python一键发送--当前目录下文件到邮箱

"""

目标: 把当前目录下指定类型文件发送邮件

说明:

经常需要把一些小文件如*.txt,或者*.py,*.doc之类的文件发送到邮箱里供使用或备份

用qq邮箱客户端什么的实在是太麻烦了

只要把自己需要发送的文件全部放到当前目录下,代码就可以一键发送了.

本人纯编程小白,欢迎指教

"""

 

import smtplib
import email.mime.text
import email.mime.multipart

import os
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.application import MIMEApplication

 

# 1.取得当前目录下指定文件的文件名列表,缺省为txt

def get_type_file(keyword='.txt'): # 这里可以更改扩展名如.doc,.py,.zip等等
    # 打印当前的工作目录
    print("当前目录为: ",os.getcwd())

    # 列举当前工作目录下的文件名
    files=os.listdir()
    keyword=keyword
    filelist=[]

    i=0
    for file in files:
        if keyword in file:
            i=i+1
            print(i,file)
            filelist.append(file)

    return filelist

 

# 2. 把列表内的文件发送到指定邮箱

def send_email(filelist, content=""):
    smtpHost = 'smtp.139.com'   # 139邮箱SMTP服务器
    sendAddr = '发送人邮箱'
    password = '邮箱密码'    # 163邮箱,则为授权码
    receiver = '收件人邮箱'
    subject = "邮件标题"
    content = '正文内容'

    msg = MIMEMultipart()
    msg['from'] = sendAddr
    msg['to'] = receiver
    msg['Subject'] = subject

    txt = MIMEText(content, 'plain', 'utf-8')
    msg.attach(txt)  # 添加邮件正文

    # 添加附件,传送filelist列表里的文件
    filename = ""
    i = 0
    for file in filelist:
        i = i + 1
        filename = file
        # print(str(i),filename)
        part = MIMEApplication(open(filename, 'rb').read())
        part.add_header('Content-Disposition', 'attachment', filename=filename)
        msg.attach(part)

    server = smtplib.SMTP(smtpHost, 25)  # SMTP协议默认端口为25
    # server.set_debuglevel(1)  # 出错时可以查看

    server.login(sendAddr, password)
    server.sendmail(sendAddr, receiver, str(msg))
    print("\n"+ str(len(filelist)) + "个文件发送成功")
    server.quit()

3 主程序

def main():
    filelist = get_type_file()
    send_email(filelist)


main()