Question

单独尝试多个to和多个cc，这很好但是当我尝试两个时我得到一个错误：

文件

“路径\连续\ anaconda2 \ ENVS \ mypython \ lib中\ smtplib.py”，第870行，在sendmail发件人[每个] =（code，resp）TypeError：不可用的类型：'list'“

代码：

from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.image import MIMEImage
from email.mime.application import MIMEApplication

strFrom = 'fasdf@dfs.com'

cc='abc.xyz@dfa.com, sdf.xciv@lfk.com'

to='sadf@sdfa.com,123.lfadf@fa.com'

msg = MIMEMultipart('related')
msg['Subject'] = 'Subject'
msg['From'] = strFrom
msg['To'] =to
msg['Cc']=cc

#msg['Bcc']= strBcc

msg.preamble = 'This is a multi-part message in MIME format.'


msgAlternative = MIMEMultipart('alternative')
msg.attach(msgAlternative)

msgText = MIMEText('This is the alternative plain text message.')
msgAlternative.attach(msgText)


msgText = MIMEText('''<html>

<body><p>Hello<p>
        </body>
        </html> '''.format(**locals()), 'html')
msgAlternative.attach(msgText)



import smtplib
smtp = smtplib.SMTP()
smtp.connect('smtp address')
smtp.ehlo()
smtp.sendmail(strFrom, to, msg.as_string())
smtp.quit()

Answer 1

附加To或from应为字符串，sendmail应始终采用列表形式。

cc=['abc.xyz@dfa.com', 'sdf.xciv@lfk.com']

to=['sadf@sdfa.com','123.lfadf@fa.com']

msg['To'] =','.join(to)

msg['Cc']=','.join(cc)   

toAddress = to + cc    

smtp.sendmail(strFrom, toAddress, msg.as_string())

Answer 2

to参数应该是您希望将邮件发送到的所有地址的列表。 To:和Cc:中的划分基本上仅用于显示目的; SMTP只有一个收件人序列，可以为每个地址转换为一个RCPT TO命令。

def addresses(addrstring):
    """Split in comma, strip surrounding whitespace."""
    return [x.strip() for x in addrstring.split(',')]

smtp.sendmail(strFrom, addresses(to) + addresses(cc), msg.as_string())

使用python

2 个答案: