使用 smtp 发送带有多个附件的电子邮件

Question

我有这个脚本可以向多个用户发送带有多个附件的电子邮件。但是，附件的文件名被设置为它们的路径。

收到的文件

终端输出

如何将它们的名称设置为实际的文件名，谢谢。

"""

import os 
import smtplib
from email.MIMEMultipart import MIMEMultipart
from email.MIMEText import MIMEText
from email.MIMEImage import MIMEImage
from email.MIMEBase import MIMEBase
from email import Encoders


#Set up crap for the attachments
files = "/tmp/test/dbfiles"
filenames = [os.path.join(files, f) for f in os.listdir(files)]
#print filenames


#Set up users for email
gmail_user = "joe@email.com"
gmail_pwd = "somepasswd"
recipients = ['recipient1','recipient2']

#Create Module
def mail(to, subject, text, attach):
   msg = MIMEMultipart()
   msg['From'] = gmail_user
   msg['To'] = ", ".join(recipients)
   msg['Subject'] = subject

   msg.attach(MIMEText(text))

   #get all the attachments
   for file in filenames:
      part = MIMEBase('application', 'octet-stream')
      part.set_payload(open(file, 'rb').read())
      Encoders.encode_base64(part)
      part.add_header('Content-Disposition', 'attachment; filename="%s"' % file)
      msg.attach(part)

   mailServer = smtplib.SMTP("smtp.gmail.com", 587)
   mailServer.ehlo()
   mailServer.starttls()
   mailServer.ehlo()
   mailServer.login(gmail_user, gmail_pwd)
   mailServer.sendmail(gmail_user, to, msg.as_string())
   # Should be mailServer.quit(), but that crashes...
   mailServer.close()

#send it
mail(recipients,
   "Todays report",
   "Test email",
   filenames)

"""

Answer 1

此链接可能有解决方案：

Python email MIME attachment filename

基本上，根据该帖子的解决方案是更改您的这一行：

part.add_header('Content-Disposition', 'attachment; filename="%s"' % file)

到这一行：

part.add_header('Content-Disposition', 'attachment', filename=AFileName)

这归结为最后的变化：

  #get all the attachments
  for file in filenames:
  
  part = MIMEBase('application', 'octet-stream')
  part.set_payload(open(file, 'rb').read())
  Encoders.encode_base64(part)
  
  ***part.add_header('Content-Disposition', 'attachment', filename=file)***

  msg.attach(part)

Documentation on how to use add_header

希望对您有所帮助！ :D

更新

在你的 for 循环中有这个应该给你文件名：

for file in filenames:

    actual_filenames = os.path.basename(file)

    #Your code

    part.add_header('Content-Disposition', 'attachment', filename=actual_filenames)

Answer 2

如果在同一个目录：

import os                                  
for f in os.listdir('.'):                                                                                               
    fn, fext = os.path.splitext(f)