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

就你的代码而言,它似乎没有任何根本性的错误,除了,不清楚你实际上是如何调用这个函数的。我能想到的是,当您的服务器没有响应时,您将得到这个SMTPServerDisconnected错误。如果您查找smtplib中的getreply()函数(摘自下面),您将得到一个概念。

def getreply(self):
    """Get a reply from the server.

    Returns a tuple consisting of:

      - server response code (e.g. '250', or such, if all goes well)
        Note: returns -1 if it can't read response code.

      - server response string corresponding to response code (multiline
        responses are converted to a single, multiline string).

    Raises SMTPServerDisconnected if end-of-file is reached.
    """

查看https://github.com/rreddy80/sendEmails/blob/master/sendEmailAttachments.py上的一个例子,它也使用了一个函数调用来发送电子邮件,如果这就是你想要做的(DRY方法)。

其他回答

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()

当我需要在Python中发送邮件时,我使用mailgun API,它在发送邮件时遇到了很多麻烦。他们有一个很棒的应用程序/api,可以让你每月发送5000封免费电子邮件。

发送电子邮件是这样的:

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信息,参见快速入门指南。

只是为了补充答案,以便您的邮件传递系统可以扩展。

我建议有一个配置文件(可以是.json, .yml, .ini等),包含发件人的电子邮件配置,密码和收件人。

通过这种方式,您可以根据需要创建不同的可定制项目。

下面是一个包含3个文件,config, functions和main的小示例。纯文本邮件。

config_email.ini

[email_1]
sender = test@test.com
password = XXXXXXXXXXX
recipients= ["email_2@test.com", "email_2@test.com"]

[email_2]
sender = test_2@test.com
password = XXXXXXXXXXX
recipients= ["email_2@test.com", "email_2@test.com", "email_3@test.com"]

这些项将从main.py调用,它将返回它们各自的值。

函数functions_email.py文件:

import smtplib,configparser,json
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText

def get_credentials(item):
    parse = configparser.ConfigParser()
    parse.read('config_email.ini')
    sender = parse[item]['sender ']
    password = parse[item]['password']
    recipients= json.loads(parse[item]['recipients'])
    return sender,password,recipients

def get_msg(sender,recipients,subject,mail_body):
    msg = MIMEMultipart()
    msg['Subject'] = subject
    msg['From'] = sender 
    msg['To'] = ', '.join(recipients)       
    text = """\
    """+mail_body+""" """
    part1 = MIMEText(text, "plain")
    msg.attach(part1)
    return msg

def send_email(msg,sender,password,recipients):
    s = smtplib.SMTP('smtp.test.com')
    s.login(sender,password)
    s.sendmail(sender, recipients, msg.as_string())
    s.quit()    

文件main.py:

from functions_email import *

sender,password,recipients = get_credenciales('email_2')
subject= 'text to subject'
mail_body = 'body....................'
msg = get_msg(sender,recipients ,subject,mail_body)    
send_email(msg,sender,password,recipients)

最好的问候!

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。

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,但我觉得它真的能帮助!)

你的整个代码将是:

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

注意,我为所有参数提供了默认值,例如,如果你想发送给自己,你可以省略to,如果你不想要一个主题,你也可以省略它。

此外,我们的目标还在于使附加html代码或图像(以及其他文件)变得非常容易。

在你放置内容的地方,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,发送附件是多么简单啊!如果没有yagmail,这大概需要20行;)

此外,如果你设置了一次,你就永远不必再输入密码(并安全地保存密码)。在你的情况下,你可以这样做:

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

这样更简洁!

我建议你看看github,或者直接用pip install yagmail安装它。