AttributeError: 'list' object 没有属性 'encode'。 我如何通过 Excel 发送 Email？ Python

Question

所以我试图从 Excel 和 Python 发送一封电子邮件。现在我只在 A1 列中设置了所有电子邮件，用分号分隔。 如果我直接将电子邮件放入 Python，它们会发送出去，但是当我使用 Python 调用时，它给我这个错误吓坏了。 我究竟怎样才能使这项工作？

import smtplib
from email.mime.image import MIMEImage
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart
import pandas
from datetime import datetime

server = smtplib.SMTP('relay.xxx.com')
email_df = pandas.read_excel(f"E-mails.xlsx", sheet_name="Emails", nrows=1, usecols = "A")
sender = 'email@email.com'

receivers = 'Email'
receivers = list(email_df)

print(receivers)
msg = MIMEMultipart('related')

msg['Subject'] = 'Test mail'
msg['From'] = sender

with smtplib.SMTP('relay.xxx.com') as server:
    msg['To'] = receivers
    server.sendmail(sender, receivers, bytes(msg.as_string()))
    print(f"Successfully sent email to: {receivers}")

我不想发送多封电子邮件。 我想要一封电子邮件，地址栏中有一堆电子邮件。

或者，如果有一种方法可以按列读取（a1 中的 addr1，a2 中的 addr），那也会更好。

欣赏它。

Answer 1

如果您可以将您的 email 地址（来自您的 df）放入所需格式的列表中，例如：

print(receivers)    
['email1@email1.com', 'email2@email2.com', 'email3@email3.com']

然后您可以在分配后添加它以将其转换为 send_mail 的字符串（因为它需要字符串而不是列表）：

receivers = ", ".join(receivers)

您提到，如果您的电子表格中 A 列下每行有一个 email 地址，那将是理想的。 因此，如果您可以将您的 Excel 电子表格转换为这种格式（列标题为“电子邮件”），那么您可以这样做以获得上述列表：

email_df = pandas.read_excel(f"E-mails.xlsx", sheet_name="Emails", usecols = "A")
receivers = email_df['Email'].tolist()

因此，为了完整性（使用新格式的 Excel 电子表格，A 列下每行一个 email 地址），您的代码将更改为：

import smtplib
from email.mime.image import MIMEImage
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart
import pandas
from datetime import datetime

server = smtplib.SMTP('relay.xxx.com')

email_df = pandas.read_excel(f"E-mails.xlsx", sheet_name="Emails")
receivers = email_df['Email'].tolist()

print(receivers)  # Check list is one valid email address per element

receivers = ", ".join(receivers)

sender = 'email@email.com'
msg = MIMEMultipart('related')

msg['Subject'] = 'Test mail'
msg['From'] = sender

with smtplib.SMTP('relay.xxx.com') as server:
    msg['To'] = receivers
    server.sendmail(sender, receivers, bytes(msg.as_string(), encoding='utf-8'))
    print(f"Successfully sent email to: {receivers}")

Answer 2

如果构建消息，最好使用send_message接口。 我建议使用较新的EmailMessage界面，它比旧的MimeMultipart界面更干净、更简单：

import smtplib
from email.message import EmailMessage

...
msg = EmailMessage()

msg['Subject'] = 'Test mail'
msg['From'] = sender

msg.set_payload('Message text.\nLine 2...')
# or eventually add parts with msg.add_related(...), msg.add_alternative(..)
#  or msg.add_attachment(...)


with smtplib.SMTP('relay.xxx.com') as server:
    msg['To'] = receivers
    server.send_messages(msg)
    print(f"Successfully sent email to: {receivers}")