[Asterisk-Dev] ACD calls to busy agents

[BJ Weschke]

 This has been discussed at length already today in the asterisk-users
forum. That is really the forum where continuing discussions on it belong.
  pqm and upqm really does work for agents not receiving queue calls when
they are on outbound calls. I have clients using this in production today
and I'd be happy to assist you with working through your dialplan to get it
working for you as well.
  As far as it being "cumbersome", I'm not quite sure how else you would
propose that the queue application know that an agent is busy when you're
often dialing out on a device that is at least one layer away from the agent
channel itself which then become members of the queues. While it may seem
somewhat odd at first, pqm and upqm is really a very straight forward
approach to the task at hand. If you have some other thoughts on this,
please take it back to asterisk-users or you can email me directly and we
can talk through the logic involved with this. If you have an approach that
hasn't yet been thought about and is straightforward in its implementation,
we'd like to hear from you.

 On 17 Oct 2005 18:16:56 -0400, J Thomas <jthomas at cequip.com> wrote:
>
> This is the scenario:
> 10 agents in a queue, strategy: rrmemory
>
> ACD works great. However, when an agent makes an outgoing call, she is
> still presented with an ACD call. We want the queue to recognize that
> the agent is busy and not give her any ACD call.
>
> Taking out call waiting is not an option, as we want a direct dialed
> call come to her even when she is busy.
>
> Someone suggested to use PauseQueueMember when she makes an outgoing
> call and UnPauseQueueMember on completion of call. This is very
> cumbersome and not a good solution IMO (if it works).
>
> Basically, we want the queue to present an ACD call only to an agent who
> is free.
>
> Now it may already be possible and I do not know how to do it. In that
> case I will appreciate the solution. However, to the best of my
> knowledge I think it is a feature request.
>
> Thanks,
> -- jt
>
>
>
>
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