Python Blowfish 加密

Question

由于我对 Java 的不完全了解，我正在努力将此加密代码转换为 Python 代码。两者应该有完全相同的结果。帮助将不胜感激。

Java 函数

import javax.crypto.Cipher;
import javax.crypto.spec.SecretKeySpec;
import java.security.Key;


class Main
{
    public static void main (String[] args) throws java.lang.Exception
    {
        String s = "testings";
        Cipher cipher = Cipher.getInstance("Blowfish/ECB/PKCS5Padding");
        Key key = new SecretKeySpec("6#26FRL$ZWD".getBytes(), "Blowfish");
        cipher.init(1, key);
        byte[] enc_bytes = cipher.doFinal(s.getBytes());
        System.out.println(enc_bytes);
    }
}

Python 等价物

def PKCS5Padding(string):
    byteNum = len(string)
    packingLength = 8 - byteNum % 8
    if packingLength == 8:
        return string
    else:
        appendage = chr(packingLength) * packingLength
        return string + appendage

def PandoraEncrypt(string):
    from Crypto.Cipher import Blowfish
    key = b'6#26FRL$ZWD'
    c1  = Blowfish.new(key, Blowfish.MODE_ECB)
    packedString = PKCS5Padding(string)
    return c1.encrypt(packedString)

结果

Java 函数：“??¾ô”

Python 函数：“È4A-¾`*ã”

Answer 1

使用您的示例，我得到了 python 和 Java 的相同输出。

爪哇：

import java.math.BigInteger;
import java.security.Key;

import javax.crypto.Cipher;
import javax.crypto.spec.SecretKeySpec;

public class Blowfish1 {

    public static void main(String[] args) throws Exception {
        String s = "testings";
        Cipher cipher = Cipher.getInstance("Blowfish/ECB/PKCS5Padding");
        Key key = new SecretKeySpec("6#26FRL$ZWD".getBytes(), "Blowfish");
        cipher.init(Cipher.ENCRYPT_MODE, key);
        byte[] enc_bytes = cipher.doFinal(s.getBytes());
        System.out.printf("%x%n", new BigInteger(1, enc_bytes));
    }

}

蟒蛇：

from Crypto.Cipher import Blowfish
import binascii

# See @falsetru answer for the following method
#
def PKCS5Padding(string):
    byteNum = len(string)
    packingLength = 8 - byteNum % 8
    appendage = chr(packingLength) * packingLength
    return string + appendage

def PandoraEncrypt(string):
    key = b'6#26FRL$ZWD'
    c1  = Blowfish.new(key, Blowfish.MODE_ECB)
    packedString = PKCS5Padding(string)
    return c1.encrypt(packedString)

if __name__ == '__main__':
    s = 'testings'
    c = PandoraEncrypt(s)
    print(binascii.hexlify(c))

在这两种情况下，输出都是

223950ff19fbea872fce0ee543692ba7

Answer 2

def PKCS5Padding(string):
    byteNum = len(string)
    packingLength = 8 - byteNum % 8
    appendage = chr(packingLength) * packingLength
    return string + appendage

使用 chr 而不是 str.

>>> chr(1)
'\x01'
>>> str(1)
'1'

str * int -> 重复的 str

>>> '!' * 5
'!!!!!'

Answer 3

https://stackoverflow.com/a/17139643/1645017 python2 的正确答案。

对于 python 3https://pycryptodome.readthedocs.io/en/latest/index.html`

from Crypto.Cipher import Blowfish
import binascii

# See @falsetru answer for the following method
#
def PKCS5Padding(string):
    byteNum = len(string)
    packingLength = 8 - byteNum % 8
    appendage = chr(packingLength).encode() * packingLength
    return string + appendage

def PandoraEncrypt(string):
    key = b'6#26FRL$ZWD'
    c1  = Blowfish.new(key, Blowfish.MODE_ECB)
    packedString = PKCS5Padding(string)
    return c1.encrypt(packedString)

if __name__ == '__main__':
    s = b'testings'
    c = PandoraEncrypt(s)
    print(binascii.hexlify(c))`