如何使用python smtplib发送电子邮件到多个收件人?

小开

你需要理解电子邮件的可见地址和交付地址之间的区别。

msg["To"]本质上是打印在字母上的内容。它实际上没有任何影响。除了你的电子邮件客户端，就像普通的邮政人员一样，会假设这是你想要发送电子邮件的人。

然而，实际的交付可能会有很大的不同。所以你可以将电子邮件(或副本)放入完全不同的人的邮箱。

这有很多原因。例如转发。To:报头字段在转发时不会改变，但电子邮件会被放入另一个邮箱。

smtp.sendmail命令现在负责实际的传递。email.Message只是信件的内容，而不是投递。

在低级的SMTP中，你需要一个接一个地给接收者，这就是为什么地址列表(不包括名字!)是合理的API。

对于头文件，它还可以包含例如名称，例如To: First Last <email@addr.tld>, Other User <other@mail.tld>。因此，不推荐使用您的代码示例，因为它将无法传递此邮件，因为仅通过在,上拆分它，您仍然没有有效地址!

小开

我在几个月前弄清楚了这个在博客上写过。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，use email.Message.get_all('To')将消息发送给所有收件人。抄送和密送收件人也是如此。

小开

这个真正起作用，我花了很多时间尝试多个变体。

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())

小开

我提出了这个可导入模块函数。本例中使用的邮件服务器为gmail。它分为头和消息，所以你可以清楚地看到发生了什么:

import smtplib


def send_alert(subject=""):


to = ['email@one.com', 'email2@another_email.com', 'a3rd@email.com']
gmail_user = 'me@gmail.com'
gmail_pwd = 'my_pass'
smtpserver = smtplib.SMTP("smtp.gmail.com", 587)
smtpserver.ehlo()
smtpserver.starttls()
smtpserver.ehlo
smtpserver.login(gmail_user, gmail_pwd)
header = 'To:' + ", ".join(to) + '\n' + 'From: ' + gmail_user + '\n' + 'Subject: ' + subject + '\n'
msg = header + '\n' + subject + '\n\n'
smtpserver.sendmail(gmail_user, to, msg)
smtpserver.close()

小开

msg['To']需要是一个字符串:

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

而sendmail(sender, recipients, message)中的recipients需要是一个列表:

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

小开

这对我很管用。

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'] = recipients
s.sendmail(sender, recipients.split(','), msg.as_string())

小开

好吧，这个说方法中的方法对我不起作用。我不知道，也许这是一个Python3(我使用3.4版本)或gmail相关的问题，但经过一些尝试，对我有效的解决方案是行

s.send_message(msg)

而不是

s.sendmail(sender, recipients, msg.as_string())

小开

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())

小开

您可以在将收件人电子邮件写入文本文件时尝试此操作

from email.mime.text import MIMEText
from email.header import Header
import smtplib


f =  open('emails.txt', 'r').readlines()
for n in f:
emails = n.rstrip()
server = smtplib.SMTP('smtp.uk.xensource.com')
server.ehlo()
server.starttls()
body = "Test Email"
subject = "Test"
from = "me@example.com"
to = emails
msg = MIMEText(body,'plain','utf-8')
msg['Subject'] = Header(subject, 'utf-8')
msg['From'] =  Header(from, 'utf-8')
msg['To'] = Header(to, 'utf-8')
text = msg.as_string()
try:
server.send(from, emails, text)
print('Message Sent Succesfully')
except:
print('There Was An Error While Sending The Message')

小开

实际上问题在于SMTP。发送邮件和电子邮件。MIMEText需要两个不同的东西。

电子邮件。MIMEText为电子邮件正文设置了“To:”标头。它仅用于向另一端的人显示结果，并且像所有电子邮件标题一样，必须是单个字符串。(请注意，它实际上不必与实际接收消息的人有任何关系。)

SMTP。另一方面，sendmail为SMTP协议设置消息的“信封”。它需要一个Python字符串列表，每个字符串都有一个地址。

所以，你需要做的就是将收到的两个回复结合起来。将msg['To']设置为单个字符串，但将原始列表传递给sendmail:

emails = ['a.com','b.com', 'c.com']
msg['To'] = ', '.join( emails )
....
s.sendmail( msg['From'], emails, msg.as_string())

小开

我使用python 3.6和以下代码为我工作

email_send = 'xxxxx@xxx.xxx,xxxx@xxx.xxx'
server.sendmail(email_user,email_send.split(','),text)

小开

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


def sender(recipients):


body = 'Your email content here'
msg = MIMEMultipart()


msg['Subject'] = 'Email Subject'
msg['From'] = 'your.email@gmail.com'
msg['To'] = (', ').join(recipients.split(','))


msg.attach(MIMEText(body,'plain'))


server = smtplib.SMTP('smtp.gmail.com', 587)
server.starttls()
server.login('your.email@gmail.com', 'yourpassword')
server.send_message(msg)
server.quit()


if __name__ == '__main__':
sender('email_1@domain.com,email_2@domain.com')

