Question

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

credentials = get_credentials()
http = credentials.authorize(httplib2.Http())
service = discovery.build('gmail', 'v1', http=http)

results = service.users().messages().list(userId='me', labelIds='UNREAD').execute()
messages = results.get('messages', [])

for msg in messages:
    id = msg['id']
    msgobj = GetMessage(service, 'me', id)

    print msgobj['snippet']

当您通过Gmail回复时，您会收到消息，然后是之前回复中的消息。消息的“片段”部分也会收到回复。我想在电子邮件中找到最新的文本（而不是回复部分）

yeah that sounds good

On Sun, Feb 21, 2016 at 7:36 PM, <apple@gmail.com> wrote:
you want to play ball? ________________________________________ From:         banana@gmail.com

除了片段之外还有其他任何内容可以获取最新的电子邮件文本吗？

Answer 1

遗憾的是，没有什么可以做到的。 snippet是消息的一小部分，（基于API Explorer观察）通常约为100 + - 字符。如果消息太短，它将获得前面消息的回复部分。

我观察到的是，该片段的消息部分通常具有On [date]语句。我认为如果有一个类似的短语，你可以检查和分割消息（这会有点棘手，因为短语本身也可以是部分的）。

Answer 2

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)
    for part in mime_msg.walk():
        if part.get_content_type() == 'text/plain':
            print part.get_payload()
    return mime_msg
  except errors.HttpError, error:
    print 'An error occurred: %s' % error

我决定改为使用Mimemessage。如果有的话，这似乎总是有回复部分。这样可以更容易地查找并解析它们

Gmail API：消息代码段，用于忽略正文的回复部分

2 个答案: