22 Oct
2012
22 Oct
'12
5:28 a.m.
This is impossible, by definition, with a proxy. That is because a
proxy is a proxy, not a back-to-back user agent (B2BUA), so it does not
create two different, logically independent call legs and bridge them.
Instead, one call leg A comes in, and that same call leg B goes out.
On 10/22/2012 05:12 AM, pars3c wrote:
Hi ,
i don’t know how to realize this:
A --- INVITE---->proxy ----INVITE----> B
<--- 200 OK
---BYE---->
---INVITE ----->C
<--- 200 OK proxy <--- 200 OK
-------ACK ---> -------ACK --->
When B answer the call, the proxy should immediately hang up the call
and make a new invite. For user A is always the same call.
I tried this :
onreply_route {
################################ connect
if (is_method("INVITE") && status == "200") {
dlg_bye("callee");
drop();
}
}
But it doesn’t work. When arrive 200 , the proxy drop the message to
the user a , wait for a while, and then send a cancel to b (it should
send a BYE, instead send a CANCEL, probably because it not have a ACK).
How can I try this?
Thanks
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