python smtplib在linux上發送郵件失敗解決方式

通過zabbix配置郵件告警的時候，發現郵件訪問不了

之后將腳本copy到window上執行發現可以正常發送

linux卻不行，以為是python版本問題，經過檢驗並不是

新建send.py文件 代碼如下

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

# configure your own parameters here
#下面郵件地址的smtp地址
mail_host = 'smtp.163.com'
#用來發郵件的郵箱,在發件人抬頭顯示(不然你的郵件會被當成是垃圾郵件)
mail_user = '[e-mail]'
# 客戶端授權碼
mail_auth = '********'
# 發送方顯示的名稱
send_name = mail_user
# 接收方顯示的名稱
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('[e-mail]', 'chyhis is title', 'this is content')

在windows下，通過python send.py 執行之后即可收到郵件

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

Traceback (most recent call last):
  File "aa.py", line 22, in <module>
    excute('[e-mail]', '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)

之后參考https://www.jianshu.com/p/fc55404b6db7

解決方式：

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

修改為

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

之后在linux上執行python send.py，嗯，完美！