Trying to use simple variables with my bash script for ffmpeg but getting errors

When I run this like this, it works:

ffmpeg -re -y -i /home/video.webm -ss 2 \ -i /home/audio.ogg -map 0:0 -map 1:0 \ -c:v libx264 -preset veryfast -b:v 3000k -maxrate 3000k \ -bufsize 6000k -pix_fmt yuv420p -g 50 -c:a aac -b:a 160k -ac 2 -ar 44100 -async 1 \ -f flv "rtmps://live-api-s.facebook.com:443/rtmp/xxx" 

However, when I introduce a variable I have a problem:

STREAM=-f flv \"rtmps://live-api-s.facebook.com:443/rtmp/xxx\" ffmpeg -re -y -i /home/video.webm -ss 2 \ -i /home/audio.ogg -map 0:0 -map 1:0 \ -c:v libx264 -preset veryfast -b:v 3000k -maxrate 3000k \ -bufsize 6000k -pix_fmt yuv420p -g 50 -c:a aac -b:a 160k -ac 2 -ar 44100 -async 1 \ $STREAM 

When I run this, I get:

At least one output file must be specified 

I've tried a number of different ways of doing this, but they all have errors.

1 Answer

This line:

STREAM=-f flv \"rtmps://live-api-s.facebook.com:443/rtmp/xxx\" 

tries to run flv \"rtmps://live-api-s.facebook.com:443/rtmp/xxx\" with STREAM environment variable set to -f. Didn't you get bash: flv: command not found?

Compare STREAM=-f env.

The variable is not set for the current shell and $STREAM expands to an empty string later. But even if you did:

# but don't STREAM='-f flv "rtmps://live-api-s.facebook.com:443/rtmp/xxx"' 

it wouldn't work. Unquoted $STREAM later undergoes word splitting and filename generation and quotes that appear from variable expansion are not special to the shell that expanded the variable.

See this: How can we run a command stored in a variable? In your case a solution with an array is like:

STREAM=(-f flv "rtmps://live-api-s.facebook.com:443/rtmp/xxx") ffmpeg … "${STREAM[@]}" 

Also consider lowercase names for variables.