Question

我正在尝试为DFP使用服务帐户，但我收到此错误

OpenSSL.crypto.Error：[（'asn1 encoding routine'，'ASN1_D2I_READ_BIO'，'数据不够'）]

我在dfp中添加了服务帐户，并检查了已创建的服务帐户的email_id，但仍然无法解决问题

这是我的代码，有人可以帮我解决这个问题

from googleads import dfp
from googleads import oauth2


import logging
logging.basicConfig(filename='debug.log',level=logging.DEBUG)
# OAuth 2.0 credential information. In a real application, you'd probably be
# pulling these values from a credential storage.
SERVICE_ACCOUNT_EMAIL = '600632524044-6g7l2njp5g5ldd21j7eaolf1k15pdm1u@developer.gserviceaccount.com'
KEY_FILE = 'D:\DFP\GoogleDFP.p12'

# DFP API information.
APPLICATION_NAME = 'New Srevice Account'


def main(service_account_email, key_file, application_name):

    # with open(key_file, 'rb') as f:
    #   private_key = f.read()
    # f.close()
    oauth2_client = oauth2.GoogleServiceAccountClient(
    oauth2.GetAPIScope('dfp'), service_account_email, key_file)

    client = dfp.DfpClient(oauth2_client, application_name)

    inventory_service = client.GetService('InventoryService', version='v201505')
    statement = dfp.FilterStatement()

  # Get ad units by statement.
    while True:
        response = inventory_service.getAdUnitsByStatement(
        statement.ToStatement())
        if 'results' in response:
      # Display results.
            for ad_unit in response['results']:
                print ('Ad unit with ID \'%s\' and name \'%s\' was found.'
                % (ad_unit['id'], ad_unit['name']))
                statement.offset += dfp.SUGGESTED_PAGE_LIMIT
        else:
            break
  # networks = dfp_client.GetService('NetworkService').getAllNetworks()
  # for network in networks:
  #   print ('Network with network code \'%s\' and display name \'%s\' was found.'
  #          % (network['networkCode'], network['displayName']))


if __name__ == '__main__':
  main(SERVICE_ACCOUNT_EMAIL, KEY_FILE, APPLICATION_NAME)

Answer 1

嘿，我收到错误，oauth2.py

中的googleads中有bug

密钥文件应以二进制模式打开 oauth2.py中的第208行 open（key_file）为f：

它应该以open（key_file，＆＃39; rb＆＃39;）作为f：

使用DFP广告管理系统的Google服务帐户

1 个答案: