Relay rtp stream

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Relay rtp stream

morrisford
What would be the simplest way to relay an rtp stream? Looks like I can receive/depayload/payload/broadcast the stream. Is that the easiest way?

Thanks

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Re: Relay rtp stream

Olivier Crête-2
On Thu, 2010-05-06 at 11:38 -0400, Morris Ford wrote:
> What would be the simplest way to relay an rtp stream? Looks like I
> can receive/depayload/payload/broadcast the stream. Is that the
> easiest way?

If the encoding and the payload types are exactly the same. You could
just so "recvfrom(..); sendto(...);" on the same buffer without
modifying it at all.

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Olivier Crête
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Re: Relay rtp stream

Marc Leeman
In reply to this post by morrisford
> What would be the simplest way to relay an rtp stream? Looks like I can
> receive/depayload/payload/broadcast the stream. Is that the easiest way?

The simplest way would be:

udpsrc ! udpsink

You can use gstrtpbin for cleaning up or basic filtering if needed.



--
  greetz, marc
What is wanted is not the will to believe, but the will to find out,
which is the exact opposite.
                -- Bertrand Russell, "Skeptical Essays", 1928
crichton 2.6.26 #1 PREEMPT Tue Jul 29 21:17:59 CDT 2008 GNU/Linux

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Re: Relay rtp stream

Marc Leeman
In reply to this post by Olivier Crête-2
> If the encoding and the payload types are exactly the same. You could
> just so "recvfrom(..); sendto(...);" on the same buffer without
> modifying it at all.

Unless you are working with Bosch encoders; in that case, you would want
to clean up their interpretation of RTP.

Just venting some steam ;-)

--
  greetz, marc
Isn't it strange that the same people that laugh at gypsy fortune
tellers take economists seriously?
crichton 2.6.26 #1 PREEMPT Tue Jul 29 21:17:59 CDT 2008 GNU/Linux

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Re: Relay rtp stream

morrisford
I fear that I was not clear enough. I want to set up a relay server that will pick up an rtsp stream (rtsp://1.1.1.1:6666/test) and broadcast it as another rtsp stream (rtsp://2.2.2.2:6666/test). I am not clear that any of the suggestions would do that.

On Thu, May 6, 2010 at 12:10 PM, Marc Leeman <[hidden email]> wrote:
> If the encoding and the payload types are exactly the same. You could
> just so "recvfrom(..); sendto(...);" on the same buffer without
> modifying it at all.

Unless you are working with Bosch encoders; in that case, you would want
to clean up their interpretation of RTP.

Just venting some steam ;-)

--
 greetz, marc
Isn't it strange that the same people that laugh at gypsy fortune
tellers take economists seriously?
crichton 2.6.26 #1 PREEMPT Tue Jul 29 21:17:59 CDT 2008 GNU/Linux

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Re: Relay rtp stream

Tristan Matthews-2
In reply to this post by Olivier Crête-2
Le 6 mai 2010 11:51, Olivier Crête <[hidden email]> a écrit :
On Thu, 2010-05-06 at 11:38 -0400, Morris Ford wrote:
> What would be the simplest way to relay an rtp stream? Looks like I
> can receive/depayload/payload/broadcast the stream. Is that the
> easiest way?

If the encoding and the payload types are exactly the same. You could
just so "recvfrom(..); sendto(...);" on the same buffer without
modifying it at all.

What would be the best way of calling those on the buffer? In an identity element's handoff?
 

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Olivier Crête
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web: http://tristanswork.blogspot.com

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Re: Relay rtp stream

Marc Leeman
In reply to this post by morrisford
> I fear that I was not clear enough. I want to set up a relay server that
> will pick up an rtsp stream (rtsp://1.1.1.1:6666/test) and broadcast it as
> another rtsp stream (rtsp://2.2.2.2:6666/test). I am not clear that any of
> the suggestions would do that.

rtsp in not a 'broadcasting' service; but a on-request service. You
might happen to find a multicast that is started and kept alive by
another client; but that would be sheer luck.

Creating a rtsp session (rtspsrc); capturing the rtp uni- or multicast
and sending it back out on pure RTP multicast is trivial to do

/* simplified */
rtspsrc ! gstrtpbin ! udpsink

This assumes that your re-broadcast stream is the same RTP as you
received from your encoder.

--
  greetz, marc
Every nonzero finite dimensional inner product space has an orthonormal basis.

It makes sense, when you don't think about it.
crichton 2.6.26 #1 PREEMPT Tue Jul 29 21:17:59 CDT 2008 GNU/Linux

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