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Python相当于PHP的MCRYPT_RIJNDAEL_256 CBC

作者:互联网

我需要这个函数的Python实现 – 我想在appengine上使用它.

我在Python方面不太好,所以请帮忙.

function encrypt($data) {
  return base64_encode(mcrypt_encrypt(MCRYPT_RIJNDAEL_256 ,'oqufXQ(?bc=6_hR2I3sMZChDpb6dDlw4', $data , MCRYPT_MODE_CBC, utf8_encode('fOaiIOkD8*9Xeu_s4_bb87Ox_UG+D9GA')));
}

解决方法:

你试过this one(也包括在下面)?它实现了16,44或32字节的Rijndael分组密码.您正在使用块密码的256位(32字节)版本.

"""
A pure python (slow) implementation of rijndael with a decent interface

To include -

from rijndael import rijndael

To do a key setup -

r = rijndael(key, block_size = 16)

key must be a string of length 16, 24, or 32
blocksize must be 16, 24, or 32. Default is 16

To use -

ciphertext = r.encrypt(plaintext)
plaintext = r.decrypt(ciphertext)

If any strings are of the wrong length a ValueError is thrown
"""

# ported from the Java reference code by Bram Cohen, April 2001
# this code is public domain, unless someone makes 
# an intellectual property claim against the reference 
# code, in which case it can be made public domain by 
# deleting all the comments and renaming all the variables

import copy
import string

shifts = [[[0, 0], [1, 3], [2, 2], [3, 1]],
          [[0, 0], [1, 5], [2, 4], [3, 3]],
          [[0, 0], [1, 7], [3, 5], [4, 4]]]

# [keysize][block_size]
num_rounds = {16: {16: 10, 24: 12, 32: 14}, 24: {16: 12, 24: 12, 32: 14}, 32: {16: 14, 24: 14, 32: 14}}

A = [[1, 1, 1, 1, 1, 0, 0, 0],
     [0, 1, 1, 1, 1, 1, 0, 0],
     [0, 0, 1, 1, 1, 1, 1, 0],
     [0, 0, 0, 1, 1, 1, 1, 1],
     [1, 0, 0, 0, 1, 1, 1, 1],
     [1, 1, 0, 0, 0, 1, 1, 1],
     [1, 1, 1, 0, 0, 0, 1, 1],
     [1, 1, 1, 1, 0, 0, 0, 1]]

# produce log and alog tables, needed for multiplying in the
# field GF(2^m) (generator = 3)
alog = [1]
for i in range(255):
    j = (alog[-1] << 1) ^ alog[-1]
    if j & 0x100 != 0:
        j ^= 0x11B
    alog.append(j)

log = [0] * 256
for i in range(1, 255):
    log[alog[i]] = i

# multiply two elements of GF(2^m)
def mul(a, b):
    if a == 0 or b == 0:
        return 0
    return alog[(log[a & 0xFF] + log[b & 0xFF]) % 255]

# substitution box based on F^{-1}(x)
box = [[0] * 8 for i in range(256)]
box[1][7] = 1
for i in range(2, 256):
    j = alog[255 - log[i]]
    for t in range(8):
        box[i][t] = (j >> (7 - t)) & 0x01

B = [0, 1, 1, 0, 0, 0, 1, 1]

# affine transform:  box[i] <- B + A*box[i]
cox = [[0] * 8 for i in range(256)]
for i in range(256):
    for t in range(8):
        cox[i][t] = B[t]
        for j in range(8):
            cox[i][t] ^= A[t][j] * box[i][j]

# S-boxes and inverse S-boxes
S =  [0] * 256
Si = [0] * 256
for i in range(256):
    S[i] = cox[i][0] << 7
    for t in range(1, 8):
        S[i] ^= cox[i][t] << (7-t)
    Si[S[i] & 0xFF] = i

