sending invite request to myself says error: "cannot call to myself" about drachtio-fsmrf HOT 5 CLOSED

Yes, for loop detection purposes the drachtio server will reject any attempt to send a SIP message for itself.

What are you trying to accomplish that you think requires this?

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Hi,
thank you for the response.

all i want to do is to send request to my drachtio server using the request function (like for now i am sending using zoiper) or by creating the srf.createUAC function to connect the call to the dialogflow agent so the request will hit the srf.invite function then it will execute and forward the call to the outbound number... as in the code i have pasted above.

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in short, I want to invoke srf.invite function via node.js app or possibly http request ..

kindly guide me to accomplish this.

i look forward for the response.

from drachtio-fsmrf.

You want to dial out and have the answered party connected to dialogflow it sounds like. You are going about it the wrong way. You should instead do this:

1. connect to the media server (mrf.connect)
2. create an endpoint (ms.createEndpoint)
3. Send an INVITE offering the SDP of the endpoint (srf.createUAC)
4. Modify the endpoint to stream to the caller once he or she answers (ep.modify)

At that point you can issue whatever commands you want to the endpoint, as far as connecting to dialogflow etc

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Thanks @davehorton It works now

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