获取Mime消息是不是返回了基本的64解码版本？ （Gmail API）

Question

在我的脚本中，我需要提取一组与某些查询匹配的电子邮件。我决定使用GMail的API python客户端。现在，我的理解是GetMimeMessage（）应该返回一组解码的 base 64消息。这是我的代码：

def GmailInput():

    credentials = get_credentials()
    http = credentials.authorize(httplib2.Http())
    service = discovery.build('gmail', 'v1', http=http)
    defaultList= ListMessagesMatchingQuery(service, 'me', 'subject:infringement label:unread ')
    print(defaultList)
    for msg in defaultList:
        currentMSG=GetMimeMessage(service, 'me', msg['id'])
        ....then I parse the text of the emails and extract some things

问题是，我无法实际解析消息体，因为GetMimeMessage没有返回base64解码消息。因此，我实际解析的内容最终会被人类完全看不懂。

我发现这个特殊，因为GetMimeMessage（为方便起见，下面复制）字面上对消息数据进行了url-safe base 64解码。有人有什么建议吗？我真的很难过。

def GetMimeMessage(service, user_id, msg_id):
  """Get a Message and use it to create a MIME Message.

  Args:
    service: Authorized Gmail API service instance.
    user_id: User's email address. The special value "me"
    can be used to indicate the authenticated user.
    msg_id: The ID of the Message required.

  Returns:
    A MIME Message, consisting of data from Message.
  """
  try:
    message = service.users().messages().get(userId=user_id, id=msg_id, format='raw').execute()

    print ('Message snippet: %s' % message['snippet'])

    msg_str = base64.urlsafe_b64decode(message['raw'].encode('ASCII'))

    mime_msg = email.message_from_string(msg_str)

    return mime_msg
  except errors.HttpError, error:
    print ('An error occurred: %s' % error)

Answer 1

您可以使用User.messages:get。此请求需要authorization且至少具有以下范围之一。

HTTP请求

GET https://www.googleapis.com/gmail/v1/users/userId/messages/id

import base64
import email
from apiclient import errors

def GetMessage(service, user_id, msg_id):
"""Get a Message with given ID.

Args:
service: Authorized Gmail API service instance.
user_id: User's email address. The special value "me"
can be used to indicate the authenticated user.
msg_id: The ID of the Message required.

Returns:
A Message.
"""
try:
message = service.users().messages().get(userId=user_id, id=msg_id).execute()

print 'Message snippet: %s' % message['snippet']

return message
except errors.HttpError, error:
print 'An error occurred: %s' % error


def GetMimeMessage(service, user_id, msg_id):
"""Get a Message and use it to create a MIME Message.

Args:
service: Authorized Gmail API service instance.
user_id: User's email address. The special value "me"
can be used to indicate the authenticated user.
msg_id: The ID of the Message required.

Returns:
A MIME Message, consisting of data from Message.
"""
try:
message = service.users().messages().get(userId=user_id, id=msg_id,
format='raw').execute()

print 'Message snippet: %s' % message['snippet']

msg_str = base64.urlsafe_b64decode(message['raw'].encode('ASCII'))

mime_msg = email.message_from_string(msg_str)

return mime_msg
except errors.HttpError, error:
print 'An error occurred: %s' % error