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• ffmpeg converted video from s3 bucket downloads before playing the video

  12 octobre 2020, par Rutu

  I making a video streaming application with react and node js.
I am converting video into different resolution with ffmpeg and storing directly to S3 bucket with the help of piping.

  


  I am streaming uploaded resolution video from S3 bucket directly through cloudfront in HTML5 tag with video.js

  


  When i tried to play original video through cloudfront in player its working fine, But as soon i play ffmpeg converted video to video player in cloudfront i am facing issue :

  


  Video takes a lot load (Downloads in browser) before playing in player.

  


  Below is my ffmpeg command

  


  await loadProcess( [&#xA;&#x27;i&#x27;,<s3 url="url" of="of" original="original" video="video">,&#xA;  &#x27;-movflags&#x27;,&#xA;  &#x27;frag_keyframe&#x2B;empty_moov&#x27;,&#x27;-vf&#x27;, &#x27;scale=-2:360&#x27;,&#x27;-c:v&#x27;,&#x27;h264&#x27;,&#x27;-profile:v&#x27;,&#x27;baseline&#x27;,&#x27;-r&#x27;,30,&#x27;-g&#x27;, 60,&#x27;-b:v&#x27;,&#x27;1M&#x27;,&#x27;-f&#x27;,&#x27;mp4&#x27;,&#x27;-&#x27;]&#xA;, outPath,&#x27;video/mp4&#x27;)&#xA;&#xA;</s3>

  


  this is my loadProcess function : i am using aws cli to direct upload resolution video to S3 bucket :

  


  export function loadProcess(ffmpegOptions, outPath,contentType) {
    return new Promise((resolve, reject) => {
        const  videoPath = outPath.replace(/\\/g, "/");
        let ffmpeg = spawn(conf.ffmpeg, ffmpegOptions);
        let awsPipe = spawn('aws', ['s3', 'cp', '--content-type', `${contentType}`, '-', `s3://${process.env.AWS_S3_BUCKET}/${process.env.AWS_S3_VIDEOS_FOLDER}${videoPath}` ])
        ffmpeg.stdout.pipe(awsPipe.stdin)

        // ffmpeg write stream flow
        let ffmpegData = ''
        ffmpeg.stderr.on('data', (_data) => {
            ffmpegData += _data.toString();
        })
        ffmpeg.on('close', (code) => {
            if (code === 0) {
                resolve(outPath)
            }
            else {
                let _dataSplit=ffmpegData.split('\n');
                _dataSplit.pop();
                console.log({action: 'close', message:_dataSplit.pop(), more:[conf.ffmpeg].concat(ffmpegOptions).join(' ')+'\n'+ffmpegData, code})
            }
        });
        ffmpeg.on('error', (err) => {
            reject({action: ' ffmpeg error', message: err});
        });
        
        // aws s3 cli read stream pipe flow
        let awsPipeData = ''
        awsPipe.stderr.on('data', _data => {
            awsPipeData += _data.toString()
        });
        awsPipe.on('close', (code) => {
            if (code === 0) {
                resolve(outPath)
            }else{
                console.log({action: 'close', message: awsPipeData})
            }
        });
        awsPipe.on('error', (err) => {
            reject({action: 'awsPipe error', message: err});
        })

    })
}


  


  I have tried using +faststart option in ffmpeg but getting command failed error.
Can someone please help me this ?

  


  Thanks in advance

  


• Bash script for splitting video using ffmpeg outputs wrong video lengths

  19 juin 2018, par jonny

  I have a video that’s x seconds long. I want to split that video up into equal segments, where each segment is no longer than a minute. To do that, I cobbled together a fairly simple bash script which uses ffmprobe to get the duration of the video, find out how long each segment should be, and then iteratively split the video up using ffmpeg :

  INPUT_FILE=$1
  
  INPUT_DURATION="$(./bin/ffprobe.exe -i "$INPUT_FILE" -show_entries format=duration -v quiet -of csv="p=0")"
  
  
  
  NUM_SPLITS="$(perl -w -e "use POSIX; print ceil($INPUT_DURATION/60), qq{\n}")"
  
  
  
  printf "\nVideo duration: $INPUT_DURATION; "
  
  printf "Number of videos to output: $NUM_SPLITS; "
  
  printf "Approximate length of each video: $(echo "$INPUT_DURATION" "$NUM_SPLITS" | awk '{print ($1 / $2)}')\n\n"
  
  
  
  for i in `seq 1 "$NUM_SPLITS"`; do
  
      START="$(echo "$INPUT_DURATION" "$NUM_SPLITS" "$i" | awk '{print (($1 / $2) * ($3 - 1))}')"
  
      END="$(echo "$INPUT_DURATION" "$NUM_SPLITS" "$i" | awk '{print (($1 / $2) * $3)}')"
  
  
  
      echo ./bin/ffmpeg.exe -v quiet -y -i "$INPUT_FILE" \
  
      -vcodec copy -acodec copy -ss "$START" -t "$END" -sn test_${i}.mp4
  
  
  
      ./bin/ffmpeg.exe -v quiet -y -i "$INPUT_FILE" \
  
      -vcodec copy -acodec copy -ss "$START" -t "$END" -sn test_${i}.mp4
  
  done
  
  
  
  printf "\ndone\n"

  If I run that script on the 30MB / 02:50 duration Big Buck Bunny sample from here, the output of the program would suggest the videos should all be of equal length :

  λ bash split.bash .\media\SampleVideo_1280x720_30mb.mp4
  
  
  
  Video duration: 170.859000; Number of videos to output: 3; Approximate length of each video: 56.953
  
  
  
  ./bin/ffmpeg.exe -v quiet -y -i .\media\SampleVideo_1280x720_30mb.mp4 -vcodec copy -acodec copy -ss 0 -t 56.953 -sn test_1.mp4
  
  ./bin/ffmpeg.exe -v quiet -y -i .\media\SampleVideo_1280x720_30mb.mp4 -vcodec copy -acodec copy -ss 56.953 -t 113.906 -sn test_2.mp4
  
  ./bin/ffmpeg.exe -v quiet -y -i .\media\SampleVideo_1280x720_30mb.mp4 -vcodec copy -acodec copy -ss 113.906 -t 170.859 -sn test_3.mp4
  
  
  
  done

  As the duration of each partial video, i.e. the time between -ss and -t, are equal for each subsequent ffmpeg command. But the durations I get are closer to :

  test_1.mp4 = 00:56
  
  test_2.mp4 = 01:53
  
  test_3.mp4 = 00:56

  Where the contents of each partial video overlap. What am I missing here ?

• how to set two video side by side and first video 60% and second video is 40% using ffmpeg

  19 mai 2017, par Jignesh.Parmar

  I try following command.its working fine.but its set 50-50% video.
  and i got 60-40%. please help me. thanks in advance.

  ffmpeg -i demo1.mp4 -i demo2.mp4 -filter_complex ’[0:v]pad=iw*2:ih[int] ;[int][1:v] overlay=W/2:0[vid]’ -map [vid] -map 1:a -c:v libx264 -crf 23 -preset veryfast output.mp4