eli*_*eli 31 python email smtplib
我正在使用Python电子邮件和smtplib从Python发送电子邮件.我是通过Gmail SMTP服务器使用我的Gmail凭据执行此操作的.这工作正常,但是我想指定一个与Reply-to地址不同的电子邮件from地址,以便回复转到单独的地址(非Gmail).
我试过reply to像这样创建一个参数:
msg = MIMEMultipart()
msg['From'] = "email@gmail.com"
msg['To'] = to
msg['Subject'] = subject
msg['Reply-to'] = "email2@domain2.com"
但这不起作用.在Python文档中找不到任何关于此的信息.
谢谢.
mat*_*ski 38
这是我的看法.我认为应该明确设置"Reply-To"标题.可能的原因是它比标题更常用,例如"主题","收件人"和"发件人".
python
Python 2.6.6 (r266:84292, May 10 2011, 11:07:28)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> MAIL_SERVER = 'smtp.domain.com'
>>> TO_ADDRESS = 'you@gmail.com'
>>> FROM_ADDRESS = 'email@domain.com'
>>> REPLY_TO_ADDRESS = 'email2@domain2.com'
>>> import smtplib
>>> import email.mime.multipart
>>> msg = email.mime.multipart.MIMEMultipart()
>>> msg['to'] = TO_ADDRESS
>>> msg['from'] = FROM_ADDRESS
>>> msg['subject'] = 'testing reply-to header'
>>> msg.add_header('reply-to', REPLY_TO_ADDRESS)
>>> server = smtplib.SMTP(MAIL_SERVER)
>>> server.sendmail(msg['from'], [msg['to']], msg.as_string())
{}
tur*_*fan 11
我有同样的问题,我必须做的就是将其设置为小写,如下所示:
msg['reply-to'] = "email2@domain2.com"
对于 2021 年的 Python3，我建议使用以下内容来构建消息：
from email.message import EmailMessage
from email.utils import formataddr
msg = EmailMessage()
msg['Subject'] = "Message Subject"
msg['From'] = formataddr(("Sender's Name", "email@gmail.com"))
msg['Reply-To'] = formataddr(("Name of Reply2", "email2@domain2.example"))
msg['To'] = formataddr(("John Smith", "john.smith@gmail.com"))
msg.set_content("""\
<html>
<head></head>
<body>
<p>A simple test email</p>
</body>
</html>
""", subtype='html')
然后，为了发送消息，我在邮件服务器上使用以下命令，该服务器在端口 587 上使用 StartTLS：
from smtplib import SMTP
from ssl import create_default_context as context
with SMTP('smtp.domain.example', 587) as server:
server.starttls(context=context())
server.login('email@domain.example', password)
server.send_message(msg)
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