如何发送电子邮件附件？

Question

我在理解如何使用Python向附件发送电子邮件时遇到问题.我已成功通过电子邮件发送简单邮件smtplib.有人可以解释如何在电子邮件中发送附件.我知道网上还有其他帖子,但作为Python初学者,我觉得很难理解.

Answer 1

这是另一个:

import smtplib
from os.path import basename
from email.mime.application import MIMEApplication
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.utils import COMMASPACE, formatdate


def send_mail(send_from, send_to, subject, text, files=None,
              server="127.0.0.1"):
    assert isinstance(send_to, list)

    msg = MIMEMultipart()
    msg['From'] = send_from
    msg['To'] = COMMASPACE.join(send_to)
    msg['Date'] = formatdate(localtime=True)
    msg['Subject'] = subject

    msg.attach(MIMEText(text))

    for f in files or []:
        with open(f, "rb") as fil:
            part = MIMEApplication(
                fil.read(),
                Name=basename(f)
            )
        # After the file is closed
        part['Content-Disposition'] = 'attachment; filename="%s"' % basename(f)
        msg.attach(part)


    smtp = smtplib.SMTP(server)
    smtp.sendmail(send_from, send_to, msg.as_string())
    smtp.close()

它与第一个例子大致相同......但它应该更容易投入.

Answer 2

这是我最终使用的代码:

import smtplib
from email.MIMEMultipart import MIMEMultipart
from email.MIMEBase import MIMEBase
from email import Encoders


SUBJECT = "Email Data"

msg = MIMEMultipart()
msg['Subject'] = SUBJECT 
msg['From'] = self.EMAIL_FROM
msg['To'] = ', '.join(self.EMAIL_TO)

part = MIMEBase('application', "octet-stream")
part.set_payload(open("text.txt", "rb").read())
Encoders.encode_base64(part)

part.add_header('Content-Disposition', 'attachment; filename="text.txt"')

msg.attach(part)

server = smtplib.SMTP(self.EMAIL_SERVER)
server.sendmail(self.EMAIL_FROM, self.EMAIL_TO, msg.as_string())

代码与Oli的帖子大致相同.谢谢大家

代码基于二进制文件电子邮件附件问题的帖子.

Answer 3

这是Olipython 3 的修改版本

import smtplib
import os.path as op
from email.mime.multipart import MIMEMultipart
from email.mime.base import MIMEBase
from email.mime.text import MIMEText
from email.utils import COMMASPACE, formatdate
from email import encoders


def send_mail(send_from, send_to, subject, message, files=[],
              server="localhost", port=587, username='', password='',
              use_tls=True):
    """Compose and send email with provided info and attachments.

    Args:
        send_from (str): from name
        send_to (str): to name
        subject (str): message title
        message (str): message body
        files (list[str]): list of file paths to be attached to email
        server (str): mail server host name
        port (int): port number
        username (str): server auth username
        password (str): server auth password
        use_tls (bool): use TLS mode
    """
    msg = MIMEMultipart()
    msg['From'] = send_from
    msg['To'] = COMMASPACE.join(send_to)
    msg['Date'] = formatdate(localtime=True)
    msg['Subject'] = subject

    msg.attach(MIMEText(message))

    for path in files:
        part = MIMEBase('application', "octet-stream")
        with open(path, 'rb') as file:
            part.set_payload(file.read())
        encoders.encode_base64(part)
        part.add_header('Content-Disposition',
                        'attachment; filename="{}"'.format(op.basename(path)))
        msg.attach(part)

    smtp = smtplib.SMTP(server, port)
    if use_tls:
        smtp.starttls()
    smtp.login(username, password)
    smtp.sendmail(send_from, send_to, msg.as_string())
    smtp.quit()

Answer 4

from email.MIMEMultipart import MIMEMultipart
from email.MIMEText import MIMEText
from email.MIMEImage import MIMEImage
import smtplib

msg = MIMEMultipart()
msg.attach(MIMEText(file("text.txt").read()))
msg.attach(MIMEImage(file("image.png").read()))

# to send
mailer = smtplib.SMTP()
mailer.connect()
mailer.sendmail(from_, to, msg.as_string())
mailer.close()

改编自这里.

Answer 5

因为这里有很多关于 Python 3 的答案，但没有一个显示如何使用emailPython 3.6 中的大修库，所以这里是当前示例文档中的快速复制+粘贴email。 （我对它进行了一些删节，以消除诸如猜测正确的 MIME 类型之类的多余内容。）

针对 Python >3.5 的现代代码不应再使用email.message.MessageAPI（包括各种MIMEText、MIMEMultipart等MIMEBase类）或更旧的mimetypes胡言乱语。

from email.message import EmailMessage
import smtplib
from pathlib import Path

msg = EmailMessage()
msg["Subject"] = "Our family reunion"
msg["From"] = "me <sender@example.org>"
msg["To"] = "recipient <victim@example.net>"
# definitely don't mess with the .preamble

msg.set_content("Hello, victim! Look at these pictures")

picpath = Path("path/to/attachment.png")
with picpath.open("rb") as fp:
    msg.add_attachment(
        fp.read(),
        maintype="image", subtype="png",
        filename=picpath.name)

