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I am using Windows Vista, and only have 1 NIC. I know that in Linux, I can create IP aliasing such as:

  • eth0 - 192.168.1.1
  • eth0:1 - 192.168.2.1
  • eth0:2 - 192.168.3.1

and access all 3 subnets with just one eth interface.

Is there such an equivalent in Windows Vista?

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1 Answer 1

up vote 2 down vote accepted

This is possible in Windows Vista, but you might have to use all-static IPs (no DHCP) if I'm remembering correctly. Follow these steps:

  1. Open the Control Panel
  2. Open the "Network and Internet" applet
  3. Click on "Network and Sharing Centre"
  4. Click on "Manage network connections" (below the "List of tasks" heading)
  5. Right-click on the appropriate NIC
  6. Click on "Properties" (usually at the bottom of the pop-up menu that will appear)
  7. Double-click on "Internet Protocol Version 4 (TCP/IPv4)" protocol
  8. Click on the "Advanced" button

In the Advanced settings, you will want to find the screen that looks like this one from XP (the one in Windows Vista looks almost exactly like this but with the current Windows Vista GUI theme), which will let you specify multiple IP addresses by clicking on the "Add" button:

enter image description here

As you can probably see, you can specify a subnet mask. This is a useful feature because different subnets could have different masks (for the most-part, the mask reveals the size of the subnet that's directly accessible from your computer).

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1  
Thanks. I tried what you suggested, but wasn't able to connect to the servers on the 2 subnets. –  Rayne Sep 2 '11 at 10:00
    
But you can set up both IPs, right? If so, then there's either a routing problem at the ethernet switch (I've encountered some cheaper models that don't handle multiple subnets very well), or there's a problem with Windows Vista (which I suspect is more likely). Try turning off your firewall temporarily to rule out interference from Windows' security features. See also the "ROUTE PRINT" command, which will output Windows' routing table to the screen (and not the printer, despite the unintuitive parameter). –  Randolf Richardson Sep 2 '11 at 15:24

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