如何使用Python在电子邮件中发送HTML内容?我可以发送简单的短信。
当前回答
你可以试试我的邮件模块。
from mailer import Mailer
from mailer import Message
message = Message(From="me@example.com",
To="you@example.com")
message.Subject = "An HTML Email"
message.Html = """<p>Hi!<br>
How are you?<br>
Here is the <a href="http://www.python.org">link</a> you wanted.</p>"""
sender = Mailer('smtp.example.com')
sender.send(message)
其他回答
对于python3,改进@taltman的回答:
使用email.message. emailmessage而不是email.message. message来构造email。 使用电子邮件。Set_content func, assign subtype='html'参数。而不是低级的func set_payload和手动添加头。 使用SMTP。send_message func而不是SMTP。Sendmail func发送电子邮件。 与块一起使用,可自动关闭连接。
from email.message import EmailMessage
from smtplib import SMTP
# construct email
email = EmailMessage()
email['Subject'] = 'foo'
email['From'] = 'sender@test.com'
email['To'] = 'recipient@test.com'
email.set_content('<font color="red">red color text</font>', subtype='html')
# Send the message via local SMTP server.
with smtplib.SMTP('localhost') as s:
s.login('foo_user', 'bar_password')
s.send_message(email)
实际上,yagmail采取了一点不同的方法。
它将在默认情况下发送HTML,并为无法阅读电子邮件的用户提供自动回退功能。现在已经不是17世纪了。
当然,它可以被覆盖,但如下所示:
import yagmail
yag = yagmail.SMTP("me@example.com", "mypassword")
html_msg = """<p>Hi!<br>
How are you?<br>
Here is the <a href="http://www.python.org">link</a> you wanted.</p>"""
yag.send("to@example.com", "the subject", html_msg)
有关安装说明和更多伟大的功能,请查看github。
你可以试试我的邮件模块。
from mailer import Mailer
from mailer import Message
message = Message(From="me@example.com",
To="you@example.com")
message.Subject = "An HTML Email"
message.Html = """<p>Hi!<br>
How are you?<br>
Here is the <a href="http://www.python.org">link</a> you wanted.</p>"""
sender = Mailer('smtp.example.com')
sender.send(message)
下面是示例代码。灵感来自Python Cookbook网站上的代码(找不到确切的链接)
def createhtmlmail (html, text, subject, fromEmail):
"""Create a mime-message that will render HTML in popular
MUAs, text in better ones"""
import MimeWriter
import mimetools
import cStringIO
out = cStringIO.StringIO() # output buffer for our message
htmlin = cStringIO.StringIO(html)
txtin = cStringIO.StringIO(text)
writer = MimeWriter.MimeWriter(out)
#
# set up some basic headers... we put subject here
# because smtplib.sendmail expects it to be in the
# message body
#
writer.addheader("From", fromEmail)
writer.addheader("Subject", subject)
writer.addheader("MIME-Version", "1.0")
#
# start the multipart section of the message
# multipart/alternative seems to work better
# on some MUAs than multipart/mixed
#
writer.startmultipartbody("alternative")
writer.flushheaders()
#
# the plain text section
#
subpart = writer.nextpart()
subpart.addheader("Content-Transfer-Encoding", "quoted-printable")
pout = subpart.startbody("text/plain", [("charset", 'us-ascii')])
mimetools.encode(txtin, pout, 'quoted-printable')
txtin.close()
#
# start the html subpart of the message
#
subpart = writer.nextpart()
subpart.addheader("Content-Transfer-Encoding", "quoted-printable")
#
# returns us a file-ish object we can write to
#
pout = subpart.startbody("text/html", [("charset", 'us-ascii')])
mimetools.encode(htmlin, pout, 'quoted-printable')
htmlin.close()
#
# Now that we're done, close our writer and
# return the message body
#
writer.lastpart()
msg = out.getvalue()
out.close()
print msg
return msg
if __name__=="__main__":
import smtplib
html = 'html version'
text = 'TEST VERSION'
subject = "BACKUP REPORT"
message = createhtmlmail(html, text, subject, 'From Host <sender@host.com>')
server = smtplib.SMTP("smtp_server_address","smtp_port")
server.login('username', 'password')
server.sendmail('sender@host.com', 'target@otherhost.com', message)
server.quit()
下面是发送HTML电子邮件的简单方法,只需指定Content-Type头为'text/ HTML ':
import email.message
import smtplib
msg = email.message.Message()
msg['Subject'] = 'foo'
msg['From'] = 'sender@test.com'
msg['To'] = 'recipient@test.com'
msg.add_header('Content-Type','text/html')
msg.set_payload('Body of <b>message</b>')
# Send the message via local SMTP server.
s = smtplib.SMTP('localhost')
s.starttls()
s.login(email_login,
email_passwd)
s.sendmail(msg['From'], [msg['To']], msg.as_string())
s.quit()