I guess not an issue if the purpose of the question wasn't to test how
well you understood the regular expressions :-)


--
Marko Milivojevic - CCIE #18427 (SP R&S)
Senior Technical Instructor - IPexpert

On Thu, Oct 27, 2011 at 03:03, Brian Stajkowski
<[email protected]> wrote:
> Just curious. Could you use this:
> 645(1[2-9]|[2-9][0-9])|65+[0-9]+
> Matches above the 65535 so it's not an exact match.  What would be the
> problem in doing it this way?
> On Wed, Oct 26, 2011 at 10:20 AM, Marko Milivojevic <[email protected]>
> wrote:
>>
>> To get only 16bit private ASs (so this regex won't work with 32bit
>> private ASs), I would rather use this one:
>>
>>
>> (6451[2-9]|645[2-9][0-9]|64[6-9][0-9][0-9]|65([0-4][0-9][0-9])|(5([0-2][0-9])|(53[0-5]))
>>
>> --
>> Marko Milivojevic - CCIE #18427 (SP R&S)
>> Senior Technical Instructor - IPexpert
>>
>> On Tue, Oct 25, 2011 at 22:03, imad Abdallah <[email protected]>
>> wrote:
>> >
>> > Throughout the WBs they've been using the following AS-Path filter to
>> > select private AS numbers:
>> > 64[5-9][0-9][0-9]|65[0-5][0-9][0-9]
>> > Is this correct?As per my understanding the above filter will also
>> > match:64500, 64501..... And those are not part of the private ASs which
>> > start at 6451265540-->65599. Also outside the private AS numbers; as the
>> > last private AS is 65534
>> > Can somebody please correct me if i'm wrong. And if not, what is the
>> > quickest and easiest way to select private ASs
>> >
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>> _______________________________________________
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>
>
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