smtplib bcc 密送 失败解决方案

msg.sendmail(sender, [receiver_qq,mail_bcc], msg.as_string()) # 重点是中间的收件人地址是一个列表，
将刚才定义的收件人填入列表后面即可完成密送，这才是正确的方式

 import smtplib 
 from email.mime.text import MIMEText 
 mail_host = 'smtp.126.com' 
 mail_user = 'xxx@126.com' 
 mail_pwd = 'hellopwd' 
 mail_to = 'xxao@gmail.com' 
 mail_cc = 'xx@xx.com' 
 mail_bcc = 'xx@qq.com' 
 content = 'this is a mail sent with python' 
#表头信息
 msg = MIMEText(content) 
 msg['From'] = mail_user 
 msg['Subject'] = 'this is a python test mail' 
 msg['To'] = mail_to 
 msg['Cc'] = mail_cc 
 msg['Bcc'] = mail_bcc 
 try: 
  s = smtplib.SMTP() 
  s.connect(mail_host) 
  #login 
  s.login(mail_user,mail_pwd) 
  #send mail 
  s.sendmail(mail_user,[mail_to,mail_cc,mail_bcc],msg.as_string()) 
  s.close() 
  print 'success' 
 except Exception ,e: 
  print e