如何在python中发送邮件

时间:2016-04-20 10:46:54

标签: python email smtp

我有这个使用smtplib模块发送电子邮件的python脚本。

#! /usr/bin/python

import smtplib
from email.mime.image import MIMEImage
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart

from1 = "root"
to = "ankur.kulshrestha@ericsson.com"
subject = "test mail"

msg = MIMEMultipart()
msg['From'] = from1
msg['To'] = to
msg['Subject'] = subject

text = """Hi,
This is test messgae"""

html = """\
<html>
  <head></head>
  <body>
    <p>Hi!<br>
       How are you?<br>
       Here is the <a href="http://www.python.org">link</a> you wanted.
    </p>
  </body>
</html>
"""

fi = open("dragon.jpg", 'rb')
img = MIMEImage(fi.read())
fi.close()
msg.attach(img)

part1 = MIMEText(text, 'plain')
part2 = MIMEText(html, 'html')
msg.attach(part1)
msg.attach(part2)

smtp_obj = smtplib.SMTP('rinacac-test.egi.ericsson.com')
smtp_obj.sendmail(from1, to, msg.as_string())
smtp_obj.quit()

脚本正在运行,但我的html内容'html'&amp; image'long.jpg'作为附件出现。我希望它们显示为我的邮件内容而不是附件。请帮忙

更新:  我仍然无法理解Multipart /替代品在这里扮演的角色。此外,您在下面分享的早期链接中没有显示在邮件中。

msgText = MIMEText('This is the alternative plain text message.')
msgAlternative.attach(msgText)

其次包含html字符串的对象'msgText'附加'msgAlnernative'

msgText = MIMEText('<b>Some <i>HTML</i> text</b> and an image.<br><img src="cid:image1"><br>Nifty!', 'html')
msgAlnernative.attach(msgText)

但读取图像的对象'msgImage'附带'msgRoot'。为什么会这样..

msgImage.add_header('Content-ID', '<image1>')
msgRoot.attach(msgImage)

我找不到有关MIMEMultipart /相关和替代的任何帮助文档。请帮忙

1 个答案:

答案 0 :(得分:1)

我认为这可以解决您的问题 How to display base64 images in html

如果您的mailprovider支持电子邮件中的html标记

很抱歉,因为这是一个特定的python问题,链接中的示例似乎是php这里的python解决方案

import base64
img = 'test.jpg'
with open(img, 'rb') as imgfile:
    img64 = base64.encodestring(imgfile.read())

###

html = """\
<html>
  <head></head>
  <body>
    <p>Hi!<br>
       How are you?<br>
       Here is the <a href="http://www.python.org">link</a> you wanted.
    </p>
    <img width="160" height="160" alt="tick" src="data:image/jpg;base64,{0}">
  </body>
</html>
""".format(img64)

###