VLC in Bash Script

Question

I am attempting to write a very basic Bash script that will accomplish the same thing as this command line:

cvlc \
  'rtsp://192.168.0.66/cam/realmonitor?channel=1&subtype=0&authbasic=xxxxxxxxx?tcp' \
   --sout file/mp4:/mnt/recordings/camera16_2014_10_03_13.mp4 \
   --run-time=60 vlc://quit

Entering this into the command line works and I get the expected 60 second MP4 file. Putting this into a bash script, I cannot seem to get all of the arguments passed to VLC correctly.

#!/bin/bash
camname="CAMERA16"
token="_"
ipadd="192.168.0.66"
runtime="60"
cvlc "rtsp://$ipadd/cam/realmonitor?channel=1&subtype=0&authbasic=xxxxxxxxx?tcp --sout file/mp4:/mnt/recordings/$camname$token$(date +$Y_%m_%d_$H_$M).mp4 --run-time=$runtime vlc://quit"

Running this script launches VLC, not headless, and ignores the other arguments. It does not write to a file and it never quits. It simply connects VLC to the stream and plays the stream.

I have tried several different ways of quoting the arguments with no success. All suggestions appreciated.

Mark Reed · Accepted Answer · 2014-10-03 19:52:30Z

1

Those need to be separate arguments to the cvlc command. By putting them all in one big double-quoted string, you turn them into a single argument, which won't work. Put each of the arguments in its own set of double-quotes; don't put the spaces inside the quotes:

cvlc "rtsp://$ipadd/cam/realmonitor?channel=1&subtype=0&authbasic=xxxxxxxxx?tcp" \
     --sout "file/mp4:/mnt/recordings/$camname$token$(date +$Y_%m_%d_$H_$M).mp4" \
     --run-time="$runtime" vlc://quit

I used quoted newlines to split across multiple lines for legibility; it can also be just one long line without the \s.

edited Oct 3, 2014 at 19:52

answered Oct 3, 2014 at 19:47

Mark Reed

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