无法使用 python smtp 发送电子邮件

Question

我正在使用 python 开发一个应用程序，我需要通过邮件发送文件。我写了一个程序来发送邮件，但不知道有什么问题。代码贴在下面。请任何人帮助我使用这个 smtp 库。有什么我想念的吗？还有谁能告诉我 smtp 中的主机是什么！我正在使用 smtp.gmail.com。 也有人可以告诉我如何通过电子邮件发送文件（.csv 文件）。感谢您的帮助！

#!/usr/bin/python

import smtplib

sender = 'someone@yahoo.com'
receivers = ['someone@yahoo.com']

message = """From: From Person <someone@yahoo.com>
To: To Person <someone@yahoo.com>
Subject: SMTP e-mail test

This is a test e-mail message.
"""

try:
   smtpObj = smtplib.SMTP('smtp.gmail.com')
   smtpObj.sendmail(sender, receivers, message)         
   print "Successfully sent email"
except:
   print "Error: unable to send email"

Answer 1

您没有登录。还有几个原因可能导致您无法通过，包括被您的 ISP 阻止、如果无法在您身上获得反向 DNS 时 gmail 将您退回等等。

try:
   smtpObj = smtplib.SMTP('smtp.gmail.com', 587) # or 465
   smtpObj.ehlo()
   smtpObj.starttls()
   smtpObj.login(account, password)
   smtpObj.sendmail(sender, receivers, message)         
   print "Successfully sent email"
except:
   print "Error: unable to send email"

我刚刚注意到您要求能够附加文件。这改变了事情，因为现在你需要处理编码。虽然我不这么认为，但仍然没有那么难遵循。

import os
import email
import email.encoders
import email.mime.text
import smtplib

# message/email details
my_email = 'myemail@gmail.com'
my_passw = 'asecret!'
recipients = ['jack@gmail.com', 'jill@gmail.com']
subject = 'This is an email'
message = 'This is the body of the email.'
file_name = 'C:\\temp\\test.txt'

# build the message
msg = email.MIMEMultipart.MIMEMultipart()
msg['From'] = my_email
msg['To'] = ', '.join(recipients)
msg['Date'] = email.Utils.formatdate(localtime=True)
msg['Subject'] = subject
msg.attach(email.MIMEText.MIMEText(message))

# build the attachment
att = email.MIMEBase.MIMEBase('application', 'octet-stream')
att.set_payload(open(file_name, 'rb').read())
email.Encoders.encode_base64(att)
att.add_header('Content-Disposition', 'attachment; filename="%s"' % os.path.basename(file_name))
msg.attach(att)

# send the message
srv = smtplib.SMTP('smtp.gmail.com', 587)
srv.ehlo()
srv.starttls()
srv.login(my_email, my_passw)
srv.sendmail(my_email, recipients, msg.as_string())