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最新推荐文章于 2024-09-21 17:18:17 发布

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最新推荐文章于 2024-09-21 17:18:17 发布

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# coding=gb2312

import smtplib
from email.mime.text import MIMEText

#要发给谁
mailto_list= ["418778645@qq.com", "418778645@qq.com"]

def send_mail(to_list,sub,content):
    #设置服务器，用户名、口令以及邮箱的后缀
    mail_host="smtp.qq.com"
    mail_user="418778645"
    mail_pass="***********"
    mail_postfix="qq.com"
    me=mail_user+"<"+mail_user+"@"+mail_postfix+">"

    msg = MIMEText(content, _subtype='html',_charset='gb2312')
    msg['Subject'] = sub
    msg['From']    = me
    msg['To']      = ";".join(to_list)

    try:
        s = smtplib.SMTP()
        s.connect(mail_host)
        s.login(mail_user,mail_pass)
        s.sendmail(me, to_list, msg.as_string())
        s.close()
        return True
    except Exception, e:
        print str(e)
        return False

if __name__ == '__main__':
    txt='<html> \
            <body>   \
                <p> \
                    http://hi.baidu.com/zengzhaonong \
                </p> \
                <p> \
                    http://hi.baidu.com/zengzhaonong \
                </p> \
            </body> \
         </html>'

    if send_mail(mailto_list, "博客地址", txt):
        print "发送成功"
    else:
        print "发送失败"