# T-boxes
G = [[2, 1, 1, 3],
    [3, 2, 1, 1],
    [1, 3, 2, 1],
    [1, 1, 3, 2]]

AA = [[0] * 8 for i in range(4)]

for i in range(4):
    for j in range(4):
        AA[i][j] = G[i][j]
        AA[i][i+4] = 1

for i in range(4):
    pivot = AA[i][i]
    if pivot == 0:
        t = i + 1
        while AA[t][i] == 0 and t < 4:
            t += 1
            assert t != 4, 'G matrix must be invertible'
            for j in range(8):
                AA[i][j], AA[t][j] = AA[t][j], AA[i][j]
            pivot = AA[i][i]
    for j in range(8):
        if AA[i][j] != 0:
            AA[i][j] = alog[(255 + log[AA[i][j] & 0xFF] - log[pivot & 0xFF]) % 255]
    for t in range(4):
        if i != t:
            for j in range(i+1, 8):
                AA[t][j] ^= mul(AA[i][j], AA[t][i])
            AA[t][i] = 0

iG = [[0] * 4 for i in range(4)]

for i in range(4):
    for j in range(4):
        iG[i][j] = AA[i][j + 4]

def mul4(a, bs):
    if a == 0:
        return 0
    r = 0
    for b in bs:
        r <<= 8
        if b != 0:
            r = r | mul(a, b)
    return r

T1 = []
T2 = []
T3 = []
T4 = []
T5 = []
T6 = []
T7 = []
T8 = []
U1 = []
U2 = []
U3 = []
U4 = []

for t in range(256):
    s = S[t]
    T1.append(mul4(s, G[0]))
    T2.append(mul4(s, G[1]))
    T3.append(mul4(s, G[2]))
    T4.append(mul4(s, G[3]))

    s = Si[t]
    T5.append(mul4(s, iG[0]))
    T6.append(mul4(s, iG[1]))
    T7.append(mul4(s, iG[2]))
    T8.append(mul4(s, iG[3]))

    U1.append(mul4(t, iG[0]))
    U2.append(mul4(t, iG[1]))
    U3.append(mul4(t, iG[2]))
    U4.append(mul4(t, iG[3]))

# round constants
rcon = [1]
r = 1
for t in range(1, 30):
    r = mul(2, r)
    rcon.append(r)

del A
del AA
del pivot
del B
del G
del box
del log
del alog
del i
del j
del r
del s
del t
del mul
del mul4
del cox
del iG

class rijndael:
    def __init__(self, key, block_size = 16):
        if block_size != 16 and block_size != 24 and block_size != 32:
            raise ValueError('Invalid block size: ' + str(block_size))
        if len(key) != 16 and len(key) != 24 and len(key) != 32:
            raise ValueError('Invalid key size: ' + str(len(key)))
        self.block_size = block_size

        ROUNDS = num_rounds[len(key)][block_size]
        BC = block_size // 4
        # encryption round keys
        Ke = [[0] * BC for i in range(ROUNDS + 1)]
        # decryption round keys
        Kd = [[0] * BC for i in range(ROUNDS + 1)]
        ROUND_KEY_COUNT = (ROUNDS + 1) * BC
        KC = len(key) // 4

        # copy user material bytes into temporary ints
        tk = []
        for i in range(0, KC):
            tk.append((ord(key[i * 4]) << 24) | (ord(key[i * 4 + 1]) << 16) |
                (ord(key[i * 4 + 2]) << 8) | ord(key[i * 4 + 3]))