# Notice how smtplib now includes a send_message() method
with smtplib.SMTP("localhost") as s:
    s.send_message(msg)

现在，现代email.message.EmailMessageAPI 比旧版本的库更加通用和合乎逻辑。文档中的演示仍然存在一些问题（Content-Disposition:例如，如何更改附件的格式并不明显；并且该policy模块的讨论对于大多数新手来说可能太晦涩），从根本上讲，您仍然需要一些知识MIME 结构应该是什么样子的想法（尽管该库现在终于处理了围绕它的许多细节）。也许请参阅多部分电子邮件中的“部分”是什么？进行简单介绍。

localhost显然，仅当您的本地计算机上实际运行着 SMTP 服务器时，才可以将其用作 SMTP 服务器。正确地从系统中获取电子邮件是一个相当复杂的独立问题。对于简单的要求，可以使用您现有的电子邮件帐户和提供商的电子邮件服务器（搜索使用 Google、Yahoo 或任何您拥有的端口 587 的示例 - 确切的工作原理在某种程度上取决于提供商；有些仅支持端口 465，或者旧端口 25 但现在由于垃圾邮件过滤的原因基本上不可能在面向公众的服务器上使用）。

Answer 6

Gmail版本,使用Python 3.6(请注意,您需要更改Gmail设置才能通过smtp发送电子邮件:

import smtplib
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart
from email.mime.application import MIMEApplication
from os.path import basename


def send_mail(send_from: str, subject: str, text: str, 
send_to: list, files= None):

    send_to= default_address if not send_to else send_to

    msg = MIMEMultipart()
    msg['From'] = send_from
    msg['To'] = ', '.join(send_to)  
    msg['Subject'] = subject

    msg.attach(MIMEText(text))

    for f in files or []:
        with open(f, "rb") as fil: 
            ext = f.split('.')[-1:]
            attachedfile = MIMEApplication(fil.read(), _subtype = ext)
            attachedfile.add_header(
                'content-disposition', 'attachment', filename=basename(f) )
        msg.attach(attachedfile)


    smtp = smtplib.SMTP(host="smtp.gmail.com", port= 587) 
    smtp.starttls()
    smtp.login(username,password)
    smtp.sendmail(send_from, send_to, msg.as_string())
    smtp.close()

用法:

username = 'my-address@gmail.com'
password = 'top-secret'
default_address = ['my-address2@gmail.com'] 

send_mail(send_from= username,
subject="test",
text="text",
send_to= None,
files= # pass a list with the full filepaths here...
)

要与任何其他电子邮件提供商一起使用,只需更改smtp配置即可.

Answer 7

我能得到的最简单的代码是：

#for attachment email
from django.core.mail import EmailMessage

    def attachment_email(request):
            email = EmailMessage(
            'Hello', #subject
            'Body goes here', #body
            'MyEmail@MyEmail.com', #from
            ['SendTo@SendTo.com'], #to
            ['bcc@example.com'], #bcc
            reply_to=['other@example.com'],
            headers={'Message-ID': 'foo'},
            )

            email.attach_file('/my/path/file')
            email.send()

它基于官方的Django 文档

Answer 8

其他答案非常好，但我仍然想分享一种不同的方法，以防有人正在寻找替代方案。

这里的主要区别在于，使用这种方法，您可以使用 HTML/CSS 来格式化您的消息，这样您就可以发挥创意并为您的电子邮件添加一些样式。尽管您没有被强制使用 HTML，但您仍然可以仅使用纯文本。

请注意，此功能接受将电子邮件发送给多个收件人，还允许附加多个文件。

我只在 Python 2 上尝试过这个，但我认为它在 3 上也应该可以正常工作：

import os.path
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.application import MIMEApplication

def send_email(subject, message, from_email, to_email=[], attachment=[]):
    """
    :param subject: email subject
    :param message: Body content of the email (string), can be HTML/CSS or plain text
    :param from_email: Email address from where the email is sent
    :param to_email: List of email recipients, example: ["a@a.com", "b@b.com"]
    :param attachment: List of attachments, exmaple: ["file1.txt", "file2.txt"]
    """
    msg = MIMEMultipart()
    msg['Subject'] = subject
    msg['From'] = from_email
    msg['To'] = ", ".join(to_email)
    msg.attach(MIMEText(message, 'html'))

    for f in attachment:
        with open(f, 'rb') as a_file:
            basename = os.path.basename(f)
            part = MIMEApplication(a_file.read(), Name=basename)

        part['Content-Disposition'] = 'attachment; filename="%s"' % basename
        msg.attach(part)

    email = smtplib.SMTP('your-smtp-host-name.com')
    email.sendmail(from_email, to_email, msg.as_string())

我希望这有帮助！:-)

Answer 9

使用python 3的另一种方式（如果有人正在搜索）：

import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.base import MIMEBase
from email import encoders

fromaddr = "sender mail address"
toaddr = "receiver mail address"

msg = MIMEMultipart()

msg['From'] = fromaddr
msg['To'] = toaddr
msg['Subject'] = "SUBJECT OF THE EMAIL"

body = "TEXT YOU WANT TO SEND"

msg.attach(MIMEText(body, 'plain'))

filename = "fileName"
attachment = open("path of file", "rb")

part = MIMEBase('application', 'octet-stream')
part.set_payload((attachment).read())
encoders.encode_base64(part)
part.add_header('Content-Disposition', "attachment; filename= %s" % filename)

msg.attach(part)

server = smtplib.SMTP('smtp.gmail.com', 587)
server.starttls()
server.login(fromaddr, "sender mail password")
text = msg.as_string()
server.sendmail(fromaddr, toaddr, text)
server.quit()

确保在您的Gmail帐户中允许“ 安全性较低的应用程序 ”