I need to route RDP connections to an available XP virtual machine through linux. Is there such a software?

I have a pool of Windows Machines (VM) all with different IP addresses and I want the proxy

  1. hide all these different IP addreses. Someone connecting will have only the proxy address
  2. the proxy will route to an available Windows VM (no one is using)

Which of the following situations do you mean?

  1. [RDP client on Windows machine] => [linux server] => [windows Terminal Server]

  2. [RDP client on MacOSX/Linux machine] => [linux server] => [windows terminal Server]

    • If #2 is the case, use 'sshuttle' from https://github.com/apenwarr/sshuttle It will allow you to tunnel from your MacOSX/Linux machine to the linux server and proxy your TCP connections destined for the ip address/ip address ranges through the linux server.


Based on the comment, it sounds like what is desired is:

[client(s)] => [proxy server] => [pool of windows servers]

Client only has login credentials and connect information to the proxy server. Based on the client's login credentials, they would only be able to connect to their own windows machine in the backend pool.

To that end, a login session manager and a session key-based proxy would allow for "anonymous" access to the backend pool.

This article comes to mind:


While not a complete solution, it allows for persistent connections between a remote client and a windows terminal server on the backend. With an appropriate bit of app coding to enforce a particular backend host to a particular remote client on login to the web interface, you should be able to implement what you are asking about.

|improve this answer|||||
  • thanks @Wing Tang Wong but the situation goes a bit behond that. – Guerrio Apr 22 '13 at 13:24
  • Updated my answer with a link to an RDP-cookies based HA-proxy config solution for allowing persistance from a remote client to a backend windows terminal server. Coding a front end app, which crafts and assigns the cookie on login would be required. – Wing Tang Wong Apr 22 '13 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.