解决smtplib发送多人邮件没有展示收件人的问题

Posted 2020-11-23 i-shu

 

#!/usr/bin/env python
# -*- coding: utf-8 -*-
 
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.utils import parseaddr
from email.utils import formataddr
from email.header import Header
 
 
def __format_addr__(addr):
    # 解析邮件地址，以保证邮有别名可以显示
    alias_name, addr = parseaddr(addr)
    # 防止中文问题，进行转码处理，并格式化为str返回
    return formataddr((Header(alias_name,charset="utf-8").encode(),
                       addr.encode("uft-8") if isinstance(addr, unicode) else addr))
 
 
def send_email_to(fromAdd, toAdd, subject, html_text, filename=None):
 
    SERVER = ‘mail.***.com‘
    USER = ‘******‘
    PASSWD = ‘***‘
 
    strFrom = __format_addr(fromAdd)
 
    strTo = list()
    # 原来是一个纯邮箱的list，现在如果是一个["jayzhen<[e-mail]>"]的list给他格式化
    try:
        for a in toAdd:
            strTo.append(__format_addr(a))
    except Exception as e:
        # 没有对a和toadd进行type判断，出错就直接还原
        strTo = toAdd
 
    msgRoot = MIMEMultipart(‘related‘)
    msgRoot.preamble = ‘This is a multi-part message in MIME format.‘
 
    msgAlternative = MIMEMultipart(‘alternative‘)
    msgRoot.attach(msgAlternative)
 
    # 邮件对象
    msgText = MIMEText(html_text, ‘html‘, ‘utf-8‘)
    msgRoot[‘Subject‘] = Header(subject)   # 这是邮件的主题，通过Header来标准化
    msgRoot[‘From‘] = strFrom       # 发件人也是被格式化过的
    msgRoot[‘to‘] = ‘,‘.join(strTo)   # 这个一定要是一个str,不然会报错“AttributeError: ‘list‘ object has no attribute ‘lstrip‘”
    msgAlternative.attach(msgText)
 
    smtp = smtplib.SMTP(SERVER, 11)
    smtp.set_debuglevel(0)
    # smtp.connect(SERVER)
    smtp.login(USER, PASSWD)
    # 这里要注意了，这里的fromadd和toAdd和msgRoot[‘From‘] msgRoot[‘to‘]的区别
    smtp.sendmail(fromAdd, toAdd, msgRoot.as_string())
    smtp.quit()

 

原文：https://blog.csdn.net/u013948858/article/details/82903977