1

I want to make the code run only after ffmpeg has finished rendering the videos. Currently, the code runs faster than the videos can render.

videos.forEach((vid, i) => {
  ffmpeg(`${process.cwd()}/video/tmp/${vid}`)
  .outputOptions(['-r 30', '-filter:v scale=w=720:h=1280', '-crf 20'])
  .save(`${process.cwd()}/video/tmp/${vid}`)
  .on('end', ()=> console.log(`Video ${i} rendered`));
});
console.log("Fully Completed");

The console shows:

Fully Completed
Video 0 rendered
Video 1 rendered

The execution should be opposite. How can I make the code wait for video to finish rendering before continuing?

wongz
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1 Answers1

2

the code below will help you:

async function processVideos(){
    let videos = [vid1.mp4, vid2.mp4, vid3.mp4];
    for(vid of videos){
        await processVideoSync(vid)
    }
}
function processVideoSync(vid){
    return new Promise((resolve,reject)=>{
       ffmpeg(`${process.cwd()}/video/tmp/${vid}`)
       .outputOptions(['-r 30', '-filter:v scale=w=720:h=1280', '-crf 20'])
       .save(`${process.cwd()}/video/tmp/${vid}`)
       .on('end', ()=>{
          console.log(`Video ${i} rendered`)
          return resolve()
        })
        .on('err',(err)=>{
            return reject(err)
        })
    })
}

o loop for of é o unico capaz de esperar o async await.