313

我正在尝试使用 python 发送电子邮件(Gmail),但出现以下错误。

Traceback (most recent call last):  
File "emailSend.py", line 14, in <module>  
server.login(username,password)  
File "/usr/lib/python2.5/smtplib.py", line 554, in login  
raise SMTPException("SMTP AUTH extension not supported by server.")  
smtplib.SMTPException: SMTP AUTH extension not supported by server.

Python 脚本如下。

import smtplib
fromaddr = 'user_me@gmail.com'
toaddrs  = 'user_you@gmail.com'
msg = 'Why,Oh why!'
username = 'user_me@gmail.com'
password = 'pwd'
server = smtplib.SMTP('smtp.gmail.com:587')
server.starttls()
server.login(username,password)
server.sendmail(fromaddr, toaddrs, msg)
server.quit()
4

16 回答 16

316
def send_email(user, pwd, recipient, subject, body):
    import smtplib

    FROM = user
    TO = recipient if isinstance(recipient, list) else [recipient]
    SUBJECT = subject
    TEXT = body

    # Prepare actual message
    message = """From: %s\nTo: %s\nSubject: %s\n\n%s
    """ % (FROM, ", ".join(TO), SUBJECT, TEXT)
    try:
        server = smtplib.SMTP("smtp.gmail.com", 587)
        server.ehlo()
        server.starttls()
        server.login(user, pwd)
        server.sendmail(FROM, TO, message)
        server.close()
        print 'successfully sent the mail'
    except:
        print "failed to send mail"

如果要使用端口 465,则必须创建一个SMTP_SSL对象:

# SMTP_SSL Example
server_ssl = smtplib.SMTP_SSL("smtp.gmail.com", 465)
server_ssl.ehlo() # optional, called by login()
server_ssl.login(gmail_user, gmail_pwd)  
# ssl server doesn't support or need tls, so don't call server_ssl.starttls() 
server_ssl.sendmail(FROM, TO, message)
#server_ssl.quit()
server_ssl.close()
print 'successfully sent the mail'
于 2012-09-14T12:19:26.740 回答
224

EHLO在直接运行之前你需要说STARTTLS

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

此外,您应该真正创建From:,To:Subject:消息头,用空行与消息正文分隔并CRLF用作 EOL 标记。

例如

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

笔记:

为了使其正常工作,您需要在您的 gmail 帐户配置中启用“允许安全性较低的应用程序”选项。否则,当 gmail 检测到非 Google 应用程序正在尝试登录您的帐户时,您将收到“严重安全警报”。

于 2012-04-13T19:57:54.350 回答
146

我遇到了类似的问题并偶然发现了这个问题。我收到 SMTP 身份验证错误,但我的用户名/密码正确。这是修复它的方法。我读到这个:

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

简而言之,谷歌不允许你通过 smtplib 登录,因为它已将这种登录标记为“不太安全”,所以你要做的就是在你登录到你的谷歌帐户时转到这个链接,并允许访问:

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

一旦设置好(参见下面的屏幕截图),它应该可以工作。

在此处输入图像描述

登录现在有效:

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

更改后的响应:

(235, '2.7.0 Accepted')

之前的回应:

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')

还是行不通?如果您仍然收到 SMTPAuthenticationError 但现在代码是 534,那是因为位置未知。按照这个链接:

https://accounts.google.com/DisplayUnlockCaptcha

单击继续,这应该会给您 10 分钟的时间来注册您的新应用程序。因此,现在继续进行另一次登录尝试,它应该可以工作。

更新:这似乎无法立即工作,您可能会在 smptlib 中遇到此错误一段时间:

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

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,然后再次尝试“DisplayUnlockCaptcha”链接。根据用户体验,更改可能需要一个小时才能生效。然后再次尝试登录过程。

于 2014-12-16T23:27:25.317 回答
26

这个作品

创建Gmail APP密码!

