From: George Hansen (HansenG@xxxxxxxxxxxxxxxx)
Date: Wed Dec 05 2001 - 19:38:49 GMT-3
There's one significant difference that doesn't easily show up because you are
using long AS numbers. If you substituted with AS 1 and AS 80, you'd see that ^
1_.*_80$ gives what you asked for, and ^1.*80$ gives more than you wanted.
George
>>> "Albert Lu" <albert_ccie@yahoo.com> 12/04/01 07:06PM >>>
Greg,
I've tried it with ^7474.*13606$ and ^7474_.*_13606$ and ^7474 .* 13606$
They all work. What is the most correct solution?
Albert
-----Original Message-----
From: Gregory W. Posey Jr. [mailto:gposey@uaes.org]
Sent: Wednesday, December 05, 2001 11:22 AM
To: Albert Lu; ccielab@groupstudy.com
Subject: RE: BGP Regular Expression Quizz
I would think the appropriate regular expression would be...
^7474 .* 13606$
Since .* is the same as all paths (according to Halabi)
Thank you,
Greg Posey Jr.
CCIE #7981
CCNP - Security Specialist
Cisco Voice Access Specialist
M.S. EE
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com]On Behalf Of
Albert Lu
Sent: Tuesday, December 04, 2001 6:54 PM
To: ccielab@groupstudy.com
Subject: BGP Regular Expression Quizz
Quick quizz on how well you guys know your Regular Expressions.
How can match a route that has originated from AS 13606 and ending at AS
7474
An example would be
Network Next Hop Metric LocPrf Weight Path
* i209.37.145.0 216.14.195.13 1000 0 7474 701 7018
13606 i
My best attempt is:
^7474(_[0-9]+_)13606$
It doesn't work
Thanks
Albert
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