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Médias (1)
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Publier une image simplement
13 avril 2011, par ,
Mis à jour : Février 2012
Langue : français
Type : Video
Autres articles (83)
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Mise à jour de la version 0.1 vers 0.2
24 juin 2013, parExplications des différents changements notables lors du passage de la version 0.1 de MediaSPIP à la version 0.3. Quelles sont les nouveautés
Au niveau des dépendances logicielles Utilisation des dernières versions de FFMpeg (>= v1.2.1) ; Installation des dépendances pour Smush ; Installation de MediaInfo et FFprobe pour la récupération des métadonnées ; On n’utilise plus ffmpeg2theora ; On n’installe plus flvtool2 au profit de flvtool++ ; On n’installe plus ffmpeg-php qui n’est plus maintenu au (...) -
Personnaliser en ajoutant son logo, sa bannière ou son image de fond
5 septembre 2013, parCertains thèmes prennent en compte trois éléments de personnalisation : l’ajout d’un logo ; l’ajout d’une bannière l’ajout d’une image de fond ;
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Ecrire une actualité
21 juin 2013, parPrésentez les changements dans votre MédiaSPIP ou les actualités de vos projets sur votre MédiaSPIP grâce à la rubrique actualités.
Dans le thème par défaut spipeo de MédiaSPIP, les actualités sont affichées en bas de la page principale sous les éditoriaux.
Vous pouvez personnaliser le formulaire de création d’une actualité.
Formulaire de création d’une actualité Dans le cas d’un document de type actualité, les champs proposés par défaut sont : Date de publication ( personnaliser la date de publication ) (...)
Sur d’autres sites (15429)
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pyav - cannot save stream as mono
3 août 2022, par ZvikaI'm trying to use
pyav
to convert arbitrary audio file to a low quality, mono, wave file.

I almost managed to do it, but it's stereo, and I couldn't find how to make it mono. Furthermore, I think I made some mistake here, as I had to repeat the
rate
in theoutput_container.add_stream
and in theAudioResampler
- it seems redundant, and I can't understand what would happen if those numbers won't match.

My code is :


import av
 
 input_file = 'some.mp3'
 output_file = 'new.wav'
 
 rate = 22000
 
 output_container = av.open(output_file, 'w')

 # can I tell `output_stream` to just use `resampler`'s info?
 # or, if not, how can I tell it to have only 1 channel?
 output_stream = output_container.add_stream('pcm_u8', rate) 
 
 resampler = av.audio.resampler.AudioResampler('u8p', 'mono', rate)
 
 input_container = av.open(input_file)
 for frame in input_container.decode(audio=0):
 out_frames = resampler.resample(frame)
 for out_frame in out_frames:
 for packet in output_stream.encode(out_frame):
 output_container.mux(packet)
 output_container.close()



And not related to my main question, but any comments regarding my code, or pointing out mistakes, are welcomed. I hardly could find usage examples to use a reference, and PyAV API documentation isn't very detailed...


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Convert video into mp3 format using ffmpeg in nodejs and angular and then save converted audio into the database(mongodb)
6 septembre 2021, par Amir ShahzadThis is nodejs server side file


const express = require('express');
const ffmpeg = require('fluent-ffmpeg');
const fileUpload = require('express-fileupload');
const mongoose = require('mongoose');
const cors = require('cors')
const app = express();
const Video = require('./models/video');
mongoose.connect('mongodb://localhost:27017/YoutubeApp', {
 useNewUrlParser: true,
 useUnifiedTopology: true,
});
const db = mongoose.connection;
db.on('error', console.error.bind(console, 'connection error'));
db.once('open', () => {
 console.log('Data Base Connected Successfully!');
});

app.use(fileUpload({
 useTempFiles: true,
 tempFileDir: 'temp/'
}));
app.use(express.json());
app.use(express.urlencoded({ extended: true }));
app.use(cors({ origin: 'http://localhost:4200' }));


ffmpeg.setFfmpegPath('/usr/bin/ffmpeg');

app.post('/mp4tomp3', (req, res) => {
 convertdata = req.body;
 console.log('path of innput is', req.body);
 function convert(input, output, callback) {
 ffmpeg(input)
 .output(output)
 .on('end', function() { 
 console.log('conversion ended');
 callback(null);
 }).on('error', function(err){
 console.log('error: ', err.code, err.msg);
 callback(err);
 }).run();
 }
 convert(convertdata, './temp/output.mp3', function(err){
 if(!err) {
 console.log('conversion complete');

