在摆弄了很多例子之后，例如这里 这对我来说很管用:

import smtplib
from email.mime.text import MIMEText

# SMTP sendmail server mail relay
host = 'mail.server.com'
port = 587 # starttls not SSL 465 e.g gmail, port 25 blocked by most ISPs & AWS
sender_email = 'name@server.com'
recipient_email = 'name@domain.com'
password = 'YourSMTPServerAuthenticationPass'
subject = "Server - "
body = "Message from server"

def sendemail(host, port, sender_email, recipient_email, password, subject, body):
    try:
        p1 = f'<p><HR><BR>{recipient_email}<BR>'
        p2 = f'<h2><font color="green">{subject}</font></h2>'
        p3 = f'<p>{body}'
        p4 = f'<p>Kind Regards,<BR><BR>{sender_email}<BR><HR>'
        
        message = MIMEText((p1+p2+p3+p4), 'html')  
        # servers may not accept non RFC 5321 / RFC 5322 / compliant TXT & HTML typos

        message['From'] = f'Sender Name <{sender_email}>'
        message['To'] = f'Receiver Name <{recipient_email}>'
        message['Cc'] = f'Receiver2 Name <>'
        message['Subject'] = f'{subject}'
        msg = message.as_string()

        server = smtplib.SMTP(host, port)
        print("Connection Status: Connected")
        server.set_debuglevel(1)
        server.ehlo()
        server.starttls()
        server.ehlo()
        server.login(sender_email, password)
        print("Connection Status: Logged in")
        server.sendmail(sender_email, recipient_email, msg)
        print("Status: Email as HTML successfully sent")

    except Exception as e:
            print(e)
            print("Error: unable to send email")

# Run
sendemail(host, port, sender_email, recipient_email, password, subject, body)
print("Status: Exit")

在缩进函数中的代码时(这是可以的)，还缩进了原始消息字符串的行。但是前导空白意味着标题行的折叠(连接)，如RFC 2822 - Internet Message Format的2.2.3和3.2.3节所述:

每个报头字段在逻辑上是由一行字符组成的 字段名、冒号和字段主体。为了方便 但是，为了处理每行998/78个字符的限制， 报头字段的字段主体部分可以分成多个 线表示;这叫做“折叠”。

在sendmail调用的函数形式中，所有行都以空白开始，因此是“展开的”(连接)，您正在尝试发送

From: monty@python.com    To: jon@mycompany.com    Subject: Hello!    This message was sent with Python's smtplib.

与我们的想法不同，smtplib将不再理解To:和Subject:头文件，因为这些名称只在一行的开头被识别。相反，smtplib将假设一个非常长的发送者电子邮件地址:

monty@python.com    To: jon@mycompany.com    Subject: Hello!    This message was sent with Python's smtplib.

这将不起作用，因此出现异常。

解决方案很简单:只保留原来的消息字符串。这可以通过一个函数来完成(正如Zeeshan建议的那样)，也可以直接在源代码中完成:

import smtplib

def sendMail(FROM,TO,SUBJECT,TEXT,SERVER):
    """this is some test documentation in the function"""
    message = """\
From: %s
To: %s
Subject: %s

%s
""" % (FROM, ", ".join(TO), SUBJECT, TEXT)
    # Send the mail
    server = smtplib.SMTP(SERVER)
    server.sendmail(FROM, TO, message)
    server.quit()

现在展开没有发生，你发送

From: monty@python.com
To: jon@mycompany.com
Subject: Hello!

This message was sent with Python's smtplib.

