查看Python FAQ的源代码

===发送MIME邮件附件时，收件人列表注意事项===
#!/usr/bin/env python
#coding = utf-8

import smtplib, mimetypes 

from email.mime.text import MIMEText


from email.mime.multipart import MIMEMultipart   

from email.mime.image import MIMEImage   

msg            = MIMEMultipart()   

msg['From']    = "mail1@163.com" 

#msg['To']     = ["mail1@163.com","mail1@163.com","mail1@163.com"] 会出错 

mail_to        = ["mail1@163.com","mail1@163.com","mail1@163.com"] 

msg['To']      = ', '.join(mail_to) 

msg['Subject'] = "Report " + (time.strftime('%Y-%m-%d %H:%M:%S',time.localtime(time.time()))) 

http://bytes.com/topic/python/answers/472868-sending-emails-list-recipients

The problem is that SMTP.sendmail and email.MIMEText need two different things. email.MIMEText sets up the “To:” header for the body of the e-mail. It is ONLY used for displaying a result to the human being at the other end, and like all e-mail headers, must be a single string. (Note that it does not actually have to have anything to do with the people who actually receive the message.)SMTP.sendmail, on the other hand, sets up the “envelope” of the message for the SMTP protocol. It needs a Python list of strings, each of which has a single address.So, what you need to do is COMBINE the two replies you received. Set msg‘To’ to a single string, but pass the raw list to sendmail:msg['To'] = ', '.join( emails ) s.sendmail( msg['From'], emails, msg.as_string() )