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Need a IP-block/unblock Function for Windows2003

Hello Experts,

i have a Windows2003 Server, i need to block and unblock a specific IP by my program. (Simply to block all connection attempts from one IP).

What api's are recommended, what SDK do i need?
A brief description and a example of a function call would be appreciated.

When using a ISA firewall, does a commandlinke tool exist like ip-chains ip-tables on linux?


thanks
Alex P.

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AlexPiko
Asked:
AlexPiko
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1 Solution
 
bbaoIT ConsultantCommented:
hi Alex,

you can certainly do that with any program called Firewall, either ISA or other 3rd-parties, either free or commercial. but it sounds not necessary to use them only for such a blocking function you need, except you dont care your money or computer performance.

ok. actually, you can block ONE IP with the following single command:

ROUTE ADD 1.2.3.4 mask 255.255.255.255 192.168.0.253

where 1.2.3.4 is the IP you want to block, 192.168.0.253 is a non-existing IP on your local subnet assuming it is 192.168.0.x.

this command will cause you system to reply any request from the specific IP to a null address, which is actually blocking the communication between your host and that IP, beause the remote IP to be blocked will never receive the returning packet.

hope it helps,
bbao
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AlexPikoAuthor Commented:
Thankfs for your comment bbao,
very Interesting thing. I didnt thought about routing.
How could this be done? -> BLOCK ALL IP's initially, unblock a few IP's on demand.

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bbaoIT ConsultantCommented:
> BLOCK ALL IP's initially (DO THIS FIRST)

redirect all outgoing traffic to an non-existing IP assuming 192.168.0.253:
ROUTE ADD 0.0.0.0 MASK 0.0.0.0 192.168.0.253

or

delete current default gateway to disable all outgoing traffic to other net at all:
ROUTE DELETE 0.0.0.0 MASK 0.0.0.0

> unblock a few IP's on demand. (DO THIS SECOND)

enable a group of IPs (1.2.3.0~1.2.3.15) to access the local host via 192.168.0.1 (the default gateway):
ROUTE ADD 1.2.3.0 MASK 255.255.255.240 192.168.0.1

hope it helps,
bbao
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AlexPikoAuthor Commented:
this is a very simple solution.
A good point to start.
thanks
Alex
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bbaoIT ConsultantCommented:
simple does not mean weak. right? :) thanks for you grade. enjoy, bbao
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