这就是您的旧代码所做的工作。

请注意，我还保留了标题和正文之间的空行，以适应RFC的第3.5节(这是必需的)，并根据Python风格指南PEP-0008(这是可选的)将include放在函数之外。

我建议您使用标准包email和smtplib一起发送电子邮件。请看下面的例子(摘自Python文档)。注意，如果遵循这种方法，“简单”任务确实很简单，而更复杂的任务(如附加二进制对象或发送纯/HTML多部分消息)将很快完成。

# Import smtplib for the actual sending function
import smtplib

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

# Open a plain text file for reading.  For this example, assume that
# the text file contains only ASCII characters.
with open(textfile, 'rb') as fp:
    # Create a text/plain message
    msg = MIMEText(fp.read())

# me == the sender's email address
# you == the recipient's email address
msg['Subject'] = 'The contents of %s' % textfile
msg['From'] = me
msg['To'] = you

# Send the message via our own SMTP server, but don't include the
# envelope header.
s = smtplib.SMTP('localhost')
s.sendmail(me, [you], msg.as_string())
s.quit()

要将电子邮件发送到多个目的地，您还可以遵循Python文档中的示例:

# Import smtplib for the actual sending function
import smtplib

# Here are the email package modules we'll need
from email.mime.image import MIMEImage
from email.mime.multipart import MIMEMultipart

# Create the container (outer) email message.
msg = MIMEMultipart()
msg['Subject'] = 'Our family reunion'
# me == the sender's email address
# family = the list of all recipients' email addresses
msg['From'] = me
msg['To'] = ', '.join(family)
msg.preamble = 'Our family reunion'

# Assume we know that the image files are all in PNG format
for file in pngfiles:
    # Open the files in binary mode.  Let the MIMEImage class automatically
    # guess the specific image type.
    with open(file, 'rb') as fp:
        img = MIMEImage(fp.read())
    msg.attach(img)

# Send the email via our own SMTP server.
s = smtplib.SMTP('localhost')
s.sendmail(me, family, msg.as_string())
s.quit()

如您所见，MIMEText对象中的报头To必须是由逗号分隔的电子邮件地址组成的字符串。另一方面，sendmail函数的第二个参数必须是一个字符串列表(每个字符串都是一个电子邮件地址)。

所以，如果你有三个电子邮件地址:person1@example.com, person2@example.com和person3@example.com，你可以这样做(明显的部分省略了):

to = ["person1@example.com", "person2@example.com", "person3@example.com"]
msg['To'] = ",".join(to)
s.sendmail(me, to, msg.as_string())

"，".join(to)部分从列表中生成一个单独的字符串，用逗号分隔。

从你的问题中，我猜你还没有读过Python教程——如果你想在Python中有所了解，这是必须的——标准库的文档大部分都很出色。

有压痕问题。下面的代码将工作:

import textwrap

def sendMail(FROM,TO,SUBJECT,TEXT,SERVER):
    import smtplib
    """this is some test documentation in the function"""
    message = textwrap.dedent("""\
        From: %s
        To: %s
        Subject: %s
        %s
        """ % (FROM, ", ".join(TO), SUBJECT, TEXT))
    # Send the mail
    server = smtplib.SMTP(SERVER)
    server.sendmail(FROM, TO, message)
    server.quit()

下面是Python 3的一个例子。X，比2.x简单得多:

import smtplib
from email.message import EmailMessage
def send_mail(to_email, subject, message, server='smtp.example.cn',
              from_email='xx@example.com'):
    # import smtplib
    msg = EmailMessage()
    msg['Subject'] = subject
    msg['From'] = from_email
    msg['To'] = ', '.join(to_email)
    msg.set_content(message)
    print(msg)
    server = smtplib.SMTP(server)
    server.set_debuglevel(1)
    server.login(from_email, 'password')  # user & password
    server.send_message(msg)
    server.quit()
    print('successfully sent the mail.')

调用这个函数:

send_mail(to_email=['12345@qq.com', '12345@126.com'],
          subject='hello', message='Your analysis has done!')

以下仅限中国用户使用:

If you use 126/163, NetEase Mail, you need to set" Client Authorization Password ", like below:

裁判:https://stackoverflow.com/a/41470149/2803344 https://docs.python.org/3/library/email.examples.html#email-examples

它可能会在你的信息中添加标签。在你把它传递给sendMail之前打印出消息。