尝试使用 smtplib 和 email.mime.multipart 从 python 发送电子邮件，收到错误“'Series' 对象没有属性 'encode'”

Question

我正在尝试制作一个采用 excel sheet 作为输入文件的 streamlit 应用程序，将其保存为工作文件，然后将邮件发送到存储的电子邮件地址。我收到此错误： AttributeError: 'Series' 对象没有属性 'encode'

代码如下：

import streamlit as st
import pandas as pd 
import numpy as np 
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
import smtplib,ssl

st.title("BD Triggers-Lead Generator Mail")
#uploaded_file = st.file_uploader("Choose a file")
temp_file = st.file_uploader("Enter file here!")
if temp_file: 
    temp_file_contents = temp_file.read()

if st.button("Save as working file"):
    with open("ON_DISK_FILE.extension","wb") as file_handle:
        file_handle.write(temp_file_contents)


result= st.button('Click To Send Mail')
st.write(result)
if result:

    my_email= "example@gmail.com"
    password= "abc@123"



    server = smtplib.SMTP_SSL('smtp.gmail.com' ,465)
    server.ehlo()
    server.login(my_email, password)
    email_list = pd.read_excel("ON_DISK_FILE.extension") 
    st.write(email_list)
 

 
    #defining objects
    names = email_list['Lead Generated']
    emails = email_list['Lead generator Email']
    subjects = email_list["Subject"]    
    ccs=email_list['CCs']

    for i in range(len(emails)):
        name=names[i]
        email=emails[i]
        subject=subjects[i]
        cc=ccs[i]

        msg=MIMEMultipart()
        msg['Subject']=subjects
        msg['From']=my_email
        msg["To"]=email
        msg["Cc"]=cc
        text="Hi"

        part1 = MIMEText(text, "plain")
        msg.attach(part1)
        
        server.sendmail(msg["From"], msg["To"].split(",") + msg["Cc"].split(","), msg.as_string())
    
    server.close()

这是追溯：

    File "c:\users\dell\anaconda3\lib\site-packages\streamlit\script_runner.py", line 338, in _run_script
    exec(code, module.__dict__)
File "C:\Users\DELL\Desktop\bdmail.py", line 81, in <module>
    server.sendmail(msg["From"], msg["To"].split(",") + msg["Cc"].split(","), msg.as_string())
File "c:\users\dell\anaconda3\lib\email\message.py", line 158, in as_string
    g.flatten(self, unixfrom=unixfrom)
File "c:\users\dell\anaconda3\lib\email\generator.py", line 116, in flatten
    self._write(msg)
File "c:\users\dell\anaconda3\lib\email\generator.py", line 195, in _write
    # Write the headers.  First we see if the message object wants to
File "c:\users\dell\anaconda3\lib\email\generator.py", line 222, in _write_headers
    #
File "c:\users\dell\anaconda3\lib\email\_policybase.py", line 326, in fold
    return self._fold(name, value, sanitize=True)
File "c:\users\dell\anaconda3\lib\email\_policybase.py", line 369, in _fold
    parts.append(h.encode(linesep=self.linesep, maxlinelen=maxlinelen))
File "c:\users\dell\anaconda3\lib\site-packages\pandas\core\generic.py", line 5274, in __getattr__

我无法修复错误，有人可以帮我解决这个问题吗？

Answer 1

这里的错误信息是一个提示：它说你正试图在该行中处理一个 pandas Series:

server.sendmail(msg["From"], msg["To"].split(",") + msg["Cc"].split(","), msg.as_string())

这意味着在撰写邮件时，您在某处编写了一个原始系列。罪魁祸首在这里：

msg['Subject']=subjects      # subjects is the full pandas column, ie a Series

整个问题是由打字错误引起的：您可能想要：

msg['Subject']=subject

从中学到什么：

你已经在你跌倒的地方设置了陷阱，通过使用单数和复数形式的相同变量来避免处理索引形式（subject for subjects[i]).本身没有什么不好的，但是需要非常谨慎。
Python 错误消息通常包含很多信息 前提是您尝试理解它们：引发错误的行，通常是有关类型或内容的信息有问题的变量（这里涉及 Pandas Series）

尝试使用 smtplib 和 email.mime.multipart 从 python 发送电子邮件，收到错误“'Series' 对象没有属性 'encode'”

Trying to send emails from python using smtplib and email.mime.multipart, getting the error "'Series' object has no attribute 'encode'"

python

email

smtplib

pandas

streamlit