经过大量的搜索,我无法找到如何使用smtplib。发送邮件到多个收件人。问题是每次发送邮件时,邮件标题似乎包含多个地址,但实际上只有第一个收件人会收到电子邮件。

问题似乎出在邮件上。Message模块期望与smtplb .sendmail()函数不同的内容。

简而言之,要发送给多个收件人,您应该将标题设置为一串以逗号分隔的电子邮件地址。sendmail()参数to_addr应该是一个电子邮件地址列表。

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

msg = MIMEMultipart()
msg["Subject"] = "Example"
msg["From"] = "me@example.com"
msg["To"] = "malcom@example.com,reynolds@example.com,firefly@example.com"
msg["Cc"] = "serenity@example.com,inara@example.com"
body = MIMEText("example email body")
msg.attach(body)
smtp = smtplib.SMTP("mailhost.example.com", 25)
smtp.sendmail(msg["From"], msg["To"].split(",") + msg["Cc"].split(","), msg.as_string())
smtp.quit()

当前回答

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())

其他回答

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')

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")

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。

这个方法对我没用。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())

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())