        # copy values into round key arrays
        t = 0
        j = 0
        while j < KC and t < ROUND_KEY_COUNT:
            Ke[t // BC][t % BC] = tk[j]
            Kd[ROUNDS - (t // BC)][t % BC] = tk[j]
            j += 1
            t += 1
        tt = 0
        rconpointer = 0
        while t < ROUND_KEY_COUNT:
            # extrapolate using phi (the round key evolution function)
            tt = tk[KC - 1]
            tk[0] ^= (S[(tt >> 16) & 0xFF] & 0xFF) << 24 ^  \
                     (S[(tt >>  8) & 0xFF] & 0xFF) << 16 ^  \
                     (S[ tt        & 0xFF] & 0xFF) <<  8 ^  \
                     (S[(tt >> 24) & 0xFF] & 0xFF)       ^  \
                     (rcon[rconpointer]    & 0xFF) << 24
            rconpointer += 1
            if KC != 8:
                for i in range(1, KC):
                    tk[i] ^= tk[i-1]
            else:
                for i in range(1, KC // 2):
                    tk[i] ^= tk[i-1]
                tt = tk[KC // 2 - 1]
                tk[KC // 2] ^= (S[ tt        & 0xFF] & 0xFF)       ^ \
                               (S[(tt >>  8) & 0xFF] & 0xFF) <<  8 ^ \
                               (S[(tt >> 16) & 0xFF] & 0xFF) << 16 ^ \
                               (S[(tt >> 24) & 0xFF] & 0xFF) << 24
                for i in range(KC // 2 + 1, KC):
                    tk[i] ^= tk[i-1]
            # copy values into round key arrays
            j = 0
            while j < KC and t < ROUND_KEY_COUNT:
                Ke[t // BC][t % BC] = tk[j]
                Kd[ROUNDS - (t // BC)][t % BC] = tk[j]
                j += 1
                t += 1
        # inverse MixColumn where needed
        for r in range(1, ROUNDS):
            for j in range(BC):
                tt = Kd[r][j]
                Kd[r][j] = U1[(tt >> 24) & 0xFF] ^ \
                           U2[(tt >> 16) & 0xFF] ^ \
                           U3[(tt >>  8) & 0xFF] ^ \
                           U4[ tt        & 0xFF]
        self.Ke = Ke
        self.Kd = Kd

    def encrypt(self, plaintext):
        if len(plaintext) != self.block_size:
            raise ValueError('wrong block length, expected ' + str(self.block_size) + ' got ' + str(len(plaintext)))
        Ke = self.Ke

        BC = self.block_size // 4
        ROUNDS = len(Ke) - 1
        if BC == 4:
            SC = 0
        elif BC == 6:
            SC = 1
        else:
            SC = 2
        s1 = shifts[SC][1][0]
        s2 = shifts[SC][2][0]
        s3 = shifts[SC][3][0]
        a = [0] * BC
        # temporary work array
        t = []
        # plaintext to ints + key
        for i in range(BC):
            t.append((ord(plaintext[i * 4    ]) << 24 |
                      ord(plaintext[i * 4 + 1]) << 16 |
                      ord(plaintext[i * 4 + 2]) <<  8 |
                      ord(plaintext[i * 4 + 3])        ) ^ Ke[0][i])
        # apply round transforms
        for r in range(1, ROUNDS):
            for i in range(BC):
                a[i] = (T1[(t[ i           ] >> 24) & 0xFF] ^
                        T2[(t[(i + s1) % BC] >> 16) & 0xFF] ^
                        T3[(t[(i + s2) % BC] >>  8) & 0xFF] ^
                        T4[ t[(i + s3) % BC]        & 0xFF]  ) ^ Ke[r][i]
            t = copy.copy(a)
        # last round is special
        result = []
        for i in range(BC):
            tt = Ke[ROUNDS][i]
            result.append((S[(t[ i           ] >> 24) & 0xFF] ^ (tt >> 24)) & 0xFF)
            result.append((S[(t[(i + s1) % BC] >> 16) & 0xFF] ^ (tt >> 16)) & 0xFF)
            result.append((S[(t[(i + s2) % BC] >>  8) & 0xFF] ^ (tt >>  8)) & 0xFF)
            result.append((S[ t[(i + s3) % BC]        & 0xFF] ^  tt       ) & 0xFF)
        return ''.join(map(chr, result))

    def decrypt(self, ciphertext):
        if len(ciphertext) != self.block_size:
            raise ValueError('wrong block length, expected ' + str(self.block_size) + ' got ' + str(len(ciphertext)))
        Kd = self.Kd

