Python 发送邮件，邮件可以正常接收，但收件人栏显示异常

# 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)

receivers

filepath

收件人

import

8 条回复 • 2020-07-25 23:12:20 +08:00

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

2020-07-20 17:13:32 +08:00

message['To'] = Header(",".join(receivers), 'utf-8')

Aliencn

2020-07-20 17:31:21 +08:00

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

2020-07-20 17:34:02 +08:00

第二个参数是 list 不是 str 噢

Kvip

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

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

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

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)