python - 使用Python从公钥获取RSA指数和模数

Question

我知道如何使用 openssl 从公钥获取 RSA 模数和指数，但现在我尝试用 Python 来做。我看过this并遵循步骤。

假设这是公钥:

-----BEGIN PUBLIC KEY-----
MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQCdZGziIrJOlRomzh7M9qzo4ibw
QmwORcVDI0dsfUICLUVRdUN+MJ8ELd55NKsfYy4dZodWX7AmdN02zm1Gk5V5i2Vw
GVWE205u7DhtRe85W1oR9WTsMact5wuqU6okJd2GKrEGotgd9iuAJm90N6TDeDZ4
KHEvVEE1yTyvrxQgkwIDAQAB
-----END PUBLIC KEY-----

首先，公钥从base64解码:

import base64
bytearray = base64.b64decode("""MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQCdZGziIrJOlRomzh7M9qzo4ibw
QmwORcVDI0dsfUICLUVRdUN+MJ8ELd55NKsfYy4dZodWX7AmdN02zm1Gk5V5i2Vw
GVWE205u7DhtRe85W1oR9WTsMact5wuqU6okJd2GKrEGotgd9iuAJm90N6TDeDZ4
KHEvVEE1yTyvrxQgkwIDAQAB""")
print(bytearray)

输出:

0\x81\x9f0\r\x06\t*\x86H\x86\xf7\r\x01\x01\x01\x05\x00\x03\x81\x8d\x000\x81\x89\x02\x81\x81\x00\x9ddl\xe2"\xb2N\x95\x1a&\xce\x1e\xcc\xf6\xac\xe8\xe2&\xf0Bl\x0eE\xc5C#Gl}B\x02-EQuC~0\x9f\x04-\xdey4\xab\x1fc.\x1df\x87V_\xb0&t\xdd6\xcemF\x93\x95y\x8bep\x19U\x84\xdbNn\xec8mE\xef9[Z\x11\xf5d\xec1\xa7-\xe7\x0b\xaaS\xaa$%\xdd\x86*\xb1\x06\xa2\xd8\x1d\xf6+\x80&ot7\xa4\xc3x6x(q/TA5\xc9<\xaf\xaf\x14 \x93\x02\x03\x01\x00\x01

然后，字节数组被转换为 base 16(十六进制)字符串:

bytearray.encode('hex')

输出:

30819f300d06092a864886f70d010101050003818d00308189028181009d646ce222b24e951a26ce1eccf6ace8e226f0426c0e45c54323476c7d42022d455175437e309f042dde7934ab1f632e1d6687565fb02674dd36ce6d469395798b6570195584db4e6eec386d45ef395b5a11f564ec31a72de70baa53aa2425dd862ab106a2d81df62b80266f7437a4c378367828712f544135c93cafaf1420930203010001

为了设计，让我们在字符串中的每两个字符之间添加“:”:

':'.join([ i+j for i,j in zip(bytearray[::2],bytearray[1::2])])

输出:

30:81:9f:30:0d:06:09:2a:86:48:86:f7:0d:01:01:01:05:00:03:81:8d:00:30:81:89:02:81:81:00:9d:64:6c:e2:22:b2:4e:95:1a:26:ce:1e:cc:f6:ac:e8:e2:26:f0:42:6c:0e:45:c5:43:23:47:6c:7d:42:02:2d:45:51:75:43:7e:30:9f:04:2d:de:79:34:ab:1f:63:2e:1d:66:87:56:5f:b0:26:74:dd:36:ce:6d:46:93:95:79:8b:65:70:19:55:84:db:4e:6e:ec:38:6d:45:ef:39:5b:5a:11:f5:64:ec:31:a7:2d:e7:0b:aa:53:aa:24:25:dd:86:2a:b1:06:a2:d8:1d:f6:2b:80:26:6f:74:37:a4:c3:78:36:78:28:71:2f:54:41:35:c9:3c:af:af:14:20:93:02:03:01:00:01

但是，不幸的是，我无法理解最后一步。我知道输出包括指数和模数，但还有其他字符需要过滤掉(因为它们无法转换回以 10 为底数的字符)。我还知道上面的十六进制表示是基于 ASN.1 语法的，所以在这两个值出现的十六进制表示中应该有一些特定的偏移量。

要从这个结果中获得模数和指数，需要做什么？

谢谢!

Answer 1

公钥在 ASN.1 SubjectPublicKeyInfo 中格式，在 Java 世界中也称为 X509EncodedKeySpec。许多 crypt 包可以直接导入这样的对象。

例如，使用 PyCrypto和 Cryptography.io , 以下片段

from Crypto.PublicKey import RSA
key_encoded='''-----BEGIN PUBLIC KEY-----
MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQCdZGziIrJOlRomzh7M9qzo4ibw
QmwORcVDI0dsfUICLUVRdUN+MJ8ELd55NKsfYy4dZodWX7AmdN02zm1Gk5V5i2Vw
GVWE205u7DhtRe85W1oR9WTsMact5wuqU6okJd2GKrEGotgd9iuAJm90N6TDeDZ4
KHEvVEE1yTyvrxQgkwIDAQAB
-----END PUBLIC KEY-----'''


pubkey = RSA.importKey(key_encoded)
print(pubkey.n)
print(pubkey.e)

from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives import serialization

pubkey2 = serialization.load_pem_public_key(
    key_encoded.encode('ascii'),
    backend=default_backend()
)

print(pubkey2.public_numbers().n)
print(pubkey2.public_numbers().e)

产生

110524622184298189406696366981362867320131527048683492811128204661745388510505145389459518039217549444918405620726988722254633562452576638635488354260221598432448974859895979017211032905988949400704082939941050902513120244660937339078367607684436944094809985731012813959774525636937965082155868293686780764307
65537
110524622184298189406696366981362867320131527048683492811128204661745388510505145389459518039217549444918405620726988722254633562452576638635488354260221598432448974859895979017211032905988949400704082939941050902513120244660937339078367607684436944094809985731012813959774525636937965082155868293686780764307
65537