|Subject:||[Linphone-users] No dtmf generator when attempting to play dtmf [Python]|
|Date:||Wed, 16 Sep 2015 13:39:53 +0000|
I am having trouble generating dtmf tones when I receive a call. Here is what my call_state_changed callback looks like (this is in Python):
def call_state_changed(core, call, state, message):
logging.warning("call_state_changed: " + str(state) + ", " + message)
if state == linphone.CallState.IncomingReceived:
print "***Incoming call on " + core.identity
time.sleep(10) # wait 10 seconds for audio to play
elif state == linphone.CallState.Connected:
core.play_dtmf(1,300) # press 1 for 300 ms to acknowledge the alert
print "***Call successfully answered on " + core.identity
elif state == linphone.CallState.End:
print "***Call successfully hung up on " + core.identity
elif state == linphone.CallState.Error:
logging.error("call_state_changed: " + core.identity + ", " + str(state) + ", " + message)
When I am handling the linphone.CallState.Connected case, I am starting the dtmf stream, playing a ‘1’ digit for 300 ms, and then stopping the dtmf stream. However, when I run my SIP client script and send a call to it, I get the following output in console when my client tries to generate the dtmf tone:
ERROR:root:No dtmf generator at this time !
I’ve made sure that both my client and my PBX are using the same DTMF payload type (rfc2833) (I’ve set use_rfc2833_for_dtmf to true in my core object). What is required in order for me to create a dtmf generator? I am not finding much help in the Linphone Python documentation.
- Ken C.
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