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从目录中选择一个随机文件并发送(Python、MIME)

我正在开发一个 python 程序,该程序从目录中随机选择一个文件,然后使用该email.mime模块将其发送给您。我遇到了一个问题,我可以选择随机文件,但由于此错误而无法发送:


 File "C:\Users\Mihkel\Desktop\dnak.py", line 37, in sendmemeone

    attachment  =open(filename, 'rb')

TypeError: expected str, bytes or os.PathLike object, not list

这是代码:


import smtplib

from email.mime.text import MIMEText

from email.mime.multipart import  MIMEMultipart

from email.mime.base import MIMEBase

from email import encoders

import os

import random


path ='C:/Users/Mihkel/Desktop/memes'

files = os.listdir(path)

index = random.randrange(0, len(files))

print(files[index])


def send():

    email_user = 'yeetbotmemes@gmail.com'

    email_send = 'miku.rebane@gmail.com'

    subject = 'Test'

    msg = MIMEMultipart()

    msg['From'] = email_user

    msg['To']   = email_send

    msg['Subject'] = subject

    body = 'Here is your very own dank meme of the day:'

    msg.attach(MIMEText (body, 'plain'))

    filename=files

    attachment  =open(filename, 'rb')

    part = MIMEBase('application','octet-stream')

    part.set_payload((attachment).read())

    encoders.encode_base64(part)

    part.add_header('Content-Disposition',"attachment; 

    filename= "+filename)

    msg.attach(part)

    text = msg.as_string()

    server = smtplib.SMTP('smtp.gmail.com',587)

    server.starttls()

    server.login(email_user,"MY PASSWORD")

    server.sendmail(email_user,email_send,text)

    server.quit()

我相信它只是将文件名作为选定的随机选择,我怎么能让它选择文件本身?


编辑:在进行建议的更改后,我现在收到此错误:


File "C:\Users\Mihkel\Desktop\e8re.py", line 29, in send

    part.add_header('Content-Disposition',"attachment; filename= "+filename)

TypeError: can only concatenate str (not "list") to str

似乎这部分仍在列表中,我该如何解决?


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慕姐4208626

您选择一个随机文件,然后将其丢弃(好吧,您打印它,然后将其丢弃):files = os.listdir(path)index = random.randrange(0, len(files))print(files[index])(顺便说一句,你可以用它做什么random.choice(files))并在调用open时将整个files列表传递给它:filename = filesattachment  = open(filename, 'rb')相反,传递open您选择的文件:attachment  = open(random.choice(files), 'rb')但是,这仍然不起作用,因为listdir只返回文件名而不是完整路径,因此您需要将其取回,最好使用os.path.join:attachment  = open(os.path.join(path, random.choice(files)), 'rb')
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