> '!(ether[0:2] = 0x0002 and ether[2] = 0x2d) and > !(ether[0:2] = 0x0050 and ether[2] = 0x8b)' I think. > But double check my boolean logic; it's > rusty and it's almost 2:30 AM here :-). Thanks Marco. I should be able to combine the third byte with the first two bytes. If so, my capture expression would look like: !(ether[6:3] = 0x00022d) and !(ether[6:3]=0x00508b). Is this right? (I am looking at the source addresses, so it is at an offset of 6) Thanks, rakesh __________________________________________________ Do You Yahoo!? Great stuff seeking new owners in Yahoo! Auctions! http://auctions.yahoo.com
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