Hello,
I'm a bit confused about a piece of the documentation. I'm using rtpproxy 3.0.1
+ rtpproxy module on kamailio 5.6.4.
On the rtpproxy module, I'm trying to get the "timeout_socket" param working.
From the documentation, I have
modparam("rtpproxy", "timeout_socket",
"xmlrpc:http://192.168.155.64:8888/XMLRPC";)
modparam("xmlrpc", "route", "XMLRPC_ROUTE")
modparam("xmlrpc", "mode", 1)
modparam("xmlrpc", "url_match", "^/XMLRPC")
/usr/bin/rtpproxy -i -u rtpproxy -n tcp:192.168.155.64:8888 -l 192.168.155.64
-s udp:127.0.0.1:7333 -m 29152 -M 49150 -d DBUG
However when purposely trying to timeout RTP, I'm seeing the following in my
rtpproxy logs:
Oct 10 00:31:22 localhost rtpproxy[3802129]:
DBUG:GLOBAL:rtpp_command_split:401: received command "3801518_4 USc9,0,8,18,101
219b2465decc-n7al36xva4lw 192.168.155.168 2266 C60D788E-8E14A9FA;1 w6p2ke9ttk;1
xmlrpc:http://192.168.155.64:8888/XMLHTTP";
Oct 10 00:31:22 localhost rtpproxy[3802129]:
ERR:219b2465decc-n7al36xva4lw:rtpp_command_ul_handle:603: invalid socket name
w6p2ke9ttk;1
I'm just wondering if I'm missing anything, do I need to define
"listen=tcp:192.168.155.64:8888" in my configs? I know that rtpengine is
recommended/ I would like to experiement with rtpproxy. Do I need a specific
version of rtpproxy in order for this to work -- I did try it with 2.1.1
however that didn't work.
Thanks.__________________________________________________________
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