On 3/26/10 4:25 PM, Uriel Rozenbaum wrote:
Done,
I had a lot of errors so I'll just show the final version that works OK.
=~ "192\.168\.([0-9]{1,3})\.([0-9]{1,3})
The only drawback is that I could pass as valid 192.168.999.999 but as
these IPs come from a DNS query, I assume they'll be fine.
I think you can reduce it to be just: "^192\.168\."
Since it comes from dns server or socket attributes, then you are safe.
The other option is to convert it to integer (with transformations) and
check it as net mask via bit wise operators.
Cheers,
Daniel
Cheers,
Uriel
On Fri, Mar 26, 2010 at 11:29 AM, Uriel Rozenbaum
<uriel.rozenb...@gmail.com <mailto:uriel.rozenb...@gmail.com>> wrote:
Hi Alex,
Actually what I'm trying to do is check the IPs on a request on a
Kamailio+RTPProxy acting as border of our network.
So I have the ingress IP and egress IP and need to check if I have
to bridge ii, ei, ie or ee.
I managed to obtain all IPs in AVPs, but now I have to check if
they are public or private.
So far our network uses only 192.168.x.x class for private servers.
Thanks for the quick reply
Uriel
On Fri, Mar 26, 2010 at 10:56 AM, Alex Balashov
<abalas...@evaristesys.com <mailto:abalas...@evaristesys.com>> wrote:
172.16.0.0/12 <http://172.16.0.0/12> does not line up on octet
boundaries. You will need to do something other than a
regular expression. Fortunately, 'src_ip' is a composite that
supports comparisons against subnets in shorthand CIDR notation.
It might also be that whatever you are trying to accomplish
can be done better some other way, but since you did not pose
the question in terms of the objective, I cannot speak to that.
--
Sent from mobile device
On Mar 26, 2010, at 9:46 AM, Uriel Rozenbaum
<uriel.rozenb...@gmail.com <mailto:uriel.rozenb...@gmail.com>>
wrote:
Hi guys,
Does anyone have a REGEX syntax to match a private IP on
the 192.168.x.x range?
I'm trying with:
if($avp(s:ip_origen)=~"192.168(\.([1]?\d{1,2}|2[0-4]{1}\d{1}|25[0-5]{1})){2}"
)
But all IPs pass as private, even public ones.
Thanks!
Uriel
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