Hi,
I don't understand how to register to an iax server an place
a call.
For exemple trying with testcall program gived with
sources.
Supposing i've a peer in my iax.conf :
[jf]
type=friend
secret=foo
auth=plaintext
host=dynamic
context=internal
peercontext=internal
qualify=yes
and in my extensions.conf
[internal]
exten => 888,1,Dial(IAX2/888 10.0.0.4)
When trying with testcall :
./testcall -u jf -p foo -h 10.0.0.2 888
OR :
./testcall -u jf -p foo -h 10.0.0.2 888 internal
i see in my cli :
: chan_iax2.c:6027 update_registry: Restricting registration
for peer
'jf' to 60 seconds (requested 300)
But no call.. no sound.. nothing happen.
If i try :
./testcall -u jf -p foo -h 10.0.0.2 888 10.0.0.4
It runs very well...
I don't understand why i can't use exten of my context and
how to avoid
this strange behavior.
regards,
jfg
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