python3 skip32加密解密
skip32：把给定的任意int型的数字加密成最长长度为10位的数字；
网上有python2实现的skip32源码，以下是我通过参考python2实现的skip32而修改成的python3版本
参考链接：python2 实现的skip32
import struct
class Skip32:
"""
_key is a string of len 10
"""
_ftable = [
0xa3, 0xd7, 0x09, 0x83, 0xf8, 0x48, 0xf6, 0xf4, 0xb3, 0x21, 0x15, 0x78, 0x99, 0xb1, 0xaf, 0xf9,
0xe7, 0x2d, 0x4d, 0x8a, 0xce, 0x4c, 0xca, 0x2e, 0x52, 0x95, 0xd9, 0x1e, 0x4e, 0x38, 0x44, 0x28,
0x0a, 0xdf, 0x02, 0xa0, 0x17, 0xf1, 0x60, 0x68, 0x12, 0xb7, 0x7a, 0xc3, 0xe9, 0xfa, 0x3d, 0x53,
0x96, 0x84, 0x6b, 0xba, 0xf2, 0x63, 0x9a, 0x19, 0x7c, 0xae, 0xe5, 0xf5, 0xf7, 0x16, 0x6a, 0xa2,
0x39, 0xb6, 0x7b, 0x0f, 0xc1, 0x93, 0x81, 0x1b, 0xee, 0xb4, 0x1a, 0xea, 0xd0, 0x91, 0x2f, 0xb8,
0x55, 0xb9, 0xda, 0x85, 0x3f, 0x41, 0xbf, 0xe0, 0x5a, 0x58, 0x80, 0x5f, 0x66, 0x0b, 0xd8, 0x90,
0x35, 0xd5, 0xc0, 0xa7, 0x33, 0x06, 0x65, 0x69, 0x45, 0x00, 0x94, 0x56, 0x6d, 0x98, 0x9b, 0x76,
0x97, 0xfc, 0xb2, 0xc2, 0xb0, 0xfe, 0xdb, 0x20, 0xe1, 0xeb, 0xd6, 0xe4, 0xdd, 0x47, 0x4a, 0x1d,
0x42, 0xed, 0x9e, 0x6e, 0x49, 0x3c, 0xcd, 0x43, 0x27, 0xd2, 0x07, 0xd4, 0xde, 0xc7, 0x67, 0x18,
0x89, 0xcb, 0x30, 0x1f, 0x8d, 0xc6, 0x8f, 0xaa, 0xc8, 0x74, 0xdc, 0xc9, 0x5d, 0x5c, 0x31, 0xa4,
0x70, 0x88, 0x61, 0x2c, 0x9f, 0x0d, 0x2b, 0x87, 0x50, 0x82, 0x54, 0x64, 0x26, 0x7d, 0x03, 0x40,
0x34, 0x4b, 0x1c, 0x73, 0xd1, 0xc4, 0xfd, 0x3b, 0xcc, 0xfb, 0x7f, 0xab, 0xe6, 0x3e, 0x5b, 0xa5,
0xad, 0x04, 0x23, 0x9c, 0x14, 0x51, 0x22, 0xf0, 0x29, 0x79, 0x71, 0x7e, 0xff, 0x8c, 0x0e, 0xe2,
0x0c, 0xef, 0xbc, 0x72, 0x75, 0x6f, 0x37, 0xa1, 0xec, 0xd3, 0x8e, 0x62, 0x8b, 0x86, 0x10, 0xe8,
0x08, 0x77, 0x11, 0xbe, 0x92, 0x4f, 0x24, 0xc5, 0x32, 0x36, 0x9d, 0xcf, 0xf3, 0xa6, 0xbb, 0xac,
0x5e, 0x6c, 0xa9, 0x13, 0x57, 0x25, 0xb5, 0xe3, 0xbd, 0xa8, 0x3a, 0x01, 0x05, 0x59, 0x2a, 0x46
]
def __init__(self, pack_format='I', key='1234567899'):
self._key = key
self._pack_format = pack_format
def _g(self, tmp_key, k, w):
g1 = (w >> 8) & 0xff
g2 = w & 0xff
g3 = self._ftable[g2 ^ tmp_key[(4 * k) % 10]] ^ g1
g4 = self._ftable[g3 ^ tmp_key[(4 * k + 1) % 10]] ^ g2
g5 = self._ftable[g4 ^ tmp_key[(4 * k + 2) % 10]] ^ g3
g6 = self._ftable[g5 ^ tmp_key[(4 * k + 3) % 10]] ^ g4
return ((g5 << 8) + g6)
def _skip32(self, temp_key, buf, encrypt):
if encrypt:
kstep = 1
k = 0
else:
kstep = -1
k = 23
# pack into words
wl = (buf[0] << 8) + buf[1]
wr = (buf[2] << 8) + buf[3]
# 24 feistel rounds, unrolled
wr ^= self._g(temp_key, k, wl) ^ k # 01
k += kstep
wl ^= self._g(temp_key, k, wr) ^ k # 02
k += kstep
wr ^= self._g(temp_key, k, wl) ^ k # 03
k += kstep
wl ^= self._g(temp_key, k, wr) ^ k # 04
k += kstep
wr ^= self._g(temp_key, k, wl) ^ k # 05
k += kstep
wl ^= self._g(temp_key, k, wr) ^ k # 05
k += kstep
wr ^= self._g(temp_key, k, wl) ^ k # 06
k += kstep
wl ^= self._g(temp_key, k, wr) ^ k # 08
