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.