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• Getting this error using ffmpeg concat in ReactNative "concat imposible to open"

  2 février 2021, par Josip Bogdan

  Hy, I am trying to concat multyple video using ffmpeg concat, testing on samsung j76, android version 8.1.0

  


  const textFile = await (await writeTextFileWithAllVideoFiles(filePaths))  
const outputPath = Platform.OS === 'ios' ? `${RNFS.DocumentDirectoryPath}/video-1.mp4` : `${RNFS.DocumentDirectoryPath}/video-1.mp4`
console.log(`-f concat -safe 0 -i ${textFile} -c copy ${outputPath}`)
const result = await RNFFmpeg.execute(`-f concat -safe 0 -i ${textFile} -c copy ${outputPath}`)


  


  And here is how I create the txt file

  


  const writeTextFileWithAllVideoFiles = async (filePaths) => {
var RNFS = require('react-native-fs');
var path = RNFS.DocumentDirectoryPath + '/videoList.txt';

var fileContent = ''
console.log("fileplay1");
filePaths.forEach(path => {
   fileContent += 'file ' + '\'' + path.substring(8) + '\'' + '\r\n'
});
console.log(fileContent);
return RNFS.writeFile(path, fileContent, 'utf8')
.then((success) => {
   console.log(path)
   if (RNFS.exists(path))
      console.log('FILE WRITTEN!')
   return path
})
.catch((err) => {
    console.log(err.message)
    return err.message
});  
}


  


  This is my console log output

  


  file 'data/user/0/com.videoeditorapp/cache/Camera/f27579d2-1488-4a59-8a9d-7a4e7b6fe716.mp4'
file 'data/user/0/com.videoeditorapp/cache/Camera/58130175-bd20-4002-9629-070d0cdd18ac.mp4'

[Tue Feb 02 2021 16:30:45.375]  LOG      /data/user/0/com.videoeditorapp/files/videoList.txt
[Tue Feb 02 2021 16:30:45.376]  LOG      FILE WRITTEN!
[Tue Feb 02 2021 16:30:45.376]  LOG      -f concat -safe 0 -i /data/user/0/com.videoeditorapp/files/videoList.txt -c copy /data/user/0/com.videoeditorapp/files/video-1.mp4
[Tue Feb 02 2021 16:30:45.376]  LOG      [concat @ 0xe2b22000] Impossible to open '/data/user/0/com.videoeditorapp/files/data/user/0/com.videoeditorapp/cache/Camera/f27579d2-1488-4a59-8a9d-7a4e7b6fe716.mp4'
[Tue Feb 02 2021 16:30:45.377]  LOG      /data/user/0/com.videoeditorapp/files/videoList.txt: No such file or directory


  


  And the txt file is not on my android device
My manifest is like this

  


  &#xA;  &#xA;      &#xA;&#xA;<application></application>&#xA;

  


• FFMPEG Audio Optimisation with speex

  2 février 2021, par ThinkAdvantage

  I am trying to cleanup the Audio of my Paragliding Videos down to my own voice mostly by using ffmpeg. I stumbled upon speex codec and as far as I understood it I should be able to encode my Audio Stream with Libspeex and this should cleanup the audio stream down to my voice. I am working on a current windows 10 and downloaded a precompiled Windows ffmpeg build which includes libspeex. (i could verify that) Now i tried to input my mp4 and output it with libspeex encoded audio but i always get :

  


  


  Could not find tag for codec speex in stream #1, codec not currently supported in container
Could not write header for output file #0 (incorrect codec parameters ?) : Invalid argument
Error initializing output stream 0:1 —

  


  


  ffmpeg -i D :\Cyclevision\2021.01.24\20210124-short.mp4 -c:v copy -acodec libspeex -vad 1 D :\Cyclevision\2021.01.24\20210124-shortSPEEX.mp4

  


  I also tried outputting only the Audio to mp4, mp3, aac but everything failed.

  


  My question now - how can I use libspeex properly to re-encode my Audio stream of an MP4 ?

  


  Is it even a good idea to use libspeex or is it only for Audio and not Audio/Video combinations ?
Thanks in Advance for your Answers.

  


  P.S : I am no proper Programmer - just an IT Infrastructure System engineer trying to use FFMPEG with Powershell.

  


• pydub unable to find the selected files

  4 février 2021, par big_Z_0909

  my code is as below

  


  import pydub
from pathlib import Path
import os, sys

pydub.AudioSegment.converter = r"D\\Anaconda3\\Scripts\\ffmpeg-2021-02-02-git-2367affc2c-essentials_build\\bin\\ffmpeg.exe"
pydub.AudioSegment.ffprobe = r"D\\Anaconda3\\Scripts\\ffmpeg-2021-02-02-git-2367affc2c-essentials_build\\bin\\ffprobe.exe"

my_file = "D:\\Anaconda3\\Scripts\\outcall\\luyinwenjian\\20210203.mp3"
song = pydub.AudioSegment.from_mp3(my_file)


  


  i downloaded the essential windows eddition of ffmpeg from the official website. i still cannot load the mp3 file from the document, got the FileNotFoundError : [WinError 2]. In the warning, it said that it could find ffprobe and ffmpeg, but in fact the path is correct in the code. Does anyone point out what exactly i went wrong here ?