如何用python通过163发送邮件

Posted 2023-04-03

为什么会发送失败的

from smtplib import SMTP
>>> conn = SMTP('smtp.163.com')
>>> conn.sendmail('[e-mail]','[e-mail]',
"""From:[e-mail]
To:[e-mail]
Subject: test

test
""")
Traceback (most recent call last):
File "<pyshell#8>", line 6, in <module>
""")
File "C:\Python33\lib\smtplib.py", line 759, in sendmail
raise SMTPSenderRefused(code, resp, from_addr)
smtplib.SMTPSenderRefused: (553, b'authentication is required,smtp14,EsCowEApgWDDUPdSW640Ag--.317S2 1391939937', '[e-mail]')

参考技术A 目测没有给smtp你邮箱的帐号和密码登录，发不了邮件，参考：
http://blog.csdn.net/hdutigerkin/article/details/6623234追问

成功是成功了 但是被当成垃圾邮件呢

本回答被提问者采纳 参考技术B # 导入 smtplib 和 MIMEText
import smtplib
from email.mime.text import MIMEText
# 定义发送列表
mailto_list=["[e-mail]","[e-mail]"]
# 设置服务器名称、用户名、密码以及邮件后缀
mail_host = "smtp.126.com"
mail_user = "user"
mail_pass = "pass"
mail_postfix="126.com"
# 发送邮件函数
def send_mail(to_list, sub, context):
'''''
to_list: 发送给谁
sub: 主题
context: 内容
send_mail("[e-mail]","sub","context")
'''
me = mail_user + "<"+mail_user+"@"+mail_postfix+">"
msg = MIMEText(context)
msg['Subject'] = sub
msg['From'] = me
msg['To'] = ";".join(to_list)
try:
send_smtp = smtplib.SMTP()
send_smtp.connect(mail_host)
send_smtp.login(mail_user, mail_pass)
send_smtp.sendmail(me, to_list, msg.as_string())
send_smtp.close()
return True
except (Exception, e):
print(str(e))
return False
if __name__ == '__mian__':
if (True == send_mail(mailto_list,"subject","context")):
print ("测试成功") www.codesky.net
else:
print ("测试失败")

如何用ABAP实现自动发送外部邮件

参考技术A 您好，希望以下回答能帮助您 可以在资料前面加一个 ', 就不会被自动去掉 如您还有疑问可继续追问。 参考技术B 静待大神降临