WebRTC stream from gstreamer in local netwrok with lowest possible security

0
votes

I need to connect to gstreamer multicast udp stream with WebRTC. I do not need any security, but web browsers require them. The connection is made only in local network, so no ICE, STUN or TURN servers are required. What is minimum possible configuration to connect from browser's WebRTC to any rtp video stream?

I have this gstreamer pipelin (if it'd help):

gst-launch-1.0 nvarguscamerasrc ! nvvidconv ! omxh264enc ! 'video/x-h264, stream-format=(string)byte-stream' ! h264parse ! rtph264pay mtu=1400 config-interval=1 pt=96 ! udpsink host=239.127.1.21 multicast-iface=192.168.137.83 port=5000

And this SDP file which works with VLC to play the stream:

v=0
o=- 1 0 IN IP4 192.168.137.1
s=-
c=IN IP4 239.127.1.21
t=0 0
a=sendonly
m=video 5000 RTP/AVP 96
a=sendonly
a=rtpmap:96 H264/90000

If I use the SDP in a browser, of course there is problem that I don't set some ICE stuff, but i really don't need them and I have no idea on how to generate them. I would rather ignore whole ICE thing and just tell browser to use some different sort of security if it is possible.

So how can I play the stream in a browser? Thanks.

1
streamwebrtcgstreamerrtpsdp

1 Answers

1
votes

Even if you need simplest WebRTC configuration - you will need full implementation. I.e. some signaling protocol, and ICE candidate negotiation, and streams encryption. The only difference - you will always use local ICE candidates, without use STUN or TURN - it's still possible. And I think you will have to use webrtcbin.