64.b64encode(enc_text)
def main():
msg = "hello"
passwd = "xxxxxxxxxxxxxx"
encrypted_msg = encrypt(msg, passwd)
print encrypted_msg
plain_msg = decrypt(encrypted_msg, passwd)
print plain_msg
if __name__ == "__main__":
main()
如果您在Python 3中使用较新的Cryptodome库，则还需要将plaintext_to_encrypt编码为'latin-1'，如下所示。
from Cryptodome.Hash import MD5
from Cryptodome.Cipher import DES
import base64
import re
_password = b'q1w2e3r4t5y6'
_salt = b'\x80\x40\xe0\x10\xf8\x04\xfe\x01'
_iterations = 50
plaintext_to_encrypt = 'MyP455w0rd'
# Pad plaintext per RFC 2898 Section 6.1
padding = 8 - len(plaintext_to_encrypt) % 8
plaintext_to_encrypt += chr(padding) * padding
if "__main__" == __name__:
"""Mimic Java's PBEWithMD5AndDES algorithm to produce a DES key"""
hasher = MD5.new()
hasher.update(_password)
hasher.update(_salt)
result = hasher.digest()
for i in range(1, _iterations):
hasher = MD5.new()
hasher.update(result)
result = hasher.digest()
encoder = DES.new(result[:8], DES.MODE_CBC, result[8:16])
encrypted = encoder.encrypt(plaintext_to_encrypt.encode('latin-1')) #encoded plaintext
print (str(base64.b64encode(encrypted),'utf-8'))
decoder = DES.new(result[:8], DES.MODE_CBC, result[8:])
d = str(decoder.decrypt(encrypted),'utf-8')
print (re.sub(r'[\x01-\x08]','',d))