 
 }
 })

 app.listen(3000, () => {
 console.log('Server Start On Port 3000')
 })



In Convert function i not know how i can get input by the user i'm new in angular Please anyone can solve this out thanks in advance


This is video model file


const mongoose = require("mongoose");
const Schema = mongoose.Schema;

const videoSchema = new Schema({
 mp4: String,
});
module.exports = mongoose.model("Videos", videoSchema);



**This is Typescript code in angular client side that handle user input and select video **


import { ThrowStmt } from '@angular/compiler';
import { Component, OnInit } from '@angular/core';
import { FormBuilder, FormGroup, Validators } from '@angular/forms';
import { VideoConversionService } from 'src/services/video-conversion.service';

@Component({
 selector: 'app-root',
 templateUrl: './app.component.html',
 styleUrls: ['./app.component.css']
})
export class AppComponent implements OnInit {

 submitted =false;
 form! : FormGroup
 data:any

 constructor(private formBuilder: FormBuilder,
 private videoService: VideoConversionService){}

 creatForm(){
 this.form = this.formBuilder.group({
 mp4: ['', Validators.required],
 });
 }
 ngOnInit(): void {
 this.creatForm();

 }


 convertVideo(){
 this.submitted = true
 this.videoService.conversion(this.form.value).subscribe(res => {
 this.data = res;
 })
 }

 }



I do not know how to create logic to do that (convert video into audio using angular framework)


This is app.component.html file where i want to get video from the user using input field


<div class="container">
 <h1>Video Proccessing App</h1>
 <form>
 <input type="file" formcontrolname="mp4" />
 <input type="submit" value="Convert" />
 </form>
</div>





But my code is not working and video is not converting into audio


This is my video service file where i calling nodejs api to perform the task


import { Injectable } from '@angular/core';
import { HttpClient } from '@angular/common/http';
@Injectable({
 providedIn: 'root'
})
export class VideoConversionService {

constructor(private httpClient: HttpClient) { }

conversion(data: any){
 return this.httpClient.post('http://localhost:3000/mp4tomp3', data)
}
}



Please anyone can solve my problem Thanks in advance


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In my django app, celery task converts uploaded video w/ ffmpeg, but converted video won't save to s3 ?
29 janvier 2013, par GetItDoneI use Heroku to host my website, and Amazon s3 to store my static and media files. I have a celery task that converts the video file to flv, but the flv doesn't store anywhere. There isn't any error, just there is no file uploaded to my s3 bucket. How can I force the file to save to my s3 bucket after the conversion ? I'm still pretty new web development in general, and I have been stuck trying to get my video conversion working properly for weeks. To be honest, I'm not even sure that I'm doing the right thing with my task. Here is the code in my celery task :
@task(name='celeryfiles.tasks.convert_flv')
def convert_flv(video_id):
video = VideoUpload.objects.get(pk=video_id)
filename = video.video_upload
sourcefile = "%s%s" % (settings.MEDIA_URL, filename)
vidfilename = "%s.flv" % video.id
targetfile = "%svideos/flv/%s" % (settings.MEDIA_URL, vidfilename)
ffmpeg = "ffmpeg -i %s -ar 22050 -f flv -s 320x240 %s" % (sourcefile, targetfile)
#The next lines are code that I couldn't get to work in place of the line above, and are left commented out.
#I am open to suggestions or alternatives for this also.
#ffmpeg = "ffmpeg -i %s -acodec libmp3lame -ar 22050 -f flv -s 320x240 %s" % (sourcefile, targetfile)
#ffmpeg = "ffmpeg -i %s -acodec mp3 -ar 22050 -f flv -s 320x240 %s" % (sourcefile, targetfile)
try:
ffmpegresult = commands.getoutput(ffmpeg)
print "---------------FFMPEG---------------"
print "FFMPEGRESULT: %s" % ffmpegresult
except Exception as e:
ffmpegresult = None
print("Failed to convert video file %s to %s" % (sourcefile, targetfile))
print(traceback.format_exc())
video.flvfilename = vidfilename
video.save()My view :
def upload_video(request):
if request.method == 'POST':
form = VideoUploadForm(request.POST, request.FILES)
if form.is_valid():
video_upload=form.save()
video_id=video_upload.id
video_conversion = convert_flv.delay(video_id)
return HttpResponseRedirect('/current_classes/')
else:
...Any advice, insight, or ideas in general would be greatly appreciated. Obviously I am missing something, but I can't figure out what. I have been stuck with different aspects of getting my video conversion to work with ffmpeg using a celery task for weeks. Thanks in advance.