weixin_44532019 2021-12-29 10:43 采纳率: 68.2%
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已结题

python 解密方法!3DES

对接票通电子发票,票通已给返回响应!
票通返回值中 myres['content']字段又3DES 加密。求一个解密方法。
当前返回效果:

img

下图解密后效果

img


import json
import time
import random
import pyDes
import requests
from Crypto.PublicKey import RSA
from Crypto.Signature import PKCS1_v1_5
from Crypto.Hash import MD5, SHA1, SHA256
import base64
from flask import current_app
import warnings


from Crypto.Cipher import DES3
import codecs
import base64

warnings.filterwarnings("ignore")

# 私钥(与发给票通的公钥为一对)
privateKey = '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'
# 票通公钥(票通提供)
ptPublicKey = 'MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQCJkx3HelhEm/U7jOCor29oHsIjCMSTyKbX5rpoAY8KDIs9mmr5Y9r+jvNJH8pK3u5gNnvleT6rQgJQW1mk0zHuPO00vy62tSA53fkSjtM+n0oC1Fkm4DRFd5qJgoP7uFQHR5OEffMjy2qIuxChY4Au0kq+6RruEgIttb7wUxy8TwIDAQAB'
# 3DES秘钥(票通提供)
password = 'lsBnINDxtct8HZB7KCMyhWSJ'
# 请更换请求平台简称(票通提供)
platform_alias = 'DEMO'
# 请更换请求平台编码(票通提供)
platform_code = '11111111'

#RSA签名方法
def RSA_sign(data,privateKey):
    private_keyBytes = base64.b64decode(privateKey)
    priKey = RSA.importKey(private_keyBytes)
    signer = PKCS1_v1_5.new(priKey,)
    hash_obj = SHA1.new(data.encode('utf-8'))

    signature = base64.b64encode(signer.sign(hash_obj))
    signature = signature.decode('utf8')
    return signature
#生成请求流水号
def invoiceReqSerialNo():
    return platform_alias + str(round(time.time() * 1000)) + str(random.randrange(100, 1000, 100))

#http请求
def Httppost(url, values):
    headers = {
        "Content-Type": "application/json; charset=UTF-8"
    }
    response = requests.post(url, data=json.dumps(values), headers=headers).text
    return response

#公共报文组装
def assembly(content):
    # 格式化成2016-03-20 11:45:39形式
    timestamp = time.strftime("%Y-%m-%d %H:%M:%S", time.localtime())
    serialNo = str(invoiceReqSerialNo()) + '_12345'

    contentstr = str(DES_Encode(password, content), 'utf-8')#加密位置
    wait_sign = 'content=' + str(
        contentstr) + '&format=JSON&platformCode=' + platform_code + '&serialNo=' + serialNo + '&signType=RSA&timestamp=' + timestamp + '&version=1.0'

    sign = RSA_sign(str(wait_sign), privateKey)

    postjson = {'platformCode': str(platform_code), 'signType': 'RSA', 'sign': str(sign), 'format': 'JSON',
                'timestamp': str(timestamp), 'version': '1.0', 'serialNo': str(serialNo),
                'content': contentstr}
    json.dumps(postjson)
    return postjson

#3des加密
def DES_Encode(keys,encrypt):
    k = pyDes.triple_des(keys, pyDes.ECB, "\0\0\0\0\0\0\0\0", pad=None, padmode=pyDes.PAD_PKCS5)
    d = k.encrypt(encrypt)
    return base64.b64encode(d)

#3des解密
def DES_Decrypt(keys,encrypt):
    k = pyDes.triple_des(keys, pyDes.ECB, "\0\0\0\0\0\0\0\0", pad=None, padmode=pyDes.PAD_PKCS5)
    d = k.decrypt(encrypt)
    # d = k.decrypt(binascii.a2b_hex(encrypt), padmode=pyDes.PAD_PKCS5)
    return d


class EncryptDate:
    def __init__(self, key):
        self.key = key  # 初始化密钥
        self.length = DES3.block_size  # 初始化数据块大小
        self.aes = DES3.new(self.key, DES3.MODE_CBC, b'12345678')  # 初始化AES,ECB模式的实例
        # 截断函数,去除填充的字符
        self.unpad = lambda date: date[0:-ord(date[-1])]

    def pad(self, text):
        """
        #填充函数,使被加密数据的字节码长度是block_size的整数倍
        """
        count = len(text.encode('utf-8'))
        add = self.length - (count % self.length)
        entext = text + (chr(add) * add)
        return entext

    def encrypt(self, encrData):  # 加密函数
        res = self.aes.encrypt(self.pad(encrData).encode("utf8"))
        # msg = str(base64.b64encode(res), encoding="utf8")
        msg =  res.hex()
        return msg

    def decrypt(self, decrData):  # 解密函数
        # res = base64.decodebytes(decrData.encode("utf8"))
        res = bytes.fromhex(decrData)
        msg = self.aes.decrypt(res).decode("utf8")
        return self.unpad(msg)


if __name__ == '__main__':
    url = "http://fpkj.testnw.vpiaotong.cn/tp/openapi/invoiceBlue.pt";
    taxpayerNum='500102201007206608'#测试环境用此税号,正式环境用商家税号

    itemmap1 = {'taxClassificationCode': '1010101020000000000', 'quantity': '1.00', 'goodsName': '货物名称1',
                'unitPrice': '56.64', 'invoiceAmount': '56.64', 'taxRateValue': '0.16', 'includeTaxFlag': '0',
                'zeroTaxFlag': None, 'preferentialPolicyFlag': None, 'vatSpecialManage': None}
    itemmap2 = {'taxClassificationCode': '1010101020000000000', 'quantity': '1.00', 'goodsName': '货物名称2',
                'unitPrice': '56.64', 'invoiceAmount': '56.64', 'taxRateValue': '0.16', 'includeTaxFlag': '0',
                'zeroTaxFlag': None, 'preferentialPolicyFlag': None, 'vatSpecialManage': None}
    itemlist = [itemmap1, itemmap2]

    map = {'taxpayerNum': taxpayerNum, 'invoiceReqSerialNo': invoiceReqSerialNo(), 'buyerName': '购买方名称',
           'buyerTaxpayerNum': 'XX0000000000000000', 'itemList': itemlist}
    content = json.dumps(map)

    post=assembly(content)
    # print(json.dumps(post))
    response=Httppost(url, post)
    # print(response)

    myres = json.loads(response)
    print(myres['content'])



  • 写回答

4条回答 默认 最新

  • 小僵尸打字员 2021-12-29 11:56
    关注

    解开了, 但是返回值跟示例结构有些改变

    img

    def DES_Decrypt(keys, encrypt):
        k = pyDes.triple_des(keys, pyDes.ECB, "\0\0\0\0\0\0\0\0", pad=None, padmode=pyDes.PAD_PKCS5)
        encrypt = base64.b64decode(encrypt)
        d = k.decrypt(encrypt)
        return str(d, "utf-8")
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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  • 系统已结题 1月6日
  • 已采纳回答 12月29日
  • 创建了问题 12月29日

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