从目录中选择一个随机文件并发送(Python、MIME)

Question

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email.mime模块将其发送给您。我有一个问题，我可以选择随机文件，但由于这个错误，我无法发送它：

File "C:UsersMihkelDesktopdnak.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 = '[e-mail]'
email_send = '[e-mail]'
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()

我相信它只是把文件名作为随机选择，我怎么能让它选择文件本身呢？

编辑：在做出建议的更改后，我现在得到了这个错误：

File "C:UsersMihkelDesktope8re.py", line 29, in send
part.add_header('Content-Disposition',"attachment; filename= "+filename)
TypeError: can only concatenate str (not "list") to str

这部分似乎仍在列表中，我该如何修复？

Answer 1

您选择一个随机文件，然后将其丢弃(好吧，您打印它，然后将它丢弃(：

files = os.listdir(path)
index = random.randrange(0, len(files))
print(files[index])

(你可以用random.choice(files)做哪个BTW(

当调用open时，您会将整个files列表传递给它：

filename = files
attachment  = open(filename, 'rb')

相反，将您选择的文件传递给open：

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，所以您需要将其取回，最好使用os.path.join:

attachment  = open(os.path.join(path, random.choice(files)), 'rb')