直接上代碼了

代碼如下:
import smtplib
msg = MIMEMultipart()

#構造附件1
att1 = MIMEText(open('/home/a2bgeek/develop/python/hello.py', 'rb').read(), 'base64', 'gb2312')
att1["Content-Type"] = 'application/octet-stream'
att1["Content-Disposition"] = 'attachment; filename="hello.txt"'#這里的filename可以任意寫，寫什么名字，郵件中顯示什么名字
msg.attach(att1)

#構造附件2
#att2 = MIMEText(open('/home/a2bgeek/develop/python/mail.py', 'rb').read(), 'base64', 'gb2312')
#att2["Content-Type"] = 'application/octet-stream'
#att2["Content-Disposition"] = 'attachment; filename="123.txt"'
#msg.attach(att2)

#加郵件頭
strTo = ['XXX1@139.com', 'XXX2@163.com', 'XXX3@126.com']
msg['to']=','.join(strTo)
msg['from'] = 'YYY@163.com'
msg['subject'] = '郵件主題'
#發送郵件
try:
    server = smtplib.SMTP()
    server.connect('smtp.163.com')
    server.login('YYY@163.com','yourpasswd')
    server.sendmail(msg['from'], strTo ,msg.as_string())
    server.quit()
    print '發送成功'
except Exception, e:
    print str(e)

細心的讀者會發現代碼中有這樣一句：msg['to']=','.join(strTo)，但是msg[['to']并沒有在后面被使用，這么寫明顯是不合理的，但是這就是stmplib的bug。你只有這樣寫才能群發郵件。查明原因如下：

The problem is that SMTP.sendmail and email.MIMEText need two different things.

email.MIMEText sets up the “To:” header for the body of the e-mail. It is ONLY used for displaying a result to the human being at the other end, and like all e-mail headers, must be a single string. (Note that it does not actually have to have anything to do with the people who actually receive the message.)

SMTP.sendmail, on the other hand, sets up the “envelope” of the message for the SMTP protocol. It needs a Python list of strings, each of which has a single address.

So, what you need to do is COMBINE the two replies you received. Set msg‘To' to a single string, but pass the raw list to sendmail.

好了今天就到這里。