        BC = self.block_size // 4
        ROUNDS = len(Kd) - 1
        if BC == 4:
            SC = 0
        elif BC == 6:
            SC = 1
        else:
            SC = 2
        s1 = shifts[SC][1][1]
        s2 = shifts[SC][2][1]
        s3 = shifts[SC][3][1]
        a = [0] * BC
        # temporary work array
        t = [0] * BC
        # ciphertext to ints + key
        for i in range(BC):
            t[i] = (ord(ciphertext[i * 4    ]) << 24 |
                    ord(ciphertext[i * 4 + 1]) << 16 |
                    ord(ciphertext[i * 4 + 2]) <<  8 |
                    ord(ciphertext[i * 4 + 3])        ) ^ Kd[0][i]
        # apply round transforms
        for r in range(1, ROUNDS):
            for i in range(BC):
                a[i] = (T5[(t[ i           ] >> 24) & 0xFF] ^
                        T6[(t[(i + s1) % BC] >> 16) & 0xFF] ^
                        T7[(t[(i + s2) % BC] >>  8) & 0xFF] ^
                        T8[ t[(i + s3) % BC]        & 0xFF]  ) ^ Kd[r][i]
            t = copy.copy(a)
        # last round is special
        result = []
        for i in range(BC):
            tt = Kd[ROUNDS][i]
            result.append((Si[(t[ i           ] >> 24) & 0xFF] ^ (tt >> 24)) & 0xFF)
            result.append((Si[(t[(i + s1) % BC] >> 16) & 0xFF] ^ (tt >> 16)) & 0xFF)
            result.append((Si[(t[(i + s2) % BC] >>  8) & 0xFF] ^ (tt >>  8)) & 0xFF)
            result.append((Si[ t[(i + s3) % BC]        & 0xFF] ^  tt       ) & 0xFF)
        return ''.join(map(chr, result))

def encrypt(key, block):
    return rijndael(key, len(block)).encrypt(block)

def decrypt(key, block):
    return rijndael(key, len(block)).decrypt(block)

请注意,rijndael.py文件仅实现块密码.加密/解密函数仅处理精确到块大小的明文.这意味着这些函数的调用者必须提供分组密码操作模式和零填充本身.

示例python代码(来自Java程序员,请注意):

class zeropad:

    def __init__(self, block_size):
        assert block_size > 0 and block_size < 256
        self.block_size = block_size

    def pad(self, pt):
        ptlen = len(pt)
        padsize = self.block_size - ((ptlen + self.block_size - 1) % self.block_size + 1)
        return pt + "\0" * padsize

    def unpad(self, ppt):
        assert len(ppt) % self.block_size == 0
        offset = len(ppt)
        if (offset == 0):
            return ''
        end = offset - self.block_size + 1
        while (offset > end):
            offset -= 1;
            if (ppt[offset] != "\0"):
                return ppt[:offset + 1]
        assert false

class cbc:

    def __init__(self, padding, cipher, iv):
        assert padding.block_size == cipher.block_size;
        assert len(iv) == cipher.block_size;
        self.padding = padding
        self.cipher = cipher
        self.iv = iv

    def encrypt(self, pt):
        ppt = self.padding.pad(pt)
        offset = 0
        ct = ''
        v = self.iv
        while (offset < len(ppt)):
            block = ppt[offset:offset + self.cipher.block_size]
            block = self.xorblock(block, v)
            block = self.cipher.encrypt(block)
            ct += block
            offset += self.cipher.block_size
            v = block
        return ct;

    def decrypt(self, ct):
        assert len(ct) % self.cipher.block_size == 0
        ppt = ''
        offset = 0
        v = self.iv
        while (offset < len(ct)):
            block = ct[offset:offset + self.cipher.block_size]
            decrypted = self.cipher.decrypt(block)
            ppt += self.xorblock(decrypted, v)
            offset += self.cipher.block_size
            v = block
        pt = self.padding.unpad(ppt)
        return pt;

    def xorblock(self, b1, b2):
        # sorry, not very Pythonesk
        i = 0
        r = '';
        while (i < self.cipher.block_size):
             r += chr(ord(b1[i]) ^ ord(b2[i]))
             i += 1
        return r

标签:python,php,rijndael
来源: https://codeday.me/bug/20190626/1293100.html