《数据结构、算法与应用》第一章习题1.2

一年前写的一个脚本， 通过爬虫技术，将网站页面上出现的email地址给扒下来，

然后进行重发邮件， 由于邮箱的安全机制， 可能会进行阻拦(可以考虑怎么避免这样).........

附代码：

main.py

#! /usr/bin/python
# -*- coding: utf-8 -*-

import re, urllib, sys, SendMail

patt2 = ‘\w+@\w+\.com‘
pattern = ‘[\w.-]+@[\w.-]+\.\w+‘

if len(sys.argv) < 2:
	print "print mail --help for more help"
	sys.exit(1)

if sys.argv[1].startswith("--help"):
	print ‘‘‘mail [url]
	send email to the catched mail address‘‘‘
	sys.exit(1)

else:
	
	url = sys.argv[1]
	info = urllib.urlopen(url).read()
	mailinfo = re.findall(pattern, info)

	hostname = raw_input("input the hostname you want:")
	mailsub = raw_input("input the subject of you mail:")
	mailcontext = raw_input("input the context you send:")
	
	for mailaddress in mailinfo:
		if SendMail.send_mail(hostname, mailaddress, mailcontext, mailsub) == 1:
			print "成功向%s发送邮件" % mailaddress
		else:
			print "发送到%s的邮件发生错误" %  mailaddress

其中的mail 主机 和用户密码可以手动进行改动！

#! /usr/bin/python


import smtplib, sys
from email.mime.text import MIMEText


def send_mail(mailfrom, mailto, context, sub):
	
	mail_host = "smtp.qq.com"   //可以改动
	mail_usr = "872639280@qq.com" 
	mail_passwd = "*****"
	
	msg = MIMEText(context)
	msg[‘Subject‘] = sub
	msg[‘From‘] = mailfrom
	msg[‘To‘] = mailto
	
	try:	
		s = smtplib.SMTP()
		s.connect(mail_host)
		s.login(mail_usr, mail_passwd)
		s.sendmail(mail_usr, mailto, msg.as_string())
		s.close()
		return 1
	except:
		return 0


if __name__ == "__main__":
	if send_mail("HuntingGo@Hunting.Go", "872639280@qq.com", "miss u", "miss u"):
		print "send success!"
	else:
		print "failed to send!"

《数据结构、算法与应用》第一章习题1.2,布布扣,bubuko.com

《数据结构、算法与应用》第一章习题1.2

原文：http://blog.csdn.net/oktears/article/details/22597113