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Géodiversité
9 septembre 2011, par ,
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USGS Real-time Earthquakes
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La conservation du net art au musée. Les stratégies à l’œuvre
26 mai 2011
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Podcasting Legal guide
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Autres articles (81)
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Des sites réalisés avec MediaSPIP
2 mai 2011, parCette page présente quelques-uns des sites fonctionnant sous MediaSPIP.
Vous pouvez bien entendu ajouter le votre grâce au formulaire en bas de page. -
Keeping control of your media in your hands
13 avril 2011, parThe vocabulary used on this site and around MediaSPIP in general, aims to avoid reference to Web 2.0 and the companies that profit from media-sharing.
While using MediaSPIP, you are invited to avoid using words like "Brand", "Cloud" and "Market".
MediaSPIP is designed to facilitate the sharing of creative media online, while allowing authors to retain complete control of their work.
MediaSPIP aims to be accessible to as many people as possible and development is based on expanding the (...) -
Participer à sa traduction
10 avril 2011Vous pouvez nous aider à améliorer les locutions utilisées dans le logiciel ou à traduire celui-ci dans n’importe qu’elle nouvelle langue permettant sa diffusion à de nouvelles communautés linguistiques.
Pour ce faire, on utilise l’interface de traduction de SPIP où l’ensemble des modules de langue de MediaSPIP sont à disposition. ll vous suffit de vous inscrire sur la liste de discussion des traducteurs pour demander plus d’informations.
Actuellement MediaSPIP n’est disponible qu’en français et (...)
Sur d’autres sites (10199)
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FFMPEG Merge a video that has sound with an mp3 audio file outputs audio out of sync [duplicate]
23 février 2021, par Mahmoud EidarousAn example for further explanation :
I'm recording a video singing at the same time the singer sings in a song.
So, I have a video file(Me singing) and the audio file(the song).


I tried this simple command :


-i $video-i $audio-c copy -map 0:0 -map 1:0 -shortest output.mp4



But, it gives an output with the video sound muted(my voice muted).


And I want something more advanced to control the volume of each sound, and add filter to make the output sound sounds like it was recorded in a studio.. so after a long search through the documentation and similar questions, I made up this command here..


-i $video -i $audio -filter_complex "[0:a]aformat=sample_fmts=fltp:sample_rates=44100:channel_layouts=stereo,volume=0.8[a1]; [1:a]aformat=sample_fmts=fltp:sample_rates=44100:channel_layouts=stereo,volume=4[a2]; [a1][a2]amerge,pan=stereo|c0code>


It gives a video output with audio out of sync. (My voice doesn't match the singer).


Any help would be really appreciated !


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Recursively Converting FLAC to MP3 Using FFmpeg While Maintaining Metadata and Directory Structure on Windows
20 mars 2017, par elgo2006I’m trying to use FFmpeg to convert all of the .flac’s in my music library to .mp3’s while maintaining the metadata and directory structure of the original files. I’ve got everything working so far except for maintaining the directory structure.
So far I have :
@echo off
rem Copies the original directory structure from F:\Music to D:\Music
xcopy /T %1 %2
rem Recursively converts files with .flac extension from F:\Music to .mp3 in D:\Music
for /R %1 %%i in (*.flac) do (
echo Currently converting "%%~nxi"
ffmpeg -v quiet -i "%%i" -q:a 0 -map_metadata 0 -id3v2_version 3 -write_id3v1 1 "%2\%%~ni.mp3"
)
echo Completed.
pauseI feel like my issue is coming from having my FFmpeg output as
"%2\%%~ni.mp3"
, however I can’t get it to maintain the original directory structure. I have played around with dpni as options for the output, but then it does not convert correctly.The .bat is being executed from the command line as
convert F:\Music D:\Music
All music in the original directory has the structure F :\Music\Artist Name\Album Name\Song.flac and I want it to end up as D :\Music\Artist Name\Album Name\Song.mp3.
Alternatively, is there a way to convert the .flac to an .mp3 within the same directory as the source file and then delete the flac ?
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"Cannot read property 'url' of undefined" even though it's already defined
28 novembre 2020, par Levi StanczI'm making a Discord music Bot and I'm having trouble with an error saying


TypeError: Cannot read property 'url' of undefined



I tried console logging it and it showed me the
url
, so I don't understand what is the problem with my code.