创建之后,然后创建一个名为sendgmail.py

然后添加此代码

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
# =============================================================================
# Created By  : Jeromie Kirchoff
# Created Date: Mon Aug 02 17:46:00 PDT 2018
# =============================================================================
# Imports
# =============================================================================
import smtplib

# =============================================================================
# SET EMAIL LOGIN REQUIREMENTS
# =============================================================================
gmail_user = 'THEFROM@gmail.com'
gmail_app_password = 'YOUR-GOOGLE-APPLICATION-PASSWORD!!!!'

# =============================================================================
# SET THE INFO ABOUT THE SAID EMAIL
# =============================================================================
sent_from = gmail_user
sent_to = ['THE-TO@gmail.com', 'THE-TO@gmail.com']
sent_subject = "Where are all my Robot Women at?"
sent_body = ("Hey, what's up? friend!\n\n"
             "I hope you have been well!\n"
             "\n"
             "Cheers,\n"
             "Jay\n")

email_text = """\
From: %s
To: %s
Subject: %s

%s
""" % (sent_from, ", ".join(sent_to), sent_subject, sent_body)

# =============================================================================
# SEND EMAIL OR DIE TRYING!!!
# Details: http://www.samlogic.net/articles/smtp-commands-reference.htm
# =============================================================================

try:
    server = smtplib.SMTP_SSL('smtp.gmail.com', 465)
    server.ehlo()
    server.login(gmail_user, gmail_app_password)
    server.sendmail(sent_from, sent_to, email_text)
    server.close()

    print('Email sent!')
except Exception as exception:
    print("Error: %s!\n\n" % exception)

所以,如果你成功了,会看到这样的图像:

我通过向自己发送电子邮件进行了测试。

成功发送电子邮件。

注意:我的帐户启用了两步验证。应用密码适用于此!(对于 gmail smtp 设置,您必须访问https://support.google.com/accounts/answer/185833?hl=en并按照以下步骤操作)

此设置不适用于启用两步验证的帐户。此类帐户需要特定于应用程序的密码才能访问不太安全的应用程序。

应用访问安全性较低... 此设置不适用于启用两步验证的帐户。

于 2018-08-03T01:26:14.437 回答
19

你对 OOP 失望了吗?

#!/usr/bin/env python


import smtplib

class Gmail(object):
    def __init__(self, email, password):
        self.email = email
        self.password = password
        self.server = 'smtp.gmail.com'
        self.port = 587
        session = smtplib.SMTP(self.server, self.port)        
        session.ehlo()
        session.starttls()
        session.ehlo
        session.login(self.email, self.password)
        self.session = session

    def send_message(self, subject, body):
        ''' This must be removed '''
        headers = [
            "From: " + self.email,
            "Subject: " + subject,
            "To: " + self.email,
            "MIME-Version: 1.0",
           "Content-Type: text/html"]
        headers = "\r\n".join(headers)
        self.session.sendmail(
            self.email,
            self.email,
            headers + "\r\n\r\n" + body)


gm = Gmail('Your Email', 'Password')

gm.send_message('Subject', 'Message')
于 2014-06-23T11:16:51.607 回答
17

这是一个 Gmail API 示例。虽然更复杂,但这是我发现的唯一在 2019 年有效的方法。此示例取自并修改自:

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

您需要通过他们的网站使用 Google 的 API 接口创建一个项目。接下来,您需要为您的应用启用 GMAIL API。创建凭据,然后下载这些凭据,将其保存为 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()
于 2019-09-18T17:42:58.833 回答
15

你可以在这里找到它:http: //jayrambhia.com/blog/send-emails-using-python

smtp_host = 'smtp.gmail.com'
smtp_port = 587
server = smtplib.SMTP()
server.connect(smtp_host,smtp_port)
server.ehlo()
server.starttls()
server.login(user,passw)
fromaddr = raw_input('Send mail by the name of: ')
tolist = raw_input('To: ').split()
sub = raw_input('Subject: ')

msg = email.MIMEMultipart.MIMEMultipart()
msg['From'] = fromaddr
msg['To'] = email.Utils.COMMASPACE.join(tolist)
msg['Subject'] = sub  
msg.attach(MIMEText(raw_input('Body: ')))
msg.attach(MIMEText('\nsent via python', 'plain'))
server.sendmail(user,tolist,msg.as_string())
于 2012-04-13T20:28:42.010 回答
14