它只适用于我send_message函数和使用列表中的连接函数与收件人，python 3.6。

小开

下面的解决方案对我很有效。它成功地向多个收件人发送电子邮件，包括“抄送”和“密件抄送”。

toaddr = ['mailid_1','mailid_2']
cc = ['mailid_3','mailid_4']
bcc = ['mailid_5','mailid_6']
subject = 'Email from Python Code'
fromaddr = 'sender_mailid'
message = "\n  !! Hello... !!"


msg['From'] = fromaddr
msg['To'] = ', '.join(toaddr)
msg['Cc'] = ', '.join(cc)
msg['Bcc'] = ', '.join(bcc)
msg['Subject'] = subject


s.sendmail(fromaddr, (toaddr+cc+bcc) , message)

小开

这里有很多答案在技术上或部分上是正确的。在阅读了每个人的答案后，我想出了这个更可靠/通用的电子邮件功能。我已经确认它的工作，你可以通过HTML或纯文本的主体。注意，此代码不包括附件代码:

import smtplib
import socket


# Import the email modules we'll need
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart


#
# @param [String] email_list
# @param [String] subject_line
# @param [String] error_message
def sendEmailAlert(email_list="default@email.com", subject_line="Default Subject", error_message="Default Error Message"):
hostname = socket.gethostname()
# Create message
msg = MIMEMultipart()
msg['Subject'] = subject_line
msg['From'] = f'no-reply@{hostname}'
msg['To'] = email_list
msg.attach(MIMEText(error_message, 'html'))
# Send the message via SMTP server
s = smtplib.SMTP('localhost') # Change for remote mail server!
# Verbose debugging
s.set_debuglevel(2)
try:
s.sendmail(msg['From'], msg['To'].split(","), msg.as_string())
except Exception as e:
print(f'EMAIL ISSUE: {e}')
s.quit()

这显然可以修改为使用本机Python日志记录。我只是提供了一个坚实的核心功能。我再怎么强调也不为过，sendmail()想要一个List而不是String!函数适用于Python3.6+

小开

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()

小开

这是一个老问题。我发布新答案的主要原因是解释如何在Python 3.6+中使用现代email库解决问题，以及它与旧版本有何不同;但首先，让我们回顾一下Anony-Mousse在他们的答案来自2012年。中所写的内容

SMTP根本不关心头文件中有什么。你传递给sendmail方法的接收者列表是实际上决定消息将被传递到哪里的对象。

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)和发送给谁(RCPT TO:)，然后分别传输消息本身(DATA)，其标题和正文作为一个斜字符串blob。

现代的smtplib通过提供一个send_message方法来简化Python方面的工作，该方法实际发送给消息头中指定的接收者。

现代的email库提供了一个EmailMessage对象，它替换了所有不同的MIME类型，过去你必须使用这些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:标头的最终可见性取决于邮件服务器。如果你想要真的，确保收件人彼此不可见，也许根本就不要放Bcc:标头，并在信封中单独枚举信封收件人，就像你过去必须使用sendmail那样(send_message也可以让你这样做，但如果你只是想发送给标头中指定的收件人，你就不必这样做)。

这显然是一次性向所有收件人发送一条消息。如果你要向很多人发送相同的信息，这通常是你应该做的。但是，如果每个消息都是唯一的，则需要遍历收件人并为每个收件人创建和发送新消息。(仅仅希望将收件人的姓名和地址放在To:报头中可能不足以保证发送比所需更多的消息，但当然，有时你也会在主体中为每个收件人提供唯一的内容。)

小开

对于那些希望只发送一个“to”报头的消息，下面的代码可以解决这个问题。确保你的接收者变量是一个字符串列表。

# Create message container - the correct MIME type is multipart/alternative.
msg = MIMEMultipart('alternative')
msg['Subject'] = title
msg['From'] = f'support@{config("domain_base")}'
msg['To'] =  "me"
message_content += f"""
<br /><br />
Regards,<br />
Company Name<br />
The {config("domain_base")} team
"""
body = MIMEText(message_content, 'html')
msg.attach(body)


try:
smtpObj = smtplib.SMTP('localhost')
for r in receivers:
del msg['To']
msg['To'] =  r #"Customer /n" + r
smtpObj.sendmail(f"support@{config('domain_base')}", r, msg.as_string())
smtpObj.quit()
return {"message": "Successfully sent email"}
except smtplib.SMTPException:
return {"message": "Error: unable to send email"}

小开

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。

receivers = ['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')