<div dir="ltr">Hi All,<br><br>Kamailio send 2 INVITE(lookup rewrite r-uri and append branch) request for Client behind NAT. <br>Due to this reason for one request,Client respond with 486 Busy here. we only want to send  1 INVITE request.<br><br>Following are the AOR for client at kamailio.if field(ip port) in CONTACT and RECEIVED differ then kamailio send 2 INVITE request. <br><br><br>FOLLOWING INVITE HAS BEEN GENERATED AFTER LOOKUP FUNCTION(one from contact and one from Received field).<br><br><a href="http://INVITE:sip:123@192.168.230.124:45672">INVITE:sip:123@192.168.230.124:45672</a><br><a href="http://INVITE:sip:123@204.192.206.208:45672">INVITE:sip:123@204.192.206.208:45672</a><br><br>I want only ONE INVITE request to be generated.Please suggest how do solve this issue or if there is any config change or any code change required to achieve this?<br><br><br>  AOR:: 123<br>                Contact:: <a href="http://sip:123@192.168.230.124:45672">sip:123@192.168.230.124:45672</a> Q=1<br>                        Expires:: 894<br>                        Callid:: <a href="mailto:003A8EC4-12351@192.168.230.124">003A8EC4-12351@192.168.230.124</a><br>                        Cseq:: 2<br>                        User-agent:: SIPPER for 3CX Phone<br>                        Received:: sip:<a href="http://204.192.206.208">204.192.206.208</a>: 45672<br>                        State:: CS_NEW<br>                        Flags:: 0<br>                        Cflag:: 192<br>                        Socket:: udp:<a href="http://172.12.120.88:5060">172.12.120.88:5060</a><br>                        Methods:: 4294967295<br>                        Ruid:: uloc-5aba112d-7fce-5<br>                        Reg-Id:: 0<br>                        Last-Keepalive:: 1522144583<br>                        Last-Modified:: 1522144583<br><br>thanks,<br>amit<br></div>