python3.X
1 def checkLicense(str):#str为解密的字符串 2 str_base64 = base64.b64decode(str) 3 public_key = "1qaz2wsx3edc" 4 5 de_public_key = str2key(public_key) 6 7 modulus = int(de_public_key[0], 16) 8 exponent = int(de_public_key[1], 16) 9 10 rsa_public = rsa.PublicKey(modulus, exponent) 11 12 public_rsa_key = rsa_public.save_pkcs1() 13 14 newPublic = rsa.PublicKey.load_pkcs1(public_rsa_key) 15 16 entrypted = rsa.transform.bytes2int(str_base64) 17 decryted = rsa.core.decrypt_int(entrypted, newPublic.e, newPublic.n) 18 19 decrypted_bytes = rsa.transform.int2bytes(decryted) 20 21 # if len(decrypted_bytes) > 0 and list(six.iterbytes(decrypted_bytes))[0] == 1: 22 if len(decrypted_bytes) > 0: 23 try: 24 raw_info = decrypted_bytes[decrypted_bytes.find(b'x00') + 1:] 25 print(raw_info.decode("utf-8")) 26 return (raw_info.decode("utf-8")) 27 except Exception as e: 28 print(e) 29 print ("解析失败") 30 31 return 'a' 32 33 34 def str2key(s): 35 # 对字符串解码 36 b_str = base64.b64decode(s) 37 38 if len(b_str) < 162: 39 return False 40 41 hex_str = '' 42 hex_str = b_str.hex() 43 44 # 找到模数和指数的开头结束位置 45 m_start = 29 * 2 46 e_start = 159 * 2 47 m_len = 128 * 2 48 e_len = 3 * 2 49 50 modulus = hex_str[m_start:m_start + m_len] 51 exponent = hex_str[e_start:e_start + e_len] 52 53 return modulus, exponent
上面有一个问题，就是说rsa加密密文过长，就会解析失败，20,21行注释掉就是因为这个原因，做了报错处理，要是要各位明白这里的，还望不吝指导下