On Thu, 2 May 2002, dionysus wrote:
> linux box (RedHat 7.2 (more or less stock install)):
> eth0 = 192.168.0.1
> eth1 = 192.168.0.3
>
> windows1:
> 192.168.0.2 (connected to 192.168.0.1)
>
> windows 2:
> 192.168.0.4 (connected to 192.168.0.3)
>
> and the routing table looks like
>
> windows1 192.168.0.1 255.255.255.255 UGH 0 0 0 eth0
> windows2 192.168.0.3 255.255.255.255 UGH 0 0 0 eth1
> 127.0.0.0 * 255.0.0.0 U 0 0 0 lo
>
> both of the windows machines have the correct default gateways
> installed.
>
> I'm probably missing something really obvious here, but I've tried
> pretty much every combination of options in netconf's routing section,
> and I'm damned if I can think of anything else that I should do.....
The problem is in your network design.
By having both ethernet cards in the same network {192.168.0.x}, the Linux
box is automatically asuming that any packet sent to a 192.168.0.x address
doesn't need to be passed on, because it's on the same network as the card
which receives the pcket.
What you need to do is something like this.
linux box
eth0 = 192.168.0.1
eth1 = 192.168.1.1
windows1:
192.168.0.2 (connected to 192.168.0.1)
windows 2:
192.168.1.2 (connected to 192.168.1.1)
Then your routing _should_ default to delivering the packets to the right
place - you shouldn't need routing table modifications, but if you do,
they should come out something like this.
windows1 192.168.0.1 255.255.255.255 UGH 0 0 0 eth0
windows2 192.168.1.1 255.255.255.255 UGH 0 0 0 eth1
127.0.0.0 * 255.0.0.0 U 0 0 0 lo
DaZZa
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