如何获取通过smtplib发送的邮件的消息ID
我想记录用户对我邮件的回复,并在我的应用程序中以线程的形式展示出来。为此,我使用了邮件头中的消息ID。当我发送邮件时,可以看到消息ID在屏幕上打印出来,我该如何获取这个消息ID呢?另外,我自己创建的消息ID被覆盖了。我的代码如下。
import smtplib
from email.mime.text import MIMEText
subject = 'Hello!'
message = 'hiii!!!'
email = 'someone@somewhere.com'
send_from = 'me@example.com'
msg = MIMEText(message, 'html', 'utf-8')
msg['Subject'] = subject
msg['From'] = send_from
msg['To'] = email
msg['Message-ID'] = '01234567890123456789abcdefghijklmnopqrstuvwxyz'
send_to = [email]
smtp_server = 'email-smtp.us-east-1.amazonaws.com'
smtp_port = 587
user_name = 'abcd'
password = 'abcd'
try:
server = smtplib.SMTP(smtp_server, smtp_port)
server.set_debuglevel(True)
server.starttls()
server.ehlo()
server.login(user_name,password)
server.sendmail(send_from, send_to, msg.as_string())
except Exception, e:
print e
2 个回答
5
我觉得上面的回答非常让人困惑。希望下面的内容能帮助到其他人:
import smtplib
import email.message
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.base import MIMEBase
from email import encoders
import email.utils as utils
def send_html_email(subject, msg_text,
toaddrs=['foo@gmail.com']):
fromaddr = 'me@mydomain.com'
msg = "\r\n".join([
"From: " + fromaddr,
"To: " + ",".join(toaddrs),
"Subject: " + subject,
"",
msg_text
])
msg = email.message.Message()
msg['message-id'] = utils.make_msgid(domain='mydomain.com')
msg['Subject'] = subject
msg['From'] = fromaddr
msg['To'] = ",".join(toaddrs)
msg.add_header('Content-Type', 'text/html')
msg.set_payload(msg_text)
username = fromaddr
password = 'MyGreatPassword'
server = smtplib.SMTP('mail.mydomain.com',25)
#server.ehlo() <- not required for my domain.
server.starttls()
server.login(username, password)
server.sendmail(fromaddr, toaddrs, msg.as_string())
server.quit()
5
使用 email.utils.make_msgid
来创建符合RFC 2822标准的消息ID头部:
msg-id = [CFWS] "<" id-left "@" id-right ">" [CFWS]