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• Piping output of youtube-dl to a script using ffmpeg looks ok using echo but returns an error when executing

  22 February 2016, by user556068

  I am trying to use youtube-dl to get the urls of some videos and then pipe the resulting urls into the input of my script. So in my terminal I do

  youtube-dl --ignore-config -iga ~/Desktop/youtube/videolist.txt | myscript.sh

  In my script I define things as

  command='ffmpeg'        
  
  inputArgs='-i'              
  
  outputArgs='-c:v libx264 -preset ultrafast -qp 0'       
  
  directory="${HOME}/Desktop/Videos/"
  
  output="video${count}"      
  
  extension='mp4'         

  I test it with echo to make sure everything appears in the correct order.

  echo "${command}" "${inputArgs}" "${input}" "${outputArgs}" \
  
  "${directory}""${output}${count}"."${extension}"

  And the output from that looks correct. But when I try to run the same thing without the preceding echo command, i.e.,

  "${command}" "${inputArgs}" "${input}" "${outputArgs}" \
  
  "${directory}""${output}${count}"."${extension}"

  I get an error message that says

  At least one output file must be specified.

  So it seems pretty obvious to me that I’m doing something wrong when attempting to execute it.

  I have tried:

  • quoting the entire line as a whole
  • quoting different sections together
  • using the exec command in front of everything

  No matter what I do, an error occurs at some point in the process. I know it’s something simple I’m doing wrong. Would someone please enlighten me as to what that might be?

  I feel very strongly that the . shouldn’t just be in the middle of everything like that, but I really don’t know.

  Again, everything looks as it should when I run echo before the string of shell parameters.

  If more of the script I’m using is needed to understand what I’m talking about, that is not a problem.

• Piping output of youtube-dl to a script using ffmpeg looks ok using echo but returns an error when executing

  29 May 2020, by I0_ol

  I am trying to use youtube-dl to get the urls of some videos and then pipe the resulting urls into the input of my script. So in my terminal I do

  



  youtube-dl --ignore-config -iga ~/Desktop/youtube/videolist.txt | myscript.sh


  



  In my script I define things as

  



  command='ffmpeg'        
inputArgs='-i'              
outputArgs='-c:v libx264 -preset ultrafast -qp 0'       
directory="${HOME}/Desktop/Videos/"
output="video${count}"      
extension='mp4'         


  



  I test it with echo to make sure everything appears in the correct order.

  



  echo "${command}" "${inputArgs}" "${input}" "${outputArgs}" \
"${directory}""${output}${count}"."${extension}"


  



  And the output from that looks correct. But when I try to run the same thing without the preceding echo command, i.e.,

  



  "${command}" "${inputArgs}" "${input}" "${outputArgs}" \
"${directory}""${output}${count}"."${extension}"


  



  I get an error message that says

  



  


  At least one output file must be specified.

  


  



  So it seems pretty obvious to me that I'm doing something wrong when attempting to execute it.

  



  I have tried:

  



  

  • quoting the entire line as a whole

  


  • quoting different sections together

  


  • using the exec command in front of everything

  


  



  No matter what I do, an error occurs at some point in the process. I know it's something simple I'm doing wrong. Would someone please enlighten me as to what that might be?

  



  I feel very strongly that the . shouldn't just be in the middle of everything like that, but I really don't know.

  



  Again, everything looks as it should when I run echo before the string of shell parameters.

  



  If more of the script I'm using is needed to understand what I'm talking about, that is not a problem.

  


• Converting video with ffmpeg from a script not in the root folder

  30 October 2017, by Daniel Ormerod

  I am using ffmpeg to convert videos, create thumbnails etc.

  I made the below function to convert a video and when I run it in a script in the root folder it converts ok.

  However, I am using ajax to upload the video and the script is located /ajax/upload-video.php and when I run it in that folder it fails to convert.

  I have tried sending the $src and $dest up with ../ before the filenames to go back a folder but it isn’t working.

  Can anybody help?

  Do I need to add something to the $command line to tell it to go back a folder?

  These are the variables being used in the function:

  $ffmpegpath = 'ffmpeg.exe';
  
  $src = 'videos/video1.wmv';
  
  $dest = 'videos/video1.mp4';

  I have tried:

  $ffmpegpath = 'ffmpeg.exe';
  
  $src = '../videos/video1.wmv';
  
  $dest = '../videos/video1.mp4';

  And the function:

  function convert_video($src, $dest) {
  
  
  
      global $ffmpegpath;
  
  
  
      if (!file_exists($src)) {
  
  
  
          echo '<p>The file does not exist</p>';
  
  
  
      } else {
  
  
  
          $command = "$ffmpegpath -i $src $dest";
  
  
  
          @exec($command, $ret);
  
  
  
          if (!file_exists($dest)) {
  
  
  
              echo '<p>The file failed to create</p>';
  
  
  
          } else {
  
  
  
              if (filesize($dest)==0) {
  
  
  
                  echo '<p>The file created but has not filesize</p>';
  
  
  
              } else {
  
  
  
                  echo '<p>File Created Successfully</p>';
  
  
  
              }
  
  
  
          }
  
  
  
      }
  
  
  
  }