Re: Weird TCP flags?

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On Fri, 2003-12-12 at 08:14, Ian Hunter wrote:
> > > > Dec 11 22:58:52 lucy kernel: Fwd DMZ->Internet DROP: IN=eth1 OUT=ppp0
> > > > SRC=192.168.254.242 DST=204.157.6.223 LEN=60 TOS=0x00 PREC=0x00 TTL=63
> > > > ID=56169 DF PROTO=TCP SPT=80 DPT=56319 WINDOW=32476 RES=0x00 ACK SYN
> URGP=0
> > >
> > Is there any chance that port 80 traffic is being allowed in under a
> > rule that does not enter the connection in conntrack?
> > In other words, if you have a rule such as:
> >
> > iptables -A FORWARD -d 192.168.254.242 -p tcp --dport 80 -j ACCEPT
> >
> > The SYN packet will be sent to your web server and the web server will
> > respond with a SYN, ACK packet but it will not be related to anything in
> > the conntrack table.  The connection may be being allowed by a more
> > general rule than one to the web services.
> >
> > So, I suppose we should ask what your ACCEPT rules look like - John
> 
> I'm absolutely doing that -- here are the relevant ACCEPTs:
> 
> iptables -A FORWARD -i ppp0 -o eth1 -p tcp --dport 80 -d 192.168.254.242 -j
> ACCEPT
> iptables -t nat -A PREROUTING -i ppp0 -d <my.public.ip.here> -p tcp --dport
> 80 -j DNAT --to-destination 192.168.254.242
> 
> If I want to forward port 80 to that box, isn't that the way to do it?  How
> would a connection escape conntrack?
> 
> Thanks MUCH!
> 
> Ian

You never told it to make an entry in conntrack.  Try this:
iptables -A FORWARD -i ppp0 -o eth1 -p tcp --dport 80 -d 192.168.254.242
-m state --state NEW -j
ACCEPT
-- 
John A. Sullivan III
Chief Technology Officer
Nexus Management
+1 207-985-7880
john.sullivan@xxxxxxxxxxxxx
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