[Asterisk-Users] How to generate "ringing tone" to a calling party.

Adam Goryachev mailinglists at websitemanagers.com.au
Wed Nov 17 19:46:06 MST 2004


On Thu, 2004-11-18 at 11:28, Daniel wrote:
> >On Thu, 2004-11-18 at 12:05, Chad Scott wrote:
> >> On Nov 17, 2004, at 1:30 PM, Eric Wieling wrote:
> >You *can* play a welcome message without >answering the line, however,
> >this doesn't always work. eg, I tried this >config on my PRI in Australia
> >(Telstra) and:
> >
> >a) Calling from a standard analog line I got my >welcome message as
> >expected
> >b) Calling from a mobile (Optus), I heard >silence until the call was
> >answered.
> 
> >From the point of view of Telcos, situation "a" is not "as expected". If
> this occurs you could transmit a message without paying for the call. Of
> course it will be one-way but it will be useful. Imagine, you should set a
> short message (for about 1 minute) informing everything you want!
> 

Actually, this is required to work for telco's (I would think this is
the same in most countries). Consider premium rate phone services (in
Australia, 1-900 xxx xxx) where you are charged $x per 'time unit'. eg,
$5/minute etc... The service operator is required to tell you how much
the call will cost, and allow you to dis-connect and avoid all charges.
Of course, the way this is done is to play the message before answering
the call. Once you answer the call, the telco begins charging, and you
begin providing the 'service'.

As I said, I would be interested in being able to deploy this properly,
as it means that people calling from long distance/mobiles don't have to
pay while they are waiting for a real person to answer. I imagine this
would also have a rather positive effect for people who have inbound
1300/1800 numbers (where you are charged for the incoming call, I assume
from the time you answer the call).

PS, also, premium rate numbers (1900) you are not permitted to have the
caller on hold while you are charging them etc, so again, I assume you
would need to not answer the call until they are actually connected to
the operator...

This may work differently in other countries, I may be totally wrong,
etc...

Regards,
Adam




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