Question

我打开文件：

File.Copy(openFileDialog.FileName, @"C:\");

如何立即将此文件保存在光盘中？我试过了：

# -*- coding: utf-8 -*-

"""
Mail sender
"""

from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText

import smtplib
import pystache
import codecs
import time

import sys
reload(sys)
sys.setdefaultencoding('utf-8')


HOST = 'smtp.exmail.qq.com'
PORT = 587
USER = 'your@mail.com'
PASS = 'yourpass'
FROM = 'your@mail.com'

SUBJECT = 'subject'
HTML_NAME = 'tpl.html'
CSV_NAME = 'list.txt'
FAILED_LIST = []


def send(mail_receiver, mail_to):
    # text = mail_text
    html = render(mail_receiver)

    # msg = MIMEMultipart('alternative')
    msg = MIMEMultipart('mixed')
    msg['From'] = FROM
    msg['To'] = mail_to.encode()
    msg['Subject'] = SUBJECT.encode()

    # msg.attach(MIMEText(text, 'plain', 'utf-8'))
    msg.attach(MIMEText(html, 'html', 'utf-8'))

    try:
        _sender = smtplib.SMTP(
            HOST,
            PORT
        )
        _sender.starttls()
        _sender.login(USER, PASS)
        _sender.sendmail(FROM, mail_to, msg.as_string())
        _sender.quit()
        print "Success"
    except smtplib.SMTPException, e:
        print e
        FAILED_LIST.append(mail_receiver + ',' + mail_to)


def render(name):
    _tpl = codecs.open(
        './html/' + HTML_NAME,
        'r',
        'utf-8'
    )
    _html_string = _tpl.read()
    return pystache.render(_html_string, {
        'receiver': name
    })


def main():
    ls = open('./csv/' + CSV_NAME, 'r')
    mail_list = ls.read().split('\r')

    for _receiver in mail_list:
        _tmp = _receiver.split(',')
        print 'Mail: ' + _tmp[0] + ',' + _tmp[1]
        time.sleep(20)
        send(_tmp[0], _tmp[1])

    print FAILED_LIST


main()

Answer 1

您需要为File.Copy()提供绝对路径，包括其保存的文件的名称。 "C:\"不是有效的文件名，因此无法保存。尝试类似：

string fName = "myPhoto";
File.Copy(openFileDialog.FileName, @"C:\" + fName + ".jpg");

来源： MSDN

PaulF在我发布此答案之前的评论中也提到了这一点。

如何在C＃中保存打开的文件？

1 个答案: