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。我也不能强调这一点，sendmail()需要一个列表，而不是一个字符串!函数适用于Python3.6+

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())

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')

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()

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())

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())