Python 如何获取从smtplib发送的电子邮件的消息id
我想记录用户对我的邮件的回复，并在我的应用程序中将其显示为线程。为此，我在电子邮件头中使用了消息id的帮助。当我发送邮件时，我可以看到屏幕上打印的邮件id如何获取此邮件id。此外，我创建的邮件id也被覆盖。我的代码如下Python 如何获取从smtplib发送的电子邮件的消息id,python,django,email,smtplib,Python,Django,Email,Smtplib,我想记录用户对我的邮件的回复，并在我的应用程序中将其显示为线程。为此，我在电子邮件头中使用了消息id的帮助。当我发送邮件时，我可以看到屏幕上打印的邮件id如何获取此邮件id。此外，我创建的邮件id也被覆盖。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@
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
msg-id = [CFWS] "<" id-left "@" id-right ">" [CFWS]
msg id=[CFWS]“”[CFWS]
消息ID消息ID消息ID=[CFWS]“”[CFWS]