python smtplib在linux上发送邮件失败解决方式

python

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!

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