Here's my code :


//musicBOT
const Discord = require('discord.js');
const client = new Discord.Client();
const ytdl = require('ytdl-core');
const mcPrefix = '.';

const queue = new Map();

client.on('ready', () => console.log('Music bot ready!'));

client.on('message', async message => {
 if(message.author.bot) return;
 if(!message.content.startsWith(mcPrefix)) return;

 const args = message.content.substring(mcPrefix.length).split(" ");
 const serverQueue = queue.get(message.guild.id);

 if(message.content.startsWith(`${mcPrefix}play`)) {

 const voiceChannel = message.member.voice.channel;
 if(!voiceChannel) return message.channel.send("Hang-szobában kell lenned zenelejátszáshoz.");
 const permissions = voiceChannel.permissionsFor(message.client.user);
 if(!permissions.has('CONNECT')) return message.channel.send("Nincs jogosultságom csatlakozni a hangszobához.");
 if(!permissions.has('SPEAK')) return message.channel.send("Nincs jogosultságom megszólalni ebben a hangszobában.");

 const songInfo = await ytdl.getInfo(args[1])
 const song = {
 title: songInfo.title,
 url: songInfo.videoDetails.video_url
 }

 if(!serverQueue) {
 const queueConstruct = {
 textChannel: message.channel,
 voiceChannel: voiceChannel,
 connection: null,
 songs: [],
 volume: 5,
 playing: true
 }
 queue.set(message.guild.id, queueConstruct)

 queueConstruct.songs.push(song)

 try{
 var connection = await voiceChannel.join();
 message.channel.send(`${song.title} lejátszása.`)
 queueConstruct.connection = connection
 play(message.guild, queueConstruct.songs[0])
 }catch(e){
 console.log(`Hiba csatlakozás közben itt: ${e}`);
 queue.delete(message.guild.id)
 return message.channel.send(`Hiba volt a csatlakozás közben itt: ${e}`)
 }
 } else{
 serverQueue.songs.push(song)
 return message.channel.send(`**${song.title}** hozzáadva a lejátszási listához.`)
 }
 return undefined
 

 
 }else if (message.content.startsWith(`${mcPrefix}stop`)) {
 if(!message.member.voice.channel) return message.channel.send("Hang-szobában kell lenned ahhoz, hogy leállítsd a zenét.")
 if(!serverQueue) return message.channel.send("There is nothing playing")
 serverQueue.songs= []
 serverQueue.connection.dispatcher.end()
 message.channel.send("Sikeresen megálltottad a zenét.")
 return undefined
 }else if(message.content.startsWith(`${mcPrefix}skip`)){
 if(!message.member.voice.channel) return message.channel.send("Hang-szobában kell lenned a skip parancshoz.")
 if(!serverQueue) return message.channel.send("There is nothing playing")
 serverQueue.connection.dispatcher.end()
 message.channel.send("Zene továbbléptetve.")
 message.channel.send(`${song.title} játszása.`)
 
 return undefined
 }

 function play(guild, song) {
 const serverQueue = queue.get(guild.id)
 
 if(!serverQueue.songs){
 serverQueue.voiceChannel.leave()
 queue.delete(guild.id)
 return
 }
 
 const dispatcher = serverQueue.connection.play(ytdl(song.url))
 .on('finish', () => {
 serverQueue.songs.shift()
 play(guild, serverQueue.songs[0])
 })
 .on('error', error => {
 console.log(error)
 })
 dispatcher.setVolumeLogarithmic(serverQueue.volume / 5)
 }

})
//musicBOT



and here is the full error :


const dispatcher = serverQueue.connection.play(ytdl(songInfo.url))
 ^
TypeError: Cannot read property 'url' of undefined
 at play (C:\Users\Levi\Desktop\Discord BOT Javascript\bot.js:97:70)
 at StreamDispatcher.<anonymous> (C:\Users\Levi\Desktop\Discord BOT Javascript\bot.js:100:17)
 at StreamDispatcher.emit (node:events:388:22)
 at finish (node:internal/streams/writable:734:10)
 at processTicksAndRejections (node:internal/process/task_queues:80:21)
</anonymous>


I started searching on the internet but found nothing about it, I guess my basic javascript knowledge is just not enough.