[Asterisk-Users] queues that do not play music

[Andre Ruiz]

Is there a way to send a caller directly to the queue member if one is
available, without answering it and putting him on MOH? Note that
playing a fake ringing tone instead of music does not suffice, I
really want to not answer it.

The problem: it´s very hard to convince a client that a caller should
go all the way to queue, thankyou, position in line, music and being
picked up, even when the queue is empty. They want the caller to just
ring the queue member´s phone, directly, and only go to queue() if all
of them is busy.

What I made so far is:

1 - try all phones before calling queue()
2 - if one is busy try the next, but if one rings until timeout, call queue()
3 - if all are busy, call queue()

It works very well, but has one major flaw: the calls that get to the
queue will be distributed using the queue´s strategy (random for
example), but the calls that goes directly to the extensions before
being queued go in a static order (roundrobin without memory) and so
they will overload the first person.

It would be very nice if the queue itself would distribute also this
calls, using the same strategy, but passing it directly without
queuing/thankyou/position-in-queue/music/etc.

Part of the code I did:

exten => 333,7,Dial(SIP/3000,10,)
exten => 333,8,Goto(11)
exten => 333,9,Dial(SIP/3027,10,)
exten => 333,10,Goto(11)
exten => 333,11,Dial(SIP/3001,10,)
exten => 333,12,Goto(13)
; atende a fila normalmente
exten => 333,13,Wait(2)
exten => 333,14,Answer()
exten => 333,15,Playback(queue-welcome)
exten => 333,16,Queue(333|t|||0)
exten => 333,108,Goto(9)
exten => 333,110,Goto(11)
exten => 333,112,Goto(13)

Thank you,
andre

--
Andre Ruiz  <andre.ruiz at gmail.com>
Curitiba, PR, Brasil