python - 如何将 msg 中的附件附加到 Mime 以在 Python 中作为电子邮件发送？

Question

我想打开一个.msg文件，获取附件，然后将它们发送到特定的邮件地址。我使用 extract_msg，我似乎无法仅将附件中的附件附加.msg到我的邮件中。

到目前为止我所拥有的：

mail = f"{path}/{attachment}"
msg = extract_msg.Message(mail)
msg_sender = str(msg.sender)
msg_subj = str(msg.subject)
msg_message = str(msg.body)

email_sender = '[e-mail]'
password = 'XXX'
send_to_email = send_to
print("----------------------------------------------------------------------------------------------------")
print(f"Send to: {send_to}\nAttachement: {attachment}\n")

msgMime = MIMEMultipart()
msgMime['From'] = msg_sender
msgMime['To'] = send_to_email
msgMime['Subject'] = msg_subj
body = MIMEText(msg_message)
msgMime.attach(body)

resources_dir = "resources"
attachments_dir = os.path.join(resources_dir, "attachments")
part = MIMEBase('application', 'octet-stream')

if msg.attachments:
    with tempfile.TemporaryDirectory() as tmp_dir_name:
        for att in msg.attachments:
            att_save_path = os.path.join(tmp_dir_name, att.longFilename)
            att.save(customPath=tmp_dir_name)

            attachment_stream = open(att_save_path, 'rb')
            part.set_payload(attachment_stream.read())
            encoders.encode_base64(part)
            part.add_header('Content-Disposition', "attachment; filename= %s" % att_save_path)
            msgMime.attach(part)

else: print("No attachments")

server = smtplib.SMTP('smtp.gmail.com', 587)
server.starttls()
try:
    server.login(email_sender, password)
except:
    print(f"Login failed for user: {email_sender}\nWith password: {password}")

text = msgMime.as_string()
server.sendmail(email_sender, send_to_email, text)
server.quit()

我得到错误：

     File "C:\Users\Kenneth.VanGysegem\AppData\Local\Programs\Python\Python39\lib\shutil.py", line 596, in _rmtree_unsafe
    with os.scandir(path) as scandir_it:
NotADirectoryError: [WinError 267] The directory name is invalid: 'C:\\Users\\XXX.XXX\\AppData\\Local\\Temp\\tmpguckqnkr\\image001.jpg'

.msg我写信给 temp 的唯一原因是，将原件的附件附加到 Mime似乎是不可能的。

欢迎任何帮助，包括错误或处理附件的更好方法。

编辑：堆栈跟踪：

Traceback (most recent call last):
  File "C:\Users\Kenneth.VanGysegem\AppData\Local\Programs\Python\Python39\lib\shutil.py", line 616, in _rmtree_unsafe
    os.unlink(fullname)
PermissionError: [WinError 32] The process cannot access the file because it is being used by another process: 'C:\\Users\\KENNET~1.VAN\\AppData\\Local\\Temp\\tmpguckqnkr\\image001.jpg'
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
  File "C:\Users\Kenneth.VanGysegem\AppData\Local\Programs\Python\Python39\lib\tempfile.py", line 801, in onerror
    _os.unlink(path)
PermissionError: [WinError 32] The process cannot access the file because it is being used by another process: 'C:\\Users\\KENNET~1.VAN\\AppData\\Local\\Temp\\tmpguckqnkr\\image001.jpg'
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
  File "C:\Users\Kenneth.VanGysegem\AppData\Local\Programs\Python\Python39\lib\tempfile.py", line 804, in onerror
    cls._rmtree(path)
  File "C:\Users\Kenneth.VanGysegem\AppData\Local\Programs\Python\Python39\lib\tempfile.py", line 812, in _rmtree
    _shutil.rmtree(name, onerror=onerror)
  File "C:\Users\Kenneth.VanGysegem\AppData\Local\Programs\Python\Python39\lib\shutil.py", line 740, in rmtree
    return _rmtree_unsafe(path, onerror)
  File "C:\Users\Kenneth.VanGysegem\AppData\Local\Programs\Python\Python39\lib\shutil.py", line 599, in _rmtree_unsafe
    onerror(os.scandir, path, sys.exc_info())
  File "C:\Users\Kenneth.VanGysegem\AppData\Local\Programs\Python\Python39\lib\shutil.py", line 596, in _rmtree_unsafe
    with os.scandir(path) as scandir_it:
NotADirectoryError: [WinError 267] The directory name is invalid: 'C:\\Users\\KENNET~1.VAN\\AppData\\Local\\Temp\\tmpguckqnkr\\image001.jpg'

标签： pythonemailmime-typesmimemsg

Answer 1

目前尚不清楚这Message()部分来自哪里或它究竟做了什么。我将假设它按您希望的那样工作，并推测您可能是 Outlook 受害者，他有幸找到了一个可以可接受地操纵其专有消息的库。

您的代码有两个有问题的流程：即使没有提取附件，它也会发送电子邮件，即使连接到 SMTP 服务器失败，它也会尝试发送。您想将这些重构为类似

else:
    print("No attachments")
    sys.exit(0)

在没有提取任何东西时退出，并且

try:
    server.login(email_sender, password)
except:
    print(f"Login failed for user: {email_sender}\nWith password: {password}")
    sys.exit(1)

当您无法继续时退出并出现错误（或者只是取出try/except并让脚本崩溃并带有回溯；如果仅供您个人使用，这可能是可以接受的，并且在您看到调试信息时更容易调试）。

（不是一个正确的答案，但我觉得我也需要指出这些问题。）