Hi,

I have the /proc/net/ip_conntrack file that looks like as following:

tcp 6 95 TIME_WAIT src=192.168.252.200 dst=190.68.213.193 sport=3914 dport=8080 packets=4 bytes=184 src=190.68.213.193 dst=10.128.255.10 sport=8080 dport=3914 packets=2 bytes=88 [ASSURED] use=1 mark=0


Of course there are more lines. What I am searching for is those lines that match "dst=190.68.213.193" and DOESN'T match "dport=8080" (of course without quotes).

How can I do that?

I've tried with no success using:

grep "dst=190.68.213.193.*dport=[^8080]" /proc/net/ip_conntrack

Please could you help me to write the regular expression?


Thanks in advance!

grep "dst=190\.68\.213\.193" /proc/net/ip_conntrack | grep -v "dport=8080"

I suggest alwasy starting with the simplest expressions and testing them to be sure you know how they work.

For starters:
grep "190.68.213.193" will match your example IP, but it will ALSO match "190x68g213y193". This is because "." means "any character"

Try this:

First, find all lines containing the specified IP, then---from those lines---find the ones that do NOT contain "8080"

Thanks I thought too using 2 grep but I wonder if there is way to avoid to use a pipe....

Not tested:

Acknowledging that this was posted while I was polishing the cannonball on my post.....(I'm slow...)