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<p>It can also have a different path than the INVITE, if there are
proxies that do not do record-route, so there can be less Via
headers than in the INVITE.</p>
<p>By that I am wondering if the ACK relayed via tm module has the
same Via branch as , expecting not to be. Maybe in the ACK for
300+ replies...</p>
<p>Cheers,<br>
Daniel<br>
</p>
<div class="moz-cite-prefix">On 10.02.20 15:48, Yuriy Gorlichenko
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:CABSP_Vf7DFa9HyhuX-cqigPJu9nFxd_smORRQm5P-3a2Gb3BgQ@mail.gmail.com">
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<div dir="auto">ACK for successull response is a new transaction.
It has to be different. May be it is better to point provider to
this? </div>
<br>
<div class="gmail_quote">
<div dir="ltr" class="gmail_attr">On Mon, 10 Feb 2020, 14:26
Sebastian Damm, <<a href="mailto:damm@sipgate.de"
moz-do-not-send="true">damm@sipgate.de</a>> wrote:<br>
</div>
<blockquote class="gmail_quote" style="margin:0 0 0
.8ex;border-left:1px #ccc solid;padding-left:1ex">Hi,<br>
<br>
I stumbled upon an interop problem with a carrier. We have the<br>
following scenario:<br>
<br>
Gateway --> Loadbalancer --> Carrier<br>
<br>
The loadbalancer generates a Via header for each request. But
since it<br>
is stateless, the Via tag is generated for each request. As a<br>
consequence, the Via tag in the ACK differs from the one in
the<br>
INVITE. And one carrier doesn't handle those ACKs if the Via
tag<br>
differs.<br>
<br>
Is there a way to force the creation of a "deterministic" Via
branch<br>
tag? For example, building it from a hash over call-id and
from-tag or<br>
something like that?<br>
<br>
Thanks in advance<br>
Sebastian<br>
<br>
-- <br>
Sebastian Damm<br>
Voice Engineer<br>
__________________________________________<br>
sipgate GmbH<br>
<br>
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<pre class="moz-signature" cols="72">--
Daniel-Constantin Mierla -- <a class="moz-txt-link-abbreviated" href="http://www.asipto.com">www.asipto.com</a>
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Kamailio Advanced Training - March 9-11, 2020, Berlin - <a class="moz-txt-link-abbreviated" href="http://www.asipto.com">www.asipto.com</a>
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