[英]Python - print/save the email before sending using smtplib
My objective is to save the email that my script sends using smtplib module. 我的目标是使用smtplib模块保存脚本发送的电子邮件。
Here is how I am sending the email: 这是我发送电子邮件的方式:
#Creating a multipart body
msg = MIMEMultipart()
#Attaching files to the email
for file in tc['attachments']:
try:
part = email.mime.base.MIMEBase('application', "octet-stream")
part.set_payload( open(file, "rb").read())
email.encoders.encode_base64(part)
part.add_header('Content-Disposition', 'attachment; filename="%s"' % os.path.basename(file))
msg.attach(part)
except Exception as err:
print "Exception when attaching file %s to the email: \n %s" % (file, err)
print traceback.print_exc()
#Connecting to the SMTP server
smtp = smtplib.SMTP("1.2.3.4")
#Sending the email
try:
status = smtp.sendmail(msg['From'], send_to, msg.as_string())
print status
smtp.close()
except Exception, e:
print "Unable to send the email. Exception seen: %s" % e
Now, if I save msg.as_string()
to a variable, it only saves the body of the email, but I want the whole email as it is sent. 现在,如果我将msg.as_string()
保存到一个变量中,则它仅保存电子邮件的正文,但是我希望在发送整个电子邮件。
I looked into smtplib
module's documentation but couldn't find a knob to print the headers of the email. 我查看了smtplib
模块的文档,但找不到用于打印电子邮件标题的旋钮。
Is there any hack (like using another module to monitor the traffic, etc) that I can use to save the email the way I sent from the script? 是否可以使用从脚本发送方式保存电子邮件的任何黑客手段(例如使用其他模块监视流量等)?
您可以使用str()
获得整个电子邮件的字符串版本,包括标题:
save_msg = str(msg)
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