[Freeswitch-users] Socket outbound: How to bridge two calls?
Anthony Minessale
anthony.minessale at gmail.com
Mon Nov 10 07:11:29 PST 2008
The way your script is parsing appears too simple.
The event socket has a specific protocol.
Every command has a reply and you must read it and take into account the
content len etc.
all replies, events etc have a content-type and content-len.
On Sun, Nov 9, 2008 at 7:44 AM, Dennis <odermann at googlemail.com> wrote:
> sorry for my late answer, i had to fly to a different place...
>
> we are quite sure, that we have a problem with our php script. we
> stripped down the code to the absolute minimum, to remove most
> possible error sources, but it still does not work. the code can be
> found under http://pastebin.freeswitch.org/6059.
>
> we have a very strange behavior:
> we get an inbound call over the socket to our script, we send an
> answer and then we make an originate - nothing more.
> the full fs debug can be found under http://pastebin.freeswitch.org/6060.
>
> then we change to the cli, make show channels and then uuid_bridge
> both uuid's. the result is "State RESET going to sleep".
> but, as soon as we kill the process of the script, the uuid_bridge is done.
>
> i have no idea where what the reason for the problem could be. as i
> said, everything else works, but to make an uuid_bridge, we have to
> kill the php process!?
>
>
> thanks
> dennis
>
>
> 2008/11/7 Anthony Minessale <anthony.minessale at gmail.com>:
> > Can you capture the whole log from the instant you get the inbound call
> > until you give up on the uuid bridge?
> > I don't see any of the log about your outbound call.
> >
> > are you doing
> >
> > api originate sofia/internal/user at host &park()
> >
> > For the outbound call like the cli example?
> >
> > The socket application *is* park. You do not need to tell it to park.
> >
> > as soon as you get the new inbound connection from calling socket app
> > all you need to do is answer it.
> >
> > In your code when you fork are you closing the socket in the parent side
> so
> > it's not
> > in use by both processes?
>
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--
Anthony Minessale II
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