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So, Lets say I have a Windows 2k3 server which has IIS. Lets say the address is 100.50.25.12. Now, lets say I have my stand alone RestLet webservice listening on port 1234. To hit this service externally, I can simply point to 100.50.25.12:1234. However, lets say I would rather hit my web service using services.my-domain.com and have that forward onto 100.50.25.12:1234. How would I configure this in IIS?

I think a host header setup would work, where I would just point services.my-domain.com (port 80) to localhost:1234 and it should forward my request onto my stand alone app, correct? I was under the impression that host headers would only work with websites in IIS, but I don't think this is the case? Thanks

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What you are looking for is called a reverse proxy. This isn't something that is built into IIS6, so you need to get a third party module to perform this function. Managed Fusion URL Rewriter has this capability.

Since you added the Windows 2008 tag on there, I'll talk about IIS7 as well. You could still use the Managed Fusion product - but a better option would be to get the Application Request Routing module and use that as a reverse proxy.

Another option would be to move your IIS site to a different port as well, and use a dedicated reverse proxy server, such as nginx, in front of both your IIS site and your RestLet service.

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  • Would this also be something that my domain name provider (say, GoDaddy) could do instead? Thanks for the response
    – user34975
    Feb 16, 2010 at 20:54
  • If all they are DNS services, then I don't think there is anything they can do. (They just tell the world that services.my-domain.com = 100.50.25.12 - no port information at that level.) - if you can get another IP/hostname though, then they could - you would have IIS listen on 1 ip, RestLet listen on the other (both on port 80) and then you could have services.my-domain.com point to IIS and restlet.my-domain.com point to the RestLet service.
    – MattB
    Feb 16, 2010 at 21:24
  • Thanks for the info! I'll have to dig and figure this out when I have the time, but this answer looks right.
    – user34975
    Feb 18, 2010 at 21:14

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