Hello everyone,
I'm trying to get the host part from the contact header in the sip invite packet.
Following is the contact header format:
Contact: sip:user_name@192.168.1.1:5060
What I want as result is 192.168.1.1
I searched among the pseudovairables and found $ct variable which reference to the body of contact header. (sip:user_name@192.168.1.1:5060)
Is there a way to get directly the host (192.168.1.1) without applying transformation at the level of $ct.
If not what transformation can achieve it.
Thanks in advance.
Regards,
Ali
Hello,
You can use {uri.host} http://www.kamailio.org/wiki/cookbooks/5.0.x/transformations#urihost transformation.
Regards, Szabolcs
2017-03-22 18:23 GMT+01:00 Ali Taher ataher@vanrise.com:
Hello everyone,
I’m trying to get the host part from the contact header in the sip invite packet.
Following is the contact header format:
Contact: sip:user_name@192.168.1.1:5060
What I want as result is 192.168.1.1
I searched among the pseudovairables and found $ct variable which reference to the body of contact header. (<sip:user_name@192.168.1.1:5060
)
Is there a way to get directly the host (192.168.1.1) without applying transformation at the level of $ct.
If not what transformation can achieve it.
Thanks in advance.
Regards,
Ali
SIP Express Router (SER) and Kamailio (OpenSER) - sr-users mailing list sr-users@lists.sip-router.org http://lists.sip-router.org/cgi-bin/mailman/listinfo/sr-users
Ali,
A Contact header is a "name-address pair", so you first need to fetch the URI from it before you can fetch that URI's host portion. All of that can be done using Kamailio transformations:
http://www.kamailio.org/wiki/cookbooks/5.0.x/transformations
To accomplish what you'd want, you'd use:
$(ct{nameaddr.uri}{uri.host})
-- Alex
-
Alex Balashov -
Ali Taher -
Kordován Szabolcs