如何将Userb第一条消息发送给我？下面的代码有助于获取最新消息。我希望该用户的第一条消息

Question

如何将Userb第一条消息发送给我？ 下面的代码有助于获取最新消息。 我要那个用户的第一条信息给我

用户a就是我 用户b是其他用户

我们想要用户b发送给用户a的第一条消息

我的代码获取userb的最新消息

pythoon编码新手 有什么建议吗

from __future__ import print_function
from googleapiclient.discovery import build
from httplib2 import Http
from oauth2client import file, client, tools
from apiclient import errors
import base64
import email
from pampy import match, _
# If modifying these scopes, delete the file token.json.
SCOPES = 'https://www.googleapis.com/auth/gmail.readonly'


def main():
"""Shows basic usage of the Gmail API.
Lists the user's Gmail labels.
"""
# The file token.json stores the user's access and refresh tokens, and is
# created automatically when the authorization flow completes for the first
# time.
store = file.Storage('token.json')
creds = store.get()
if not creds or creds.invalid:
    flow = client.flow_from_clientsecrets('credentials.json', SCOPES)
    creds = tools.run_flow(flow, store)
service = build('gmail', 'v1', http=creds.authorize(Http()))
#print(service.users().execute())

# Call the Gmail API
results = service.users().labels().list(userId='me').execute()
#print(results)
labels = results.get('labels', [])

if not labels:
    print('No labels found.')
else:
    print('Labels:')
    for label in labels:
        print(label['name'])
response = service.users().messages().list(userId='usera@gmail.com',
                                           q='from:userb@gmail.com').execute()
#print(response['messages'])
#print(response)
messages = []
if 'messages' in response:
    messages.extend(response['messages'])

while 'nextPageToken' in response:
    page_token = response['nextPageToken']
    response = service.users().messages().list(userId='usera@gmail.com', q='from:userb@gmail.com',
                                      pageToken=page_token).execute()
    messages.extend(response['messages'])




message = service.users().messages().get(userId='usera@gmail.com', id=messages[0]['id'],
                                         format='raw').execute()

print('Message snippet: %s' % message['snippet'])
#print(base64.urlsafe_b64decode(message['raw'].encode('ASCII')))

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

mime_msg = email.message_from_string(str(msg_str))

print(mime_msg)





if __name__ == '__main__':
   main()

Answer 1

您似乎正在使用搜索邮件

response = service.users().messages().list(userId='usera@gmail.com',
                                           q='from:userb@gmail.com',
                                           pageToken=page_token).execute()

这是正确的。这应该是将所有消息从b返回到a（在循环完成之后）

但是，如果您查看API文档中的messages.list和search syntax，您将注意到无法对这些消息进行排序。为了找到最后一个，你必须下载它们，然后在你的机器上本地排序。您可以使用^{}参数最小化请求的数据，然后仅请求所需消息的完整内容