如何用Python发送电子邮件？

Question

此代码可以正常工作并向我发送电子邮件:

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()

但是,如果我尝试将其包装在这样的函数中:

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()

并调用它我得到以下错误:

 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

谁能帮我理解为什么？

Answer 1

我建议您使用标准软件包email并smtplib一起发送电子邮件.请查看以下示例(从Python文档中再现).请注意,如果您遵循此方法,"简单"任务确实很简单,并且更复杂的任务(如附加二进制对象或发送普通/ HTML多部分消息)可以非常快速地完成.

# 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标,您还可以按照Python文档中的示例进行操作:

# 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()

正如你所看到的,头部To的MIMEText对象必须是由用逗号分隔的电子邮件地址的字符串.另一方面,sendmail函数的第二个参数必须是字符串列表(每个字符串都是一个电子邮件地址).

因此,如果您有三个电子邮件地址:person1@example.com,person2@example.com和person3@example.com,您可以执行以下操作(省略明显的部分):

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

该",".join(to)部分从列表中生成一个单独的字符串,以逗号分隔.

从你的问题我收集到你没有经过Python教程 - 如果你想在Python中的任何地方获取它是必须的 - 文档对于标准库来说非常好.

Answer 2

那么,你想要一个最新和现代的答案.

这是我的答案:

当我需要在python中邮寄时,我会使用mailgun API,因为发送邮件已经解决了很多麻烦.他们有一个很棒的应用程序/ api,允许您每月免费发送10,000封电子邮件.

发送电子邮件将是这样的:

def send_simple_message():
    return requests.post(
        "https://api.mailgun.net/v3/YOUR_DOMAIN_NAME/messages",
        auth=("api", "YOUR_API_KEY"),
        data={"from": "Excited User <mailgun@YOUR_DOMAIN_NAME>",
              "to": ["bar@example.com", "YOU@YOUR_DOMAIN_NAME"],
              "subject": "Hello",
              "text": "Testing some Mailgun awesomness!"})

您还可以跟踪事件和更多内容,请参阅快速入门指南.

希望这个对你有帮助!

Answer 3

我想通过建议yagmail包来帮助你发送电子邮件(我是维护者,抱歉广告,但我觉得它真的有帮助!).

你的整个代码是:

import yagmail
yag = yagmail.SMTP(FROM, 'pass')
yag.send(TO, SUBJECT, TEXT)

请注意,我提供了所有参数的默认值,例如,如果您想发送给自己,您可以省略TO,如果您不想要主题,也可以省略它.

此外,目标还在于使附加HTML代码或图像(以及其他文件)变得非常容易.

你放置内容的地方你可以这样做:

contents = ['Body text, and here is an embedded image:', 'http://somedomain/image.png',
            'You can also find an audio file attached.', '/local/path/song.mp3']

哇,发送附件是多么容易!这将需要20行没有yagmail;)

此外,如果您设置一次,您将永远不必再次输入密码(并安全地存储).在您的情况下,您可以执行以下操作:

import yagmail
yagmail.SMTP().send(contents = contents)

这更简洁!

我邀请您查看github或直接安装它pip install yagmail.

Answer 4

有缩进问题.以下代码将起作用:

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()

Answer 5

这是一个关于 Python 的示例3.x，比 简单得多2.x：

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!')

以下可能仅适用于中国用户：

如果你使用126/163，????，你需要设置“???????”，如下图：

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

Answer 6

试试这个:

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)
    "New part"
    server.starttls()
    server.login('username', 'password')
    server.sendmail(FROM, TO, message)
    server.quit()

它适用于smtp.gmail.com

Answer 7

确保您已授予发件人和收件人发送电子邮件以及从电子邮件帐户中的未知来源（外部来源）接收电子邮件的权限。

import smtplib

#Ports 465 and 587 are intended for email client to email server communication - sending email
server = smtplib.SMTP('smtp.gmail.com', 587)

#starttls() is a way to take an existing insecure connection and upgrade it to a secure connection using SSL/TLS.
server.starttls()

#Next, log in to the server
server.login("#email", "#password")

msg = "Hello! This Message was sent by the help of Python"

#Send the mail
server.sendmail("#Sender", "#Reciever", msg)

Answer 8

在函数中缩进您的代码（这没问题）时，您还缩进了原始消息字符串的行。但是前导空格意味着标题行的折叠（串联），如RFC 2822 - Internet Message Format 的第 2.2.3 和 3.2.3 节所述：

每个标题字段在逻辑上都是一行字符，包括字段名称、冒号和字段主体。然而，为了方便并处理每行 998/78 个字符的限制，标题字段的字段主体部分可以拆分为多行表示；这称为“折叠”。

在您sendmail调用的函数形式中，所有行都以空格开头，因此“展开”（连接）并且您正在尝试发送

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

除了我们的想法之外，smtplib将不再理解To:和Subject:标题，因为这些名称仅在一行的开头被识别。相反，smtplib将假设一个很长的发件人电子邮件地址：

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

这行不通，所以出现了您的异常。

解决方案很简单：只需保留message字符串原样即可。这可以通过函数（如 Zeeshan 建议的那样）或立即在源代码中完成：

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()

现在展开不会发生，你发送

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()

这是什么工作以及您的旧代码完成了什么。

请注意，我还保留了 headers 和 body 之间的空行以适应RFC 的第 3.5 节（这是必需的），并根据 Python 样式指南PEP-0008（这是可选的）将包含放在函数之外。