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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 (91)
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L’utiliser, en parler, le critiquer
10 avril 2011La première attitude à adopter est d’en parler, soit directement avec les personnes impliquées dans son développement, soit autour de vous pour convaincre de nouvelles personnes à l’utiliser.
Plus la communauté sera nombreuse et plus les évolutions seront rapides ...
Une liste de discussion est disponible pour tout échange entre utilisateurs. -
Websites made with MediaSPIP
2 mai 2011, parThis page lists some websites based on MediaSPIP.
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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 (...)
Sur d’autres sites (10008)
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displaying a baseline h264 frames stream in browsers
6 août 2021, par Thabet SabhaSo, I have a server that receives a live rtsp stream then generates baseline h264 frames using ffmpeg, which then are sent via an rtcDataChannel to browser, and while the frames arrive as intended, I can't figure out a way to display them on my html5 videoElement,
here is a simplified version of my current approach :


const remoteStream = new MediaSource();
myVideoElement.src = window.URL.createObjectURL(remoteStream);

// called when remoteStream.readyState === "open"
let sourceBuffer = remoteStream.addSourceBuffer('video/mp4; codecs="avc1.4d002a"');

// this gets called when ever a new frame is received from the webrtc data channel.
function onFrame(frame) {
 sourceBuffer.appendBuffer(new Uint8Array(frame));

 /*
 console.log(frame) ==> <buffer 00="00" 01="01" 41="41" 9b="9b" a0="a0" 22="22" 80="80" a5="a5" d7="d7" 42="42" ea="ea" 34="34" 14="14" 85="85" ba="ba" bc="bc" 1b="1b" f2="f2" 71="71" 0d="0d" 8b="8b" e1="e1" 3c="3c" 52="52" d5="d5" 8c="8c" ef="ef" c1="c1" 89="89" 10="10" c5="c5" 05="05" 78="78" ee="ee" 1d="1d" 03="03" 8d="8d" 2896="2896" more="more" bytes="bytes">
 */
}
</buffer>


ffmpeg options :


[
 "-rtsp_transport", "tcp",
 "-i", `${rtspCamURL}`, 
 "-framerate", "15",
 "-c:v", "libx264",
 "-vprofile", "baseline",
 "-b:v", "600k",
 "-bufsize", "600k",
 "-pix_fmt", "yuv420p",
 '-tune', 'zerolatency',
 "-preset", "ultrafast",
 "-f", "rawvideo",
 '-'
]; 



ffmpeg stream is then split using NAL delimiter (to generate individual frames) then each frame is sent via the data channel like so :

Buffer.concat([nalDelimiter, frame])
.

I am not sure if i'm missing something as i'm not getting any helpful errors due to the remoteSource closing as soon as the first frame arrives for some reason.


or does the media source just not support raw h264 frames, and if so is there a workaround to solve this issue ? (even if it has to do with changing the ffmpeg params.


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ffmpeg Error : Pattern type 'glob' was selected but globbing is not support ed by this libavformat build
14 septembre 2017, par Aryan NaimI’m trying to convert group of ".jpg" files acting as individual frames into 1 single mpeg video ".mp4"
Example parameters i used :
frame duration = 2 secs
frame rate = 30 fps
encoder = libx264 (mpeg)
input pattern = "*.jpg"
output pattern = video.mp4Based on ffmpeg wiki instructions at (https://trac.ffmpeg.org/wiki/Create%20a%20video%20slideshow%20from%20images), I issued this command :
ffmpeg -framerate 1/2 -pattern_type glob -i "*.jpg" -c:v libx264 -r 30 -pix_fmt yuv420p video.mp4
But I’m getting this error :
[image2 @ 049ab120] Pattern type 'glob' was selected but globbing is not
supported by this libavformat build *.jpg: Function not implementedWhich probably means the API pattern matching commands for my build/version have changed. By the way this my windows 32bit ffmpeg download build (
ffmpeg-20150702-git-03b2b40-win32-static
).How can I choose a group of files using pattern matching using ffmpeg ?
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Bash : sort find results using part of a filename
13 novembre 2015, par utt50I have 3 webcams set up in a building, uploading still images to a webserver. I’m using ffmpeg to encode the jpgs to mp4 video.
The directories are set up like this :
Cam1/201504
Cam1/201505
Cam2/201504
Cam2/201505
Cam3/201504
Cam3/201505I’m using the following bash loop/ffmpeg parameters to make one video per camera, per year. This works well so far (well... except that my SSD is rapidly degrading in performance - too many simultaneous read/write operations ?) :
find Cam2/2013* -name "*.jpg" -print0 | xargs -0 cat | ffmpeg -f image2pipe -framerate 30 -vcodec mjpeg -i - -vcodec libx264 -profile:v baseline -level 3.0 -movflags +faststart -crf 19 -pix_fmt yuv420p -r 30 "Cam2-2013-30fps-19crf.mp4"
The individual files are named like this (confusing ffmpeg’s built-in file sequencer) :
Cam1_2015052413543201.jpg
Cam1_2015052413544601.jpg
Cam2_2015052413032601.jpg
Cam2_2015052413544901.jpgI now need to create one video for an entire year across all 3 cameras, ordered by timestamp. To accomplish this, I need to sort the find results by the segment of the filename after the underscore.
What do I pipe the find output to to accomplish this ? For example, the files above would be ordered like this :
Cam2_2015052413032601.jpg
Cam1_2015052413543201.jpg
Cam1_2015052413544601.jpg
Cam2_2015052413544901.jpgAny help is very much appreciated !