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Using the learning platform, the subject is snort rules.

Question is:

Create a snort rule that will alert on traffic with destination ports 443 and 447.

My attempts:

alert tcp any any -> any 443 447 ( msg:"Sample alert"; sid:1; rev:1; )

alert tcp udp any any -> any 443 447 ( msg:"Sample alert"; sid:1; rev:1; )

My answer is wrong and I cant see why. Any pointers would be very much appreciated.

  • If the answer provided helped, then upvote and click the check mark to accept and the answer. – schroeder Feb 7 at 16:39
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I had to solve this exact case for Immersive Labs! This is the rule you are looking for:

alert tcp any any -> any [443, 447] ( msg:"Sample alert"; sid:1000001; rev:1; )

Also, I noticed your sid:1. All sid up to 1,000,000 are reserved. So your sid must be at least 1000001.

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I'm not familiar with snort. However, the snort documentation gives this example:

alert tcp any any -> 192.168.1.1 80 ( msg:"A ha!"; content:"attack"; sid:1; )

The structure is:

action proto source dir dest ( body )

It also states, that:

source - specifies the sending IP address and port, either of which can be the keyword any, which is a wildcard.

dir - must be either unidirectional as above or bidirectional indicated by <>.

dest - similar to source but indicates the receiving end.

There is no indication made, that you can match multiple ports at once. I'd therefore try the following rules:

alert tcp any any -> any 443 ( msg:"Sample alert 443"; sid:1; rev:1; )

alert tcp any any -> any 447 ( msg:"Sample alert 447"; sid:2; rev:1; )

Edit: If your question was how to achieve this in one rule, you might want to try:

alert tcp any any -> any [443,447] ( msg:"Sample alert"; sid:1; rev:1; )

You will also probably find this site useful.

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