这段代码工作,并向我发送电子邮件就好:

import smtplib
#SERVER = "localhost"

FROM = 'monty@python.com'

TO = ["jon@mycompany.com"] # must be a list

SUBJECT = "Hello!"

TEXT = "This message was sent with Python's smtplib."

# Prepare actual message

message = """\
From: %s
To: %s
Subject: %s

%s
""" % (FROM, ", ".join(TO), SUBJECT, TEXT)

# Send the mail

server = smtplib.SMTP('myserver')
server.sendmail(FROM, TO, message)
server.quit()

然而,如果我试图将它包装在这样一个函数中:

def sendMail(FROM,TO,SUBJECT,TEXT,SERVER):
    import smtplib
    """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()

我得到以下错误:

 Traceback (most recent call last):
  File "C:/Python31/mailtest1.py", line 8, in <module>
    sendmail.sendMail(sender,recipients,subject,body,server)
  File "C:/Python31\sendmail.py", line 13, in sendMail
    server.sendmail(FROM, TO, message)
  File "C:\Python31\lib\smtplib.py", line 720, in sendmail
    self.rset()
  File "C:\Python31\lib\smtplib.py", line 444, in rset
    return self.docmd("rset")
  File "C:\Python31\lib\smtplib.py", line 368, in docmd
    return self.getreply()
  File "C:\Python31\lib\smtplib.py", line 345, in getreply
    raise SMTPServerDisconnected("Connection unexpectedly closed")
smtplib.SMTPServerDisconnected: Connection unexpectedly closed

有人能告诉我为什么吗?


当前回答

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

其他回答

使用gmail的另一个实现让我们说:

import smtplib

def send_email(email_address: str, subject: str, body: str):
"""
send_email sends an email to the email address specified in the
argument.

Parameters
----------
email_address: email address of the recipient
subject: subject of the email
body: body of the email
"""

server = smtplib.SMTP('smtp.gmail.com', 587)
server.starttls()
server.login("email_address", "password")
server.sendmail("email_address", email_address,
                "Subject: {}\n\n{}".format(subject, body))
server.quit()

我对发送电子邮件的包选项不满意,我决定制作并开源我自己的电子邮件发送器。它易于使用,并支持高级用例。

如何安装:

pip install redmail

用法:

from redmail import EmailSender
email = EmailSender(
    host="<SMTP HOST ADDRESS>",
    port=<PORT NUMBER>,
)

email.send(
    sender="me@example.com",
    receivers=["you@example.com"],
    subject="An example email",
    text="Hi, this is text body.",
    html="<h1>Hi,</h1><p>this is HTML body</p>"
)

如果您的服务器需要用户和密码,只需将user_name和密码传递给EmailSender。

我在send方法中包含了很多特性:

包含附件 将图像直接包含到HTML主体中 金贾的模板 漂亮的HTML表格开箱即用

文档: https://red-mail.readthedocs.io/en/latest/

源代码:https://github.com/Miksus/red-mail

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。

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。

值得注意的是,SMTP模块支持上下文管理器,因此不需要手动调用quit(),这将确保即使出现异常也始终调用它。

    with smtplib.SMTP_SSL('smtp.gmail.com', 465) as server:
        server.ehlo()
        server.login(user, password)
        server.sendmail(from, to, body)