k += kstep
wr ^= self._g(temp_key, k, wl) ^ k # 09
k += kstep
wl ^= self._g(temp_key, k, wr) ^ k # 10
k += kstep
wr ^= self._g(temp_key, k, wl) ^ k # 11
k += kstep
wl ^= self._g(temp_key, k, wr) ^ k # 12
k += kstep
wr ^= self._g(temp_key, k, wl) ^ k # 13
k += kstep
wl ^= self._g(temp_key, k, wr) ^ k # 14
k += kstep
wr ^= self._g(temp_key, k, wl) ^ k # 15
k += kstep
wl ^= self._g(temp_key, k, wr) ^ k # 16
k += kstep
wr ^= self._g(temp_key, k, wl) ^ k # 17
k += kstep
wl ^= self._g(temp_key, k, wr) ^ k # 18
k += kstep
wr ^= self._g(temp_key, k, wl) ^ k # 19
k += kstep
wl ^= self._g(temp_key, k, wr) ^ k # 20
k += kstep
wr ^= self._g(temp_key, k, wl) ^ k # 21
k += kstep
wl ^= self._g(temp_key, k, wr) ^ k # 22
k += kstep
wr ^= self._g(temp_key, k, wl) ^ k # 23
k += kstep
wl ^= self._g(temp_key, k, wr) ^ k # 24
k += kstep
# implicitly swap halves while unpacking
buf[0] = wr >> 8
buf[1] = wr & 0xFF
buf[2] = wl >> 8
buf[3] = wl & 0xFF
def baseN(self, num, b):
chars = "0123456789abcdefghijklmnopqrstuvwxyz"
return ((num == 0) and "0") or (self.baseN(num // b, b).lstrip("0") + chars[num % b])
def get_base36text(self, encrypted_number):
"""
return a base36text with given number
"""
return self.baseN(encrypted_number, 36)
def _skip32_encrypt(self, number):
"""
number should be a integer (32 bit unsigned), though that can be changed by changing pack_format
self._key is a string of len 10
"""
tmp_key = str(self._key)
buf = [ord(c) for c in struct.pack(self._pack_format, number).decode('latin-1')]
if len(tmp_key) != 10:
raise ValueError("_key should be 10 bytes")
tmp_key = [ord(c) for c in tmp_key]
self._skip32(tmp_key, buf, 1)
# res
res = b''
for b in buf:
res += chr(b).encode('latin-1')
enumber = struct.unpack(self._pack_format, res)[0]
return enumber
def _skip32_decrypt(self, number):
tmp_key = str(self._key)
buf = [c for c in struct.pack(self._pack_format, number)]
if len(tmp_key) != 10:
raise ValueError("_key should be 10 bytes")
tmp_key = [ord(c) for c in tmp_key]
self._skip32(tmp_key, buf, 0)
# res
res = b''
for b in buf:
res += chr(b).encode('latin-1')
enumber = struct.unpack(self._pack_format, res)[0]
return enumber
def encrypt_id(self, user_id):
# encrypt
en_id = self._skip32_encrypt(user_id)
return en_id
def decrypt_id(self, encrypted_id):
# decrypt
de_id = self._skip32_decrypt(encrypted_id)
return de_id
测试实例
假如需求是“对用户表中的主键id加解密”，如下所示：
# create a sp object
sp = Skip32()
# method encrypt_id
en_id = sp.encrypt_id(1)
print("加密ID: ", en_id)
# method decrypt_id
de_id = sp.decrypt_id(en_id)
print("原ID: ", de_id)
- 加密之后的位数未必是十位数（比如5），但是可以考虑多次加解密
- 理论上加密之后不会重复
- 多次加解密：多次加解密会用到while循环，不会陷入死循环。(1000000000以内数字已测)
- 多次加解密：多次加密(解密)达到加密(解密)之后是十位数(原数字)的目的，需要自己按照需求改
sp.encrypt_id(sp.decrypt_id)中的代码。