You'll be required to start it up from the command line like so:
vlc -vvv http://www.example.org/your_file.mpg
This means you will need to pass in the -vvv flags as well in your array to succesfully start the stream.
Frage
Maybe it is just late, but I ran into a dead end, hoping someone can help me out.
I have a very simple program which is supposed to work like this: The user can see a list of available streams. The user picks a stream to watch. After picking a stream I then want to launch VLC media player for them and play it.
I have everything in order except from one last thing - I don't know how to make the player play the stream. I assumed it would just be something like this:
System.Diagnostics.Process.Start(pathVLC, streams[choice]);
where
PathVLC is the path to the users player, for example C:\Programs\VLC\vlc.exe
streams is an array of strings, all on the form "http://somerandomstream.m3u8"
choice is the stream the user wanted to see.
While VLC opens successfully, nothing else happens, and I am completely lost on how to actually tell VLC to play the stream. What am I missing?
Edit: Looking at Vaughan Hilts answer I figured it out!
System.Diagnostics.Process VLC = new System.Diagnostics.Process();
VLC.StartInfo.FileName = pathVLC;
VLC.StartInfo.Arguments = "-vvv " + streams[choice];
VLC.Start();
Lösung
You'll be required to start it up from the command line like so:
vlc -vvv http://www.example.org/your_file.mpg
This means you will need to pass in the -vvv flags as well in your array to succesfully start the stream.
Andere Tipps
I would start from inspecting supported command-line arguments e.g. here