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• How to write command line argument input redirection within a bash script ?

  3 septembre 2013, par Serg ikS

  I'm putting together a bash script to run movie rendering jobs. ffmpeg expects a multi-line text file as one of the arguments for concatenation filter to join several files into one. Like this :

  ffmpeg -f concat -i mylist.txt -c copy output

  There's also an option to write it all in a single line like so :

  ffmpeg -f concat -i <(printf "file '%s'\n" A.mp4 B.mp4) -c copy Output.mp4

  How do I write the latter for into a bash script, substituting the file names from other variables ? Tryed splitting the variables, but its not the one that works. $A and $B variables contains paths to input files.

  #!/bin/bash
  
  ...
  
  TMP_LIST=$(printf "file '%s'\n" ${A} ${B})
  
  APPEND="${FFMPEG} -f concat -i <<< {${TMP_LIST}} -c copy ${OUTPUT}"
  
  # run the concatenation:
  
  $APPEND

• Bash Script - Rename files that are beginning with a number to two digits [duplicate]

  9 septembre 2018, par Flavorum1

  This question already has an answer here :

  • Sorting strings with numbers in Bash [duplicate]
    
    3 answers
    

  I have a couple of folders that are audiobooks. The files are numbered and I want to convert them to one file.
  I used the following script to convert them :

      #!/bin/bash 
  
      if [ ! -d mp3 ]; then
  
      mkdir -p mp3;
  
      fi;
  
      for f in ./*.flac; do echo "file '$f'" >> mylist.txt; done
  
      ffmpeg -f concat -safe 0 -i mylist.txt -b:a 320k mp3/title.mp3
  
      [ -e mylist.txt ] && rm mylist.txt

  My problem is that I have to rename the first ten files because they are not in the right order. The files are named 1 - Title, 2 - Title, 3 - Title and so on. To get the right order I have to rename them to 01 - Title, 02 - Title, ..., 09 - Title.
  How can I do that with a bash script ? Furthermore it would be nice, if the playlist.m3u file would be changed accordingly.

  Thanks for your help.

  @Cyrus posted the right Link to solve my problem.
  The solved script ist :

  #!/bin/bash
  
  if [ ! -d mp3 ]; then
  
  mkdir -p mp3;
  
  fi;
  
  for f in ./*.flac; do echo "file '$f'" >> mylist2.txt; done
  
  sort -V mylist2.txt >> mylist.txt
  
  rm mylist2.txt
  
  ffmpeg -f concat -safe 0 -i mylist.txt -b:a 320k mp3/title.mp3
  
  [ -e mylist.txt ] && rm mylist.txt

• Strange behavior about ”ls | while read" with ffmpeg in shell script [duplicate]

  13 décembre 2018, par user3288408

  This question already has an answer here :

  • execute ffmpeg command in a loop
    
    3 answers
    

  I created a shell script to convert all wave files to mp3 files. I use Ubuntu 18.04, FFmpeg was installed using apt, and I run my script in bash.

  My script :

  #!/bin/bash
  
  
  
  ls *.wav | while read file
  
  do
  
      echo $file
  
      ffmpeg -i "$file" -codec:a libmp3lame -b:a 192k "${file%.*}.mp3"
  
  done

  The target files are followings : (include space characters. The real filenames are longer, but I simplified the filenames. Also in this case, the problem occurs)

  % ls *.wav
  
  '01 A.wav'  '02 A.wav'  '03 A.wav'

  The problem is that sometimes $file in the loop is blank or a part of filename strangely (’echo $file’ shows that), and ffmpeg says ’[broken filename] : No such file or directory’. I confirmed the following things.

  • When I comment out the ffmpeg line, ’echo’ shows what I expected. ($file not broken)
  • When I replace ffmpeg to a similar command like ’lame’, it works. ($file not broken)
  • When I replace ’ls *.wav | while read file’ to ’for file in *.wav’, it works. ($file not broken)

  So, only when I use combination of ’ls’ and ’ffmpeg’, $file is broken. What’s going on ? Or do I misunderstand something ?