我正在尝试使用python 3.2 SMTPlib.sendmail()函数 在对SMTP库进行一些修改之后发送消息(即 注释掉rset()函数,它抑制了错误消息msg) 我设法从服务器检索以下错误消息:
SendMail失败 (554,b'交易失败:由于可能滥用而无法发送消息;请访问http://postmaster.yahoo.com/abuse_smtp.html以获取更多信息')
雅虎邮件SMTP服务器认为我发送垃圾邮件,URL确实链接到任何内容 有用。我认为它与标题不合适有关,我似乎无法找到明确的标题 回答什么是合规的标题&我已经阅读过Gmail的类似问题。 模拟电子邮件已被替换为这篇文章。
任何帮助将不胜感激
我的完整代码如下:
self.message = email.message_from_string('''To: <ksmith@yahoo.co.nz>
From: <rwilson@yahoo.co.nz>
Reply-To: <rwilson@yahoo.co.nz>
Subject: Test send mail \n\n Hello''')
fromAddress = 'rwilson@yahoo.co.nz'
toAddress = 'ksmith@yahoo.co.nz'
try:
self.smtp = SMTP()
self.smtp.connect('smtp.mail.yahoo.com')
except Exception:
print('Connection Failed')
print(traceback.format_exc())
try:
self.smtp.login('rwilson','tree22')
except Exception:
print('Login Failed!')
print(traceback.format_exc())
try:
self.smtp.sendmail(fromAddress,toAddress ,self.message.as_string())
print("Message sucessfully sent!")
self.smtp.close()
except Exception as e:
print('SendMail Failed')
print(e)
答案 0 :(得分:3)
以下适用于Python 2.7和Python 3.2上的microsoft,google,yahoo帐户:
#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""Send email via smtp_host."""
import smtplib
from email.mime.text import MIMEText
from email.header import Header
####smtp_host = 'smtp.live.com' # microsoft
####smtp_host = 'smtp.gmail.com' # google
smtp_host = 'smtp.mail.yahoo.com' # yahoo
login, password = ...
recipients_emails = [login]
msg = MIMEText('body…', 'plain', 'utf-8')
msg['Subject'] = Header('subject…', 'utf-8')
msg['From'] = login
msg['To'] = ", ".join(recipients_emails)
s = smtplib.SMTP(smtp_host, 587, timeout=10)
s.set_debuglevel(1)
try:
s.starttls()
s.login(login, password)
s.sendmail(msg['From'], recipients_emails, msg.as_string())
finally:
s.quit()
答案 1 :(得分:1)
在登录之前将以下内容添加到您的代码中,然后重试;
try:
self.smtp.ehlo()
self.smtp.starttls()
self.smtp.ehlo
except:
print "No TLS :("
#do login here