My Wowza server has the following IP:192.168.1.101
Encoder’s IP: 192.168.1.105
I am sending the following ANNOUNCE request to Wowza port 554:
ANNOUNCE rtsp://192.168.1.101:1935/rtplive/myStream.sdp RTSP/1.0
Cseq: 0
Content-Type: application/sdp
Session: 12345678
Content-Length: 299
v=0
o=- 1360014152814 1360014152814 IN IP4 192.168.1.105
s=Stream-0
i=N/A
c=IN IP4 192.168.1.101
t=0 0
a=recvonly
m=video 5006 RTP/AVP 96
b=RR:0
a=rtpmap:96 H264/90000
a=fmtp:96 packetization-mode=1;profile-level-id=42801e;sprop-parameter-sets=Z0KAHukBQHsg,aM4G4g==;
a=control:trackID=0
The first line in the request should indicate the actual Wowza application, right? So my RTSP announce is correct.
After the announcement is published, I see the following in the log:
NFO rtsp connect 938250587 -
INFO server comment - sortPackets[2]: sortBufferSize:500
INFO stream create - -
INFO rtsp announce 938250587 -
WARN server comment - RTPStream.idle[938250587:rtplive/definst:]: Close RTSP/RTP session. Session timeout: 90000
INFO stream unpublish - -
INFO server comment - MediaStreamMediaCasterPlay: close
INFO stream destroy - -
INFO rtsp disconnect 938250587 -
I did some experiments with Wirecast encoder, the log is the following after publishing (to the same Wowza application):
INFO rtsp connect 344217766 -
INFO server comment - sortPackets[1]: sortBufferSize:500
INFO stream create - -
INFO rtsp announce 344217766 -
INFO server comment - RTPUDPTransport.bind[rtplive/definst]: 0.0.0.0/0.0.0.0:6970
INFO server comment - RTPUDPTransport.bind[rtplive/definst]: 0.0.0.0/0.0.0.0:6971
INFO server comment - RTPUDPTransport.bind[rtplive/definst]: 0.0.0.0/0.0.0.0:6972
INFO server comment - RTPUDPTransport.bind[rtplive/definst]: 0.0.0.0/0.0.0.0:6973
INFO stream publish myStream.sdp -
INFO rtsp publish 344217766 -
INFO server comment - UDPTransport.firstPacket: bind:0.0.0.0/0.0.0.0:6970 msg:/192.168.1.101:57954
Do I need perhaps also send RTSP PLAY request?