几个月前我发现了这一点，并在博客上发表了相关文章。总结如下:

如果您想使用smtplib向多个收件人发送电子邮件，请使用email. message。add_header('To'， eachRecipientAsString)来添加它们，然后当您调用sendmail方法时，使用email.Message.get_all('To')将消息发送给所有它们。抄送和密送收件人也是如此。

尝试声明一个包含所有收件人和cc_收件人的列表变量为字符串，而不是循环遍历它们，如下所示:

from email.MIMEMultipart import MIMEMultipart
from email.MIMEText import MIMEText
import smtplib

recipients = ["malcom@example.com","reynolds@example.com", "firefly@example.com"]
cc_recipients=["serenity@example.com", "inara@example.com"]
msg = MIMEMultipart()
msg["Subject"] = "Example"
msg["From"] = "me@example.com"
msg["To"] = ', '.join(recipients)
msg["Cc"] = ', '.join(cc_recipients)
body = MIMEText("example email body")
msg.attach(body)
smtp = smtplib.SMTP("mailhost.example.com", 25)
for recipient in recipients:
    smtp.sendmail(msg["From"], recipient, msg.as_string())
for cc_recipient in cc_recipients:
    smtp.sendmail(msg["From"], cc_recipient, msg.as_string())
smtp.quit()

这真的很管用，我花了很多时间尝试多种变体。

import smtplib
from email.mime.text import MIMEText

s = smtplib.SMTP('smtp.uk.xensource.com')
s.set_debuglevel(1)
msg = MIMEText("""body""")
sender = 'me@example.com'
recipients = ['john.doe@example.com', 'john.smith@example.co.uk']
msg['Subject'] = "subject line"
msg['From'] = sender
msg['To'] = ", ".join(recipients)
s.sendmail(sender, recipients, msg.as_string())

msg['To']需要是一个字符串:

msg['To'] = "a@b.com, b@b.com, c@b.com"

而sendmail中的收件人(sender，收件人，message)需要是一个列表:

sendmail("a@a.com", ["a@b.com", "b@b.com", "c@b.com"], "Howdy")

我尝试了下面的方法，效果很好:)

rec_list =  ['first@example.com', 'second@example.com']
rec =  ', '.join(rec_list)

msg['To'] = rec

send_out = smtplib.SMTP('localhost')
send_out.sendmail(me, rec_list, msg.as_string())

几个月前我发现了这一点，并在博客上发表了相关文章。总结如下:

如果您想使用smtplib向多个收件人发送电子邮件，请使用email. message。add_header('To'， eachRecipientAsString)来添加它们，然后当您调用sendmail方法时，使用email.Message.get_all('To')将消息发送给所有它们。抄送和密送收件人也是如此。