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...')

我遇到过类似的问题，并无意中发现了这个问题。我得到一个SMTP身份验证错误，但我的用户名/ pass是正确的。下面是解决问题的方法。我读到:

https://support.google.com/accounts/answer/6010255

简而言之，谷歌不允许您通过smtplib登录，因为它已经标记了这种登录为“不太安全”，所以你要做的是去这个链接，而你登录到你的谷歌帐户，并允许访问:

https://www.google.com/settings/security/lesssecureapps

一旦设置好了(见下面的截图)，它就可以工作了。

登录现在工作:

smtpserver = smtplib.SMTP("smtp.gmail.com", 587)
smtpserver.ehlo()
smtpserver.starttls()
smtpserver.ehlo()
smtpserver.login('me@gmail.com', 'me_pass')

更改后的回复:

(235, '2.7.0 Accepted')

响应之前:

smtplib.SMTPAuthenticationError: (535, '5.7.8 Username and Password not accepted. Learn more at\n5.7.8 http://support.google.com/mail/bin/answer.py?answer=14257 g66sm2224117qgf.37 - gsmtp')

还是不行?如果您仍然得到SMTPAuthenticationError，但现在代码是534，这是因为位置未知。点击这个链接:

https://accounts.google.com/DisplayUnlockCaptcha

点击继续，这应该给你10分钟的时间来注册你的新应用。所以现在继续进行另一次登录尝试，它应该可以工作。

更新:这似乎不能马上工作，你可能会卡在smptlib中得到这个错误:

235 == 'Authentication successful'
503 == 'Error: already authenticated'

该消息说使用浏览器登录:

SMTPAuthenticationError: (534, '5.7.9 Please log in with your web browser and then try again. Learn more at\n5.7.9 https://support.google.com/mail/bin/answer.py?answer=78754 qo11sm4014232igb.17 - gsmtp')

启用“lesssecureapps”后，去喝杯咖啡，回来后再次尝试“DisplayUnlockCaptcha”链接。从用户体验来看，更改可能需要一个小时才能生效。然后再次尝试登录过程。

你需要在直接运行到STARTTLS之前说EHLO:

server = smtplib.SMTP('smtp.gmail.com:587')
server.ehlo()
server.starttls()

此外，您应该真正创建From:、To:和Subject:消息头，用空行与消息体分隔，并使用CRLF作为EOL标记。

E.g.

msg = "\r\n".join([
  "From: user_me@gmail.com",
  "To: user_you@gmail.com",
  "Subject: Just a message",
  "",
  "Why, oh why"
  ])

注意:

为了做到这一点，你需要在gmail帐户配置中启用“允许不太安全的应用程序”选项。否则，当gmail检测到一个非谷歌应用程序试图登录你的帐户时，你会得到一个“关键安全警报”。

import smtplib
server = smtplib.SMTP('smtp.gmail.com', 587)
server.ehlo()
server.starttls()
server.login("fromaddress", "password")
msg = "HI!"
server.sendmail("fromaddress", "receiveraddress", msg)
server.quit()

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()

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...')