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Richard Stallman et le logiciel libre
19 octobre 2011, par
Mis à jour : Mai 2013
Langue : français
Type : Texte
Autres articles (91)
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Mediabox : ouvrir les images dans l’espace maximal pour l’utilisateur
8 février 2011, parLa visualisation des images est restreinte par la largeur accordée par le design du site (dépendant du thème utilisé). Elles sont donc visibles sous un format réduit. Afin de profiter de l’ensemble de la place disponible sur l’écran de l’utilisateur, il est possible d’ajouter une fonctionnalité d’affichage de l’image dans une boite multimedia apparaissant au dessus du reste du contenu.
Pour ce faire il est nécessaire d’installer le plugin "Mediabox".
Configuration de la boite multimédia
Dès (...) -
Les autorisations surchargées par les plugins
27 avril 2010, parMediaspip core
autoriser_auteur_modifier() afin que les visiteurs soient capables de modifier leurs informations sur la page d’auteurs -
Websites made with MediaSPIP
2 mai 2011, parThis page lists some websites based on MediaSPIP.
Sur d’autres sites (9930)
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How to write a unit test class that creates a new process in C#
12 mars 2023, par IceAgeI have a question about how to write unit test method that creates a new process.


At the moment I write a normal unit test to create a process, but it's very difficult to setup.


This is my code I want to test.


using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;
using System.Threading.Tasks;

namespace xxx
{
 public class FFmPeg
 {
 private Process _process;
 private string ffmpegPath { get; set; }
 private Regex _durationRegex;

 public event EventHandler<string> OutputData;
 public event EventHandler<int> ProgressChange;

 public FFmPeg()
 {
 _process = new Process();
 ffmpegPath = Path.Combine(Directory.GetCurrentDirectory(), "engine", "ffmpeg.exe");
 }

 public void Run(string Argument)
 {
 _process.StartInfo.FileName = ffmpegPath;
 _process.StartInfo.Arguments = $"{Argument} -y";
 _process.StartInfo.UseShellExecute = false;
 _process.StartInfo.RedirectStandardOutput = true;
 _process.StartInfo.RedirectStandardError = true;
 _process.StartInfo.StandardOutputEncoding = Encoding.UTF8;
 _process.StartInfo.CreateNoWindow = true;
 _process.OutputDataReceived +=(o,e) =>
 {
 };
 _process.ErrorDataReceived += (o, e) =>
 {
 if (string.IsNullOrEmpty(e.Data) == false)
 {
 OnProgressChange(OutputExtractor.GetProgress(e.Data));
 }//end if.check data is not empty
 };

 _process.Start();
 _process.BeginOutputReadLine();
 _process.BeginErrorReadLine();
 _process.WaitForExit();
 } //end method

 protected virtual void OnProgressChange(int progress)
 {
 ProgressChange?.Invoke(this,progress);
 } //end method

 protected virtual void OnOutputData(string data)
 {
 OutputData?.Invoke(this, data);
 }
 }
}
</int></string>

Thanks for any helpful answers, and sorry for my English.


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Mp4 Video File not playing after downloading using Content Disposition php but it plays when it streams directly from the server
1er juin 2021, par Razor RasshI'm trying to combine a video and audio file using the FFMPEG and it worked successfully for me. The FFMPEG successfully created the MP4 file in the target destination folder. When I'm trying to download the Video file from the destination folder using Content-Disposition in Php. The server successfully downloads the Video file from the exact target folder.


The problem is that when I'm trying to play the video using VLC or any other player, nothing happened. The VLC player just launched and do nothing. I checked the downloaded file size with the video file located on the server and found that they both are the same size.


The code I used for creating Video file and content disposition is,



 $downloadFolderPath = "/var/www/html/temp/{$downloadFolderName}";
 $result = mkdir($downloadFolderPath, 0777); 
 $downloadFileName = "$downloadFolderPath/$filename";
 //echo $downloadFileName;
 $command = $ffmpeglocation." -i $audio_link -i $link -c:v copy -c:a aac -preset fast -crf 20 $downloadFileName 2>&1";
 shell_exec($command);

 $file=fopen($downloadFileName,'r');
 header("Content-Type:video/mp4");
 
 header("Content-Disposition: attachment; filename=$filename");
 
 header('Content-Description: File Transfer');
 header('Content-Type: application/octet-stream');
 
 header('Content-Disposition: attachment; filename="'.basename($downloadFileName).'"');
 header('Expires: 0');
 header('Cache-Control: must-revalidate');
 header('Pragma: public'); 
 header('Content-Length: ' . filesize($downloadFileName)); 
 flush(); // Flush system output buffer
 readfile($downloadFileName); 
 die();


The above code did its purpose by properly downloading the video file hosted on the server. But after downloading, the video file is not playing. I tried to access the file directly through the URL of the file in the browser and it plays fluently in the browser.


Help me out. Sorry for the bad english.


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avfilter/vf_libplacebo : don't force dovi frames to bt.2020-ncl
30 octobre 2023, par Niklas Haasavfilter/vf_libplacebo : don't force dovi frames to bt.2020-ncl
This is at odds with the YUV matrix negotiation API, in which such
dynamic changes in YUV encoding are no longer easily possible. There is
also no really strong motivating reason to do this, since the choice of
YUV matrix is essentially arbitrary and not actually related to the
Dolby Vision decoding process.
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