Python 发送邮件,邮件可以正常接收,但收件人栏显示异常 - V2EX
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Kvip
V2EX    Python

Python 发送邮件,邮件可以正常接收,但收件人栏显示异常

  •  1
     
  •   Kvip 2020-07-20 16:40:42 +08:00 2977 次点击
    这是一个创建于 1961 天前的主题,其中的信息可能已经有所发展或是发生改变。

    如题,使用 python 编写了个发送邮件的 Demo 。邮件发送后,用 Foxmail 能够正常收的到。

    但是收件人一栏显示的收件邮箱很奇怪,如下图,一整个收件邮箱被按字符拆分为好多个小邮箱,小邮箱的那域名我也没见过。

    原本收件邮箱应该显示是 [email protected]

    不知道有大佬遇到过这种情况没,请教下

    邮件

    下面是源码(看着没毛病):

    # 编译日期:2019-10-14 15:01:36 # coding=utf-8 import os from email.mime.text import MIMEText from email.header import Header from email.mime.multipart import MIMEMultipart from email.mime.application import MIMEApplication # 发送邮件(附件) # email_user: 发件人 # email_pass: 密码 # receivers:收件人列表,例:['[email protected]','[email protected]'] # filepath: 文件地址 如:r'D:\分行查询支行卷别库存 Excel\作文.xlsx' def sendEmail(email_user, email_pass, receivers, filepath): mail_host = '10.10.10.10' # 设置服务器地址 sender = email_user + '@xxx.com.cn' # 发件人邮箱 # 创建一个带附件的实例 message = MIMEMultipart() # 设置发件人、收件人 message['From'] = sender message['To'] = ", ".join(receivers) # 设置邮件主题 message['Subject'] = Header('查冻扣报备文件', 'utf-8') # 邮件正文 message.attach(MIMEText('查冻扣报备文件已发送,请查看附件!', 'plain', 'utf-8')) # 构造附件 xlsx = MIMEApplication(open(filepath,'rb').read()) xlsx['Content-Type'] = 'application/octet-stream' filename = os.path.basename(filepath) xlsx.add_header('Content-Disposition', 'attachment', filename=Header(filename, 'utf-8').encode()) message.attach(xlsx) try: smtpObj = smtplib.SMTP(mail_host, 1025) smtpObj.login(email_user, email_pass) smtpObj.sendmail(sender, receivers, message.as_string()) print("邮件发送成功,收件人:" + str(receivers) + "\n 附件地址:" + str(filepath)) except smtplib.SMTPException as e: print("邮件发送失败,收件人:" + str(receivers) + "\n 附件地址:" + str(filepath)) print(e) 
    8 条回复    2020-07-25 23:12:20 +08:00
    j0hnj
        1
    j0hnj  
       2020-07-20 17:06:50 +08:00
    `sendEmail` 是怎么调用的呢?我猜可能是把 `receivers` 传成 str 了
    Vegetable
        2
    Vegetable   2020-07-20 17:13:32 +08:00
    message['To'] = Header(",".join(receivers), 'utf-8')
    Aliencn
        3
    Aliencn  
       2020-07-20 17:31:21 +08:00   1
    感觉像是一楼说的那样,调用函数的时候有问题。

    可以参考我之前写的代码: https://www.aliencn.net/archives/272.html
    BryceBu
        4
    BryceBu  
       2020-07-20 17:34:02 +08:00
    第二个参数是 list 不是 str 噢
    Kvip
        5
    Kvip  
    OP
       2020-07-21 09:34:02 +08:00
    @j0hnj @Aliencn @BryceBu 感谢几位老哥,你们说的是对的。我测试了一下,确实是 list 被传成 str 所导致。不过这点倒挺奇怪的,传错对象居然也能正常收得到。
    julyclyde
        6
    julyclyde  
       2020-07-21 10:47:02 +08:00
    @Kvip 因为其实并没有传错对象啊。sendmail 函数的参数里面,发送目标是 receivers 而不是 ", ".join(receivers)
    j0hnj
        7
    j0hnj  
       2020-07-21 14:30:47 +08:00
    @Kvip #5 决定能不能收到邮件的是 `sendmail` 的 `to_addrs` 参数,这个参数可以有特殊处理,如果是 str 的话,会变成 to_addrs = [to_addrs], 代码在这里:
    https://github.com/python/cpython/blob/master/Lib/smtplib.py#L873-L874

    而 message['To'] 是决定一封邮件里收件人那一列的内容的,这里其实是可以随便写的,并不决定邮件发给谁。
    jenas999
        8
    jenas999  
       2020-07-25 23:12:20 +08:00
    message['To'] = ", ".join(receivers)
    逗号改成分号试试,群发邮件中间貌似是分号隔开
    message['To'] = "; ".join(receivers)
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