使用Python从附件列表发送电子邮件

Question

I am trying to use a Python script I found on Github to send email with attachments. I have a list of 5 attachments and want to send one email per attachment. When I run the script it sends the first email with one attachment and the next email with 2 attachments and so on. The 5th email has all 5 attachments instead of the 5th attachment in the list. I believe that I need to iterate through the list of attachments but cannot figure out where to do so. Any help would be greatly appreciated. Script is below.

attachments = ['file1.zip', 'file2.zip', 'file3.zip', 'file4.zip', 'file5.zip']
host = 'mailer' # specify port, if required, using this notations
fromaddr = 'test@localhost' # must be a vaild 'from' address in your GApps account
toaddr = 'target@remotehost'
replyto = fromaddr # unless you want a different reply-to
msgsubject = 'Test ZIP'
htmlmsgtext = """<h2>TEST</h2>"""

######### In normal use nothing changes below this line ###############

import smtplib, os, sys
from email.MIMEMultipart import MIMEMultipart
from email.MIMEBase import MIMEBase
from email.MIMEText import MIMEText
from email import Encoders
from HTMLParser import HTMLParser

# A snippet - class to strip HTML tags for the text version of the email

class MLStripper(HTMLParser):
    def __init__(self):
        self.reset()
        self.fed = []
    def handle_data(self, d):
        self.fed.append(d)
    def get_data(self):
        return ''.join(self.fed)

def strip_tags(html):
    s = MLStripper()
    s.feed(html)
    return s.get_data()

########################################################################

try:
# Make text version from HTML - First convert tags that produce a line break to carriage returns
    msgtext = htmlmsgtext.replace('</br>',"\r").replace('<br />',"\r").replace('</p>',"\r")
# Then strip all the other tags out
    msgtext = strip_tags(msgtext)

# necessary mimey stuff
    msg = MIMEMultipart()
    msg.preamble = 'This is a multi-part message in MIME format.\n'
    msg.epilogue = ''

    body = MIMEMultipart('alternative')
    body.attach(MIMEText(msgtext))
    body.attach(MIMEText(htmlmsgtext, 'html'))
    msg.attach(body)

    if 'attachments' in globals() and len('attachments') > 0:
        for filename in attachments:
            f=filename      
            part = MIMEBase('application', "octet-stream")
            part.set_payload( open(f,"rb").read() )
            Encoders.encode_base64(part)
            part.add_header('Content-Disposition', 'attachment; filename="%s"' % f)
            msg.attach(part)
            msg.add_header('From', fromaddr)
            msg.add_header('To', toaddr)
            msg.add_header('Subject', msgsubject)
            msg.add_header('Reply-To', replyto)
            server = smtplib.SMTP(host)
            server.set_debuglevel(False) # set to True for verbose output
            server.sendmail(msg['From'], [msg['To']], msg.as_string())
            print 'Email sent with filename: "%s"' % f
            server.quit()

except:
    print ('Email NOT sent to %s successfully. %s ERR: %s %s %s ', str(toaddr), str(sys.exc_info()[0]), str(sys.exc_info()[1]), str (sys.exc_info()[2]) )

Answer 1

Each time through the loop, you add an attachment to the existing message, then send the message. So, you're going to keep accumulating a bigger and bigger message and sending each intermediate step.

It's not clear what you actually want to do, but obviously not this…

If you want to send one message with all five attachments, just move the sending code (everything from server = smtplib.SMTP(host) to server.quit() outside the loop by unindenting it.

If you want to send five messages, each with one attachment, most the message-creating code (everything from msg = MIMEMultipart() to msg.attach(body) ) into the loop by indenting it and moving it down a couple lines.

If you want something else, the answer will almost surely be similarly trivial, but nobody can tell you how to do it until you explain what it is you want.