為什么要PKCS7填充？如何獲取真實(shí)字符串？

由來(lái)

做爬蟲(chóng)逆向的時(shí)候, 經(jīng)常要用到加密解密

目錄

pkcs7填充: pad和unpad

crypto: 一個(gè)共同方法 + 一個(gè)例子

pkcs7填充: pad和unpad

1. 為什么要PKCS7填充?

因?yàn)镻KCS7是當(dāng)下各大加密算法都遵循的數(shù)據(jù)填充算法

2. 基礎(chǔ)

1個(gè)字節(jié)有8位, 所以16個(gè)字節(jié)是128位, 比如: 一個(gè)字母A就是一個(gè)字節(jié)

16進(jìn)制: 0-9, a,b,c,d,e,f, 其中a代表10,后面依次加1, 滿16進(jìn)一,所以 15對(duì)應(yīng)\x0f,16對(duì)應(yīng)\x10

轉(zhuǎn)換:

int -> unicode: chr(1) 結(jié)果是 \x01

unicode -> int: ord('\x11')值為17

3. pkcs7補(bǔ)位規(guī)則

補(bǔ)位的個(gè)數(shù): 不足16位字節(jié)的倍數(shù),補(bǔ)足變成16位的倍數(shù),如果剛好是16的倍數(shù), 補(bǔ)16個(gè)字節(jié)

補(bǔ)位的值: 等于補(bǔ)位個(gè)數(shù)的unicode碼, 比如:content有7位,那么補(bǔ)充9個(gè)字節(jié),每個(gè)字節(jié)的值都是\x09

4. 如何獲取真實(shí)字符串?

直接獲取字符串的最后一個(gè)字符,轉(zhuǎn)換為int即為補(bǔ)位的數(shù), 然后,真實(shí) = 原始字符串長(zhǎng)度 - 補(bǔ)位的數(shù)

def pad(content: str, block_size: int = 16) -> str:
p_len = block_size - len(content) % block_size
p = p_len * chr(p_len)
return content + p

def unpad(content: str) -> str:
last_char = ord(content[-1])
# 獲取真實(shí)字符串,去掉最后補(bǔ)位的數(shù)據(jù)
return content[:-last_char]

def test_pad():
# 計(jì)算字符數(shù)
assert 3 == len('11中'), 'pad china error'
assert 4 == len('abcd'), 'pad china error'

p1 = pad('1')
assert '1\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f' == p1, 'pad 1 error'

b = r'1\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f'
assert f"''" == repr(p1), 'pad 1 1 error'

p2 = pad('11')
assert '11\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e' == p2, 'pad 2 error'

p15 = pad('111111111122222')
assert '111111111122222\x01' == p15, 'pad 15 error'

p16 = pad('1111111111222222')
assert '1111111111222222\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10' == p16, 'pad 16 error'


def test_unpad():
p1 = '1\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f\x0f'
assert '1' == unpad(p1)

p2 = '11\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e\x0e'
assert '11' == unpad(p2)

p15 = '111111111122222\x01'
assert '111111111122222' == unpad(p15)

p16 = '1111111111222222\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10\x10'
assert '1111111111222222' == unpad(p16)

crypto: 一個(gè)共同方法 + 一個(gè)例子

代碼很簡(jiǎn)單, 不需要過(guò)多的解釋

from base64 import b64decode, b64encode
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad, unpad

class CryptoR():

def __init__(self, key: str, iv: str):
self.aes = AES.new(key.encode('utf8'), AES.MODE_CBC, iv.encode('utf8'))

def encrypto(self, text: str) -> str:
p = pad(text.encode('utf8'), AES.block_size)
return b64encode(self.aes.encrypt(p)).decode('utf8')

def decrypto(self, text: str) -> str:
d = self.aes.decrypt(b64decode(text))
return unpad(d, AES.block_size).decode('utf8')

def call_en():
text = 'm3u8.okjx.cc|057f1eed099f2f7e'
key = 'dvyYRQlnPRCMdQSe'
iv = '057f1eed099f2f7e'
aes = CryptoR(key, iv)
e1 = aes.encrypto(text)
assert e1 == 'ncZvyrJ2IPdKSF2EmvfZAGVrtAj9D8eXnGpjPTV1QTM=', 'encrypto error'

def call_de():
text = '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'
key = '0EAE7A71512EC09C'
iv = '675480787382e6f4'
aes = CryptoR(key, iv)
d1 = aes.decrypto(text)
d2 =
assert d1 == d2, 'decrypto error'


def main():
call_en()
call_de()


if __name__ == '__main__':
main()




審核編輯：劉清