[Freeswitch-users] Oubound ESL: unbridge

Sean Devoy sdevoy at bizfocused.com
Tue Apr 8 20:24:22 MSD 2014


Can't you just bridge the call instead of originate and uuid_bridge?

>From the WIKI:
Q: Can I bridge a call with an Outbound socket?
Yes you can, simple execute a 'Bridge'
sendmsg
call-command: execute
execute-app-name: bridge
execute-app-arg: {ignore_early_media=true}sofia/gateway/myGW/177808
event-lock: true
The event-lock is key here, if you don't have it set, other events you might have sent can terminate the call, even if you've sent them before the bridge command. Remember that if you want to hangup the call when the bridge completes, either look for the event [CHANNEL_UNBRIDGE], or send a hangup event after the bridge.


From: freeswitch-users-bounces at lists.freeswitch.org [mailto:freeswitch-users-bounces at lists.freeswitch.org] On Behalf Of Grant Bagdasarian
Sent: Tuesday, April 08, 2014 6:29 AM
To: FreeSWITCH Users Help
Subject: Re: [Freeswitch-users] Oubound ESL: unbridge

Hello,

Could anyone with Outbound ESL experience help me out with this matter?

Thanks,

Grant

From: freeswitch-users-bounces at lists.freeswitch.org<mailto:freeswitch-users-bounces at lists.freeswitch.org> [mailto:freeswitch-users-bounces at lists.freeswitch.org] On Behalf Of Grant Bagdasarian
Sent: Monday, March 31, 2014 1:11 PM
To: FreeSWITCH Users Help (freeswitch-users at lists.freeswitch.org<mailto:freeswitch-users at lists.freeswitch.org>)
Subject: [Freeswitch-users] Oubound ESL: unbridge

Hello,

I'm working on a C# application which takes control over an inbound call in Freeswitch using ESL outbound and it works like a charm! I can answer the call, play something, get digits, etc.
Now I'm at the point of implementing bridge functionality in the application.

Which one is the best approach and why?

1)      >From the Outbound ESL Application use the originate command to create a new outbound call and then use uuid_bridge to connect both incoming and outgoing?

2)      Use the bridge command from the esl application?

I would assume the first option would give me control over the b leg created as well, and the second option would not? This also brings me to my next question:

-          Is there a way to "unbridge" both call legs, do some stuff on the incoming (a) leg, and then bridge them back without disconnecting any of the legs?

Regards,

Grant
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