在您的gmail帐户上启用不太安全的应用程序，并使用(Python>=3.6):

import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText

gmailUser = 'XXXXX@gmail.com'
gmailPassword = 'XXXXX'
recipient = 'XXXXX@gmail.com'

message = f"""
Type your message here...
"""

msg = MIMEMultipart()
msg['From'] = f'"Your Name" <{gmailUser}>'
msg['To'] = recipient
msg['Subject'] = "Subject here..."
msg.attach(MIMEText(message))

try:
    mailServer = smtplib.SMTP('smtp.gmail.com', 587)
    mailServer.ehlo()
    mailServer.starttls()
    mailServer.ehlo()
    mailServer.login(gmailUser, gmailPassword)
    mailServer.sendmail(gmailUser, recipient, msg.as_string())
    mailServer.close()
    print ('Email sent!')
except:
    print ('Something went wrong...')

不直接相关，但仍然值得指出的是，我的包试图使发送gmail消息非常快速和轻松。它还试图维护一个错误列表，并试图立即指出解决方案。

它只需要下面的代码来完成你所写的:

import yagmail
yag = yagmail.SMTP('user_me@gmail.com')
yag.send('user_you@gmail.com', 'Why,Oh why!')

或者一句话:

yagmail.SMTP('user_me@gmail.com').send('user_you@gmail.com', 'Why,Oh why!')

关于包/安装，请查看git或pip，可用于Python 2和3。

下面是一个Gmail API的例子。虽然比较复杂，但这是我发现在2019年唯一有效的方法。这个例子是从以下例子中获取并修改的:

https://developers.google.com/gmail/api/guides/sending

你需要通过谷歌的网站创建一个API接口的项目。接下来，你需要为你的应用程序启用GMAIL API。创建凭证，然后下载这些凭证，保存为credentials.json。

import pickle
import os.path
from googleapiclient.discovery import build
from google_auth_oauthlib.flow import InstalledAppFlow
from google.auth.transport.requests import Request

from email.mime.text import MIMEText
import base64

#pip install --upgrade google-api-python-client google-auth-httplib2 google-auth-oauthlib

# If modifying these scopes, delete the file token.pickle.
SCOPES = ['https://www.googleapis.com/auth/gmail.readonly', 'https://www.googleapis.com/auth/gmail.send']

def create_message(sender, to, subject, msg):
    message = MIMEText(msg)
    message['to'] = to
    message['from'] = sender
    message['subject'] = subject

    # Base 64 encode
    b64_bytes = base64.urlsafe_b64encode(message.as_bytes())
    b64_string = b64_bytes.decode()
    return {'raw': b64_string}
    #return {'raw': base64.urlsafe_b64encode(message.as_string())}

def send_message(service, user_id, message):
    #try:
    message = (service.users().messages().send(userId=user_id, body=message).execute())
    print( 'Message Id: %s' % message['id'] )
    return message
    #except errors.HttpError, error:print( 'An error occurred: %s' % error )

def main():
    """Shows basic usage of the Gmail API.
    Lists the user's Gmail labels.
    """
    creds = None
    # The file token.pickle stores the user's access and refresh tokens, and is
    # created automatically when the authorization flow completes for the first
    # time.
    if os.path.exists('token.pickle'):
        with open('token.pickle', 'rb') as token:
            creds = pickle.load(token)
    # If there are no (valid) credentials available, let the user log in.
    if not creds or not creds.valid:
        if creds and creds.expired and creds.refresh_token:
            creds.refresh(Request())
        else:
            flow = InstalledAppFlow.from_client_secrets_file(
                'credentials.json', SCOPES)
            creds = flow.run_local_server(port=0)
        # Save the credentials for the next run
        with open('token.pickle', 'wb') as token:
            pickle.dump(creds, token)

    service = build('gmail', 'v1', credentials=creds)

    # Example read operation
    results = service.users().labels().list(userId='me').execute()
    labels = results.get('labels', [])

    if not labels:
        print('No labels found.')
    else:
        print('Labels:')
    for label in labels:
        print(label['name'])

    # Example write
    msg = create_message("from@gmail.com", "to@gmail.com", "Subject", "Msg")
    send_message( service, 'me', msg)

if __name__ == '__main__':
    main()

    import smtplib

    fromadd='from@gmail.com'
    toadd='send@gmail.com'

    msg='''hi,how r u'''
    username='abc@gmail.com'
    passwd='password'

    try:
        server = smtplib.SMTP('smtp.gmail.com:587')
        server.ehlo()
        server.starttls()
        server.login(username,passwd)

        server.sendmail(fromadd,toadd,msg)
        print("Mail Send Successfully")
        server.quit()

   except:
        print("Error:unable to send mail")

   NOTE:https://www.google.com/settings/security/lesssecureapps that                                                         should be enabled

在您的gmail帐户上启用不太安全的应用程序，并使用(Python>=3.6):

import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText

gmailUser = 'XXXXX@gmail.com'
gmailPassword = 'XXXXX'
recipient = 'XXXXX@gmail.com'

message = f"""
Type your message here...
"""

msg = MIMEMultipart()
msg['From'] = f'"Your Name" <{gmailUser}>'
msg['To'] = recipient
msg['Subject'] = "Subject here..."
msg.attach(MIMEText(message))

try:
    mailServer = smtplib.SMTP('smtp.gmail.com', 587)
    mailServer.ehlo()
    mailServer.starttls()
    mailServer.ehlo()
    mailServer.login(gmailUser, gmailPassword)
    mailServer.sendmail(gmailUser, recipient, msg.as_string())
    mailServer.close()
    print ('Email sent!')
except:
    print ('Something went wrong...')

似乎是老smtplib的问题。在python2.7中，一切正常。

更新:是的，server.ehlo()也可以帮助。