python smtplib在linux上发送邮件失败解决方式

通过zabbix配置邮件告警的时候，发现邮件访问不了

之后将脚本copy到window上执行发现可以正常发送

linux却不行，以为是python版本问题，经过检验并不是

新建send.py文件 代码如下

import smtplib
from email.mime.text import MIMEText
import sys

# configure your own parameters here
#下面邮件地址的smtp地址
mail_host = 'smtp.163.com'
#用来发邮件的邮箱,在发件人抬头显示(不然你的邮件会被当成是垃圾邮件)
mail_user = 'chy1559843332@163.com'
# 客户端授权码
mail_auth = '********'
# 发送方显示的名称
send_name = mail_user
# 接收方显示的名称
recv_name = mail_user

def excute(to, title, content):
    msg = MIMEText(content, 'plain', 'utf-8')
    msg['From'] = send_name
    msg['To'] = recv_name
    msg['Subject'] = title
    server = smtplib.SMTP(mail_host, 25)
    server.login(mail_user,mail_auth)
    server.sendmail(mail_user,to,msg.as_string())
    server.quit()

if __name__ == '__main__':
    excute('1559843332@qq.com', 'chyhis is title', 'this is content')

在windows下，通过python send.py 执行之后即可收到邮件

之后放在linux上却 /usr/local/python3/lib/python3.7/socket.py

Traceback (most recent call last):
  File "aa.py", line 22, in <module>
    excute('1559843332@qq.com', 'chy this is title', 'mcdh cnhk')
  File "aa.py", line 16, in excute
    server = smtplib.SMTP(mail_host, 465)
  File "/usr/local/python3/lib/python3.7/smtplib.py", line 251, in __init__
    (code, msg) = self.connect(host, port)
  File "/usr/local/python3/lib/python3.7/smtplib.py", line 338, in connect
    (code, msg) = self.getreply()
  File "/usr/local/python3/lib/python3.7/smtplib.py", line 387, in getreply
    line = self.file.readline(_MAXLINE + 1)
  File "/usr/local/python3/lib/python3.7/socket.py", line 589, in readinto
    return self._sock.recv_into(b)

之后参考https://www.jianshu.com/p/fc55404b6db7

解决方式：

smtplib.SMTP('smtp.163.com', 25)

修改为

smtplib.SMTP_SSL('smtp.163.com', 465)

之后在linux上执行python send.py，嗯，完美！