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")

这是一个老问题。我发布新答案的主要原因是解释如何用Python 3.6+中的现代电子邮件库解决问题，以及它与旧版本的区别;但首先，让我们回顾一下Anony-Mousse在2012年的回答。

SMTP根本不关心头文件中有什么。您传递给sendmail方法的收件人列表实际上决定了消息将被传递到哪里。

在SMTP术语中，这称为消息的信封。在协议级别上，您连接到服务器，然后告诉它消息来自谁(MAIL from: SMTP动词)以及将消息发送给谁(RCPT to:)，然后分别传输消息本身(DATA)，其标题和正文作为一个斜向字符串blob。

现代smtplib通过提供send_message方法简化了Python方面的工作，该方法实际发送给消息头中指定的收件人。

现代电子邮件库提供了一个EmailMessage对象，它取代了所有不同的MIME类型，过去您必须使用这些MIME类型从较小的部分组装消息。您可以添加附件，而不需要单独构造它们，如果需要，还可以构建各种更复杂的多部分结构，但通常不必这样做。只需创建一条消息并填充您想要的部分。

注意，下面有大量注释;总的来说，新的EmailMessage API比旧的API更简洁、更通用。

from email.message import EmailMessage

msg = EmailMessage()

# This example uses explicit strings to emphasize that
# that's what these header eventually get turned into
msg["From"] = "me@example.org"
msg["To"] = "main.recipient@example.net, other.main.recipient@example.org"
msg["Cc"] = "secondary@example.com, tertiary@example.eu"
msg["Bcc"] = "invisible@example.int, undisclosed@example.org.au"
msg["Subject"] = "Hello from the other side"

msg.set_content("This is the main text/plain message.")
# You can put an HTML body instead by adding a subtype string argument "html"
# msg.set_content("<p>This is the main text/html message.</p>", "html")

# You can add attachments of various types as you see fit;
# if there are no other parts, the message will be a simple
# text/plain or text/html, but Python will change it into a
# suitable multipart/related or etc if you add more parts
with open("image.png", "rb") as picture:
    msg.add_attachment(picture.read(), maintype="image", subtype="png")

# Which port to use etc depends on the mail server.
# Traditionally, port 25 is SMTP, but modern SMTP MSA submission uses 587.
# Some servers accept encrypted SMTP_SSL on port 465.
# Here, we use SMTP instead of SMTP_SSL, but pivot to encrypted
# traffic with STARTTLS after the initial handshake.
with smtplib.SMTP("smtp.example.org", 587) as server:
    # Some servers insist on this, others are more lenient ...
    # It is technically required by ESMTP, so let's do it
    # (If you use server.login() Python will perform an EHLO first
    # if you haven't done that already, but let's cover all bases)
    server.ehlo()
    # Whether or not to use STARTTLS depends on the mail server
    server.starttls()
    # Bewilderingly, some servers require a second EHLO after STARTTLS!
    server.ehlo()
    # Login is the norm rather than the exception these days
    # but if you are connecting to a local mail server which is
    # not on the public internet, this might not be useful or even possible
    server.login("me.myself@example.org", "xyzzy")

    # Finally, send the message
    server.send_message(msg)

Bcc:标题的最终可见性取决于邮件服务器。如果你真的想确保收件人之间是不可见的，也许根本就不要使用Bcc:头，并且在信封中单独列举信封中的收件人，就像你以前在sendmail中必须做的那样(send_message也允许你这样做，但如果你只是想发送给头中指定的收件人，你就不必这样做)。

This obviously sends a single message to all recipients in one go. That is generally what you should be doing if you are sending the same message to a lot of people. However, if each message is unique, you will need to loop over the recipients and create and send a new message for each. (Merely wishing to put the recipient's name and address in the To: header is probably not enough to warrant sending many more messages than required, but of course, sometimes you have unique content for each recipient in the body, too.)

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")

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()

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。

要向多个收件人发送电子邮件，请将收件人添加为电子邮件id列表。

接收者= ['user1@email.com'， 'user2@email.com'， 'user3@email.com']

import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.image import MIMEImage

smtp_server = 'smtp-example.com'
port = 26 
sender = 'user@email.com'
debuglevel = 0

# add receivers as list of email id string
receivers = ['user1@email.com', 'user2@email.com', 'user3@email.com']

message = MIMEMultipart(
    "mixed", None, [MIMEImage(img_data, 'png'), MIMEText(html,'html')])
    message['Subject'] = "Token Data"
    message['From'] = sender
    message['To'] = ", ".join(receivers)
    try:

        server = smtplib.SMTP('smtp-example.com')
        server.set_debuglevel(1)
        server.sendmail(sender, receivers, message.as_string())
        server.quit()
        # print(response)

    except BaseException:
        print('Error: unable to send email')