How to create random pass names in ffmpeg?

Question

Normally, when using ffmpeg, you define a name for the "pass" information to be stored in a file when enconding videos with 2 (or more) passes, e.g.:

ffmpeg -i "INPUT_FILE" -pass 1 -passlogfile videopass.log /dev/null -y && ffmpeg -i "INPUT_FILE" -pass 2 -passlogfile videopass.log -y "OUTPUT_FILE"

I'd love to be able to create a random pass name automatically in bash in the simplest way possible, preferably a "one-liner" using the system's default tools (Linux)... something like:

$(tr -dc A-Za-z0-9 </dev/urandom | head -c 8).log | ffmpeg -i "INPUT_FILE" -pass 1 -passlogfile > /dev/null -y && ffmpeg -i "INPUT_FILE" -pass 2 -passlogfile > -y "OUTPUT_FILE"

I'm too stupid to figure out a way of doing something similar that actually makes sense.

Thank you very much in advance!

Answer 1

After countless tries and googling, I found a way to do exactly what I was looking for... A one-liner, nothing too complicated or fancy and using my system's default tools:

printf $RANDOM | xargs -i sh -c 'ffmpeg -i "INPUT_VIDEO" -pass 1 -passlogfile {} -an -f mp4 /dev/null -y && ffmpeg -i "INPUT_VIDEO" -pass 2 -passlogfile {} -y "OUTPUT_VIDEO"'

This will create a very simple 5-digit random number, which will be used as the initial name for the pass files in ffmpeg, ffmpeg automatically adds the file extension(s). Of course, you can (and probably should) add video/audio parameters to your ffmpeg commands. The important elements are the "printf $RANDOM | xargs -i sh -c" at the begining and the "{}" curly brackets after the "-passlogfile" command.

There are probably even simpler or more elegant ways of doing it, but this works exactly as I wanted, so I'm happy.

Answer 2

Assuming the requirement is to create a random name for the password file:

$ pwdlog=$(mktemp XXXXXXXX.log)      # have mktemp create a file with 8 random characters + ".log"

$ typeset -p pwdlog
declare -- pwdlog="Em6GeMdc.log"

$ ls -l "${pwdlog}"
-rw-------+ 1 username None 0 Apr 26 15:13 Em6GeMdc.log

This file could then be referenced in the ffmpeg call like such:

ffmpeg -i "INPUT_FILE" -pass 1 -passlogfile "${pwdlog}" ...