The URI prefix is not recognized.

I have a video camera on a LAN
I have to request a video stream over this URL

VideoIQ://192.168.12.200:554/Storage/1?t=clock:20150926T04:01:02.000Z-20150926T04:01:12.000Z

however, when Its erroring with the above message on this line
 var req = System.Net.WebRequest.Create(videoUrl);

From what i read, its because it doesn't contain HTTP
I'm trying to write the video stream to file
full code is:

videoUrl = "VideoIQ://192.168.12.200:554/Storage/1?t=clock:20150926T04:01:02.000Z-20150926T04:01:12.000Z";

                const int BUFFER_SIZE = 16 * 1024;
                using (var outputFileStream = File.Create(@"c:\tempVideo\videoText.mp4", BUFFER_SIZE))
                {
                    var req = System.Net.WebRequest.Create(videoUrl);
                    using (var response = req.GetResponse())
                    {
                        using (var responseStream = response.GetResponseStream())
                        {
                            var buffer = new byte[BUFFER_SIZE];
                            int bytesRead;
                            do
                            {
                                bytesRead = responseStream.Read(buffer, 0, BUFFER_SIZE);
                                outputFileStream.Write(buffer, 0, bytesRead);
                            } while (bytesRead > 0);
                        }
                    }
                }

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websssCEOAsked:
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David Johnson, CD, MVPOwnerCommented:
how would you open this stream in a browser?
VideoIQ:

\\192.168.12.200:554 wouldn't this work for a stream?
websssCEOAuthor Commented:
No it does not
The hardware requires VideoIQ:// otherwise it won't return anything.
David Johnson, CD, MVPOwnerCommented:
so you have to "Registers a WebRequest descendant for the specified URI"
and
create the uri web request register prefix
https://msdn.microsoft.com/en-us/library/system.net.webrequest.registerprefix%28v=vs.110%29.aspx
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ste5anSenior DeveloperCommented:
I don't think your are right.

The hardware will never see the schema VideoIQ://. The schema is used to specify the kind of service. Thus I think your hardware ships with software, which needs to be installed correctly. Normally the schema defines how the connection is made to 192.168.12.200, e.g. the port. But you're also specifing a port, so I guess it's time to rtfm.

Otherwise try a search engine. E.g.

Connecting to your VideoIQ IP camera:
 
Models	Connection Type	Example URL
iCVRHD	MJPEG		http://IPADDRESS/viqcam.mjpg
iCVRHD	VLC 		rtsp://IPADDRESS:554/hiQ.sdp
iCVRHD	VLC		rtsp://IPADDRESS:554/lowQ.sdp
Other	FFMPEG 		http://IPADDRESS/videostream.asf?user=[USERNAME]&pwd=[PASSWORD]&resolution=[WIDTH]*[HEIGHT] 
Other	VLC 		rtsp://IPADDRESS:554/hiQ.sdp 

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websssCEOAuthor Commented:
Here is a snippet from the Manual which is what i'm using
sdk
I have the windows media player plug in installed, and it plays the video stream fine
The SDK project i'm using that plays this stream has a reference to AxWMPLib
The code which streams to windows media player is
                axWindowsMediaPlayer1.Ctlcontrols.stop();
                axWindowsMediaPlayer1.URL = videoUrl;
                axWindowsMediaPlayer1.Ctlcontrols.play();

However, my requirement is that I need to intercept this video stream, and write to disk as MP4 which is the format the video is sent in (as a stream not a file)
ste5anSenior DeveloperCommented:
So you can enter that URL in Windows Media Player? Seems that only that plugin translate your VideoIQ URL.

So I would use a network sniffer to see what's going on.

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