不直接相关但仍然值得指出的是,我的包试图使发送 gmail 消息变得非常快速和轻松。它还尝试维护错误列表并尝试立即指出解决方案。

从字面上看,它只需要这段代码就可以完全按照您的编写:

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

或者一个班轮:

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

对于包/安装,请查看gitpip,可用于 Python 2 和 3。

于 2015-04-18T16:58:21.917 回答
3

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...')
于 2020-02-27T16:06:27.040 回答
2

现在有一个 gmail API,它可以让你通过 REST 发送电子邮件、阅读电子邮件和创建草稿。与 SMTP 调用不同,它是非阻塞的,这对于在请求线程中发送电子邮件的基于线程的网络服务器(如 python 网络服务器)来说是一件好事。API 也很强大。

  • 当然,电子邮件应该交给非网络服务器队列,但有选项也不错。

如果您在域上拥有 Google Apps 管理员权限,则设置起来最简单,因为这样您就可以向您的客户授予一揽子权限。否则,您必须摆弄 OAuth 身份验证和权限。

这是一个证明它的要点:

https://gist.github.com/timrichardson/1154e29174926e462b7a

于 2014-10-31T04:12:18.443 回答
2

@David 的回答很好,这里是针对没有通用 try-except 的 Python 3:

def send_email(user, password, recipient, subject, body):

    gmail_user = user
    gmail_pwd = password
    FROM = user
    TO = recipient if type(recipient) is list else [recipient]
    SUBJECT = subject
    TEXT = body

    # Prepare actual message
    message = """From: %s\nTo: %s\nSubject: %s\n\n%s
    """ % (FROM, ", ".join(TO), SUBJECT, TEXT)

    server = smtplib.SMTP("smtp.gmail.com", 587)
    server.ehlo()
    server.starttls()
    server.login(gmail_user, gmail_pwd)
    server.sendmail(FROM, TO, message)
    server.close()
于 2017-08-01T22:56:11.687 回答
2

意识到通过 Python 发送电子邮件有多么痛苦,因此我为它制作了一个广泛的库。它还预先配置了 Gmail(因此您不必记住 Gmail 的主机和端口):

from redmail import gmail
gmail.user_name = "you@gmail.com"
gmail.password = "<YOUR APPLICATION PASSWORD>"

# Send an email
gmail.send(
    subject="An example email",
    receivers=["recipient@example.com"],
    text="Hi, this is text body.",
    html="<h1>Hi, this is HTML body.</h1>"
)

当然你需要配置你的Gmail账户(不用担心,很简单):

  1. 设置两步验证(如果尚未设置)
  2. 创建应用程序密码
  3. 将应用程序密码放入gmail对象并完成!

Red Mail 实际上相当广泛(包括附件、嵌入图像、使用 cc 和 bcc 发送、使用 Jinja 的模板等),并且希望能够满足您对电子邮件发件人的所有需求。它也经过了很好的测试和记录。希望对你有帮助。

安装:

pip install redmail

文档:https ://red-mail.readthedocs.io/en/latest/

源代码:https ://github.com/Miksus/red-mail

请注意,Gmail 不允许更改发件人。发件人地址始终是您。

于 2022-01-03T19:56:58.937 回答
1

似乎是旧的问题smtplib。在python2.7一切工作正常。

更新:是的,server.ehlo()也可以提供帮助。

于 2012-04-13T20:04:01.930 回答
0

2022 年 2 月更新:

尝试2 件事以使用Python发送Gmail

  1. 允许安全性较低的应用:开启↓↓↓</p>

    https://myaccount.google.com/lesssecureapps

  2. 允许访问您的 Google 帐户:(点按“继续”)↓↓↓</p>

    https://accounts.google.com/DisplayUnlockCaptcha

于 2022-02-06T15:30:43.487 回答
-1
    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
于 2016-06-14T18:52:39.793 回答
-3
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()
于 2015-12-04T14:01:16.787 回答