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• How to run a ffmpeg command in python with file paths containing spaces ?

  21 juillet 2023, par tomatochan

  I'm developing a program in Python that requires the use of FFMPEG. In one step, I need to create a video from a text file containing the list of images (path and duration) that need to be assembled to make a video. To do this, I use the following code :

  


  cmd = [ffmpeg, "-y", "-f", "concat", "-safe", "0", "-i", tmp, "-c:v", "libx264", "-r", "25", "-pix_fmt", "yuv420p", out]
subprocess.run(cmd)
# os.system(" ".join(cmd)) # returns same error


  


  What returns the error : The syntax of the file, directory or volume name is incorrect.

  


  I know that paths can be problematic when they contain spaces, and as my username has one, I've taken care to quote the paths in such a way that :

  


  ffmpeg = "C:\Users\John Doe\Documents\ffmpeg\bin\ffmpeg.exe"
tmp = "C:\Users\John Doe\Desktop\path\to\file.txt"
out = "C:\Users\John Doe\Desktop\path\to\video.mp4"


  


  When I print(" ".join(cmd)), this is what I get in my terminal :

  


  "C:\Users\John Doe\Documentsffmpeg\binffmpeg.exe" -y -f concat -safe 0 -i "C:\Users\John Doe\Desktop\path\to\file.txt" -c:v libx264 -r 25 -pix_fmt yuv420p "C:\Users\John Doe\Desktop\path\to\video.mp4"


  


  However, the problem persists.
Has anyone ever had this problem and managed to solve it ?

  


  I've also tried the method where you have to escape the spaces (taking care to replace the \ with / and then the with \ ) but nothing works... The error persists.

  


  For your information, here are the contents of one of my .txt files

  


  file C:\Users\John Doe\Desktop\path\to\image_1.png
duration 0.04
file C:\Users\John Doe\Desktop\path\to\image_2.png
duration 0.04
file C:\Users\John Doe\Desktop\path\toimage_3.png
duration 0.04


  


  When I quote the paths in the .txt files

  


  

  • subprocess gives me the error :
PermissionError: [WinError 5] Access denied

  


  • os.system gives me the error :

  


  


  C:\Users\John' is not recognized as an internal or external command, an executable program or a command file.
The syntax of the file, directory or volume name is incorrect.


  


• Can't upload .mkv video file to camtasia 2018

  11 juin 2019, par FunTech StudentsVideos

  Can’t Import .mkv video file to camtasia 2018 :

  I have made an intro for my Youtube Channel on Penzoid and then uploaded it to my laptop in .mkv file format, now it can’t be import to camtasia 2018 for rendering with the recorded video in camtasia...

  I used cloudconvert to convert my intro to .mp4 format but it give me the following error :

  ERROR Could not find tag for codec vp8 in stream #0, codec not currently supported in container
  . Last message repeated 1 times
  Last message repeated 1 times

  I also used .bat file with the following code but it didn’t help me :

  @echo off
  
  if not exist "fixed" mkdir fixed
  
  for %%f in ("*.mp4") do "C:\Users\Nadeem Ullah\Downloads\Compressed\ffmpeg-win64-static\bin\ffmpeg.exe" -i "%%f" -c:a copy -c:v copy "fixed\%%~nf.mp4"

• Converting mp4 to ogg file format results in a large file

  26 avril 2014, par parags

  I have a MP4 file of 83MB (converted from MOV of about 772MB using FFMPEG).
  For the file to be playable from all browsers from HTML5 video tag, I am converting the MP4 to OGG, again using FFMPEG command

  ffmpeg -i object-creation.mp4 -acodec libvorbis -vcodec libtheora -q:v 5 -q:a 5 object-creation-3.ogg

  The result of the above command is a very large OGG file of around 500 MB. I would certainly not want to upload such huge files to Amazon S3 (which I am using for storage, and distribution).

  Is there something I am missing here ? Is the file not compressed enough ?

  Is it possible to have the resultant file of somewhat manageable size like 80-100 MB without any appreciable loss in quality over what is seen in MP4 format ? Why is it that even the source file is 83MB, the resultant file is too big in comparison ?

  Thanks
  Parag