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Autres articles (61)
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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. -
Problèmes fréquents
10 mars 2010, parPHP et safe_mode activé
Une des principales sources de problèmes relève de la configuration de PHP et notamment de l’activation du safe_mode
La solution consiterait à soit désactiver le safe_mode soit placer le script dans un répertoire accessible par apache pour le site -
HTML5 audio and video support
13 avril 2011, parMediaSPIP uses HTML5 video and audio tags to play multimedia files, taking advantage of the latest W3C innovations supported by modern browsers.
The MediaSPIP player used has been created specifically for MediaSPIP and can be easily adapted to fit in with a specific theme.
For older browsers the Flowplayer flash fallback is used.
MediaSPIP allows for media playback on major mobile platforms with the above (...)
Sur d’autres sites (10632)
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How to extract single image (with different name each time) from multiple videos in the one folder using ffmpeg bash script
15 juillet 2014, par morriscotty_sdLet’s start with this... I’m fairly new to Linux.
I have been looking into ffmpeg to extract an image from a video using the terminal. I’ve accomplished this using this piece of code :
ffmpeg -itsoffset -4 -i oggvideo.ogg -vcodec mjpeg -vframes 1 -an -f rawvideo -s 320x240 oggThumb.jpg
I run this when I’m in the directory which holds the videos in my project. But what I want is to extract an image from all videos in that folder at the same time, naming the image corresponding with the video name it’s extracted from. My end goal is to have a folder which will hold videos that a user uploads, and at a certain time the script will run and extract images from the uploaded videos, saving the image path to a database.
I found this line of code :
for %%i in (*.mp4) do ffmpeg -ss 20 -i %%i -t 1 -s 590x340 -f image2 %%i.jpg
from this site : Trying-to-extract-1-image-from-multiple-videos-with-FFMPEG
I run that in the terminal and it gives me the error : ’bash : syntax error near unexpected token `(’’
I tried removing the ’( )’ but then it just takes me onto a new line. This is where I’m having the most problems due to being relatively new to linux commands.Any suggestions or advice would be very helpful.
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Problems with ffmpeg build in Xcode
16 décembre 2016, par jonHi I have a problem with getting ffmpeg libraries to build in my Xcode project. It is in C++ . I have installed it using homebrew and have checked that all the correct libraries have installed via terminal. I have tried with and without extern C as I know it is a c library.
#include <iostream>
#include <opencv2></opencv2>opencv.hpp>
#include <libavformat>
//extern "C" {
// #include <libavformat>
//}
</libavformat></libavformat></iostream>The linker flags have been set by using (e.g. pkg-config —libs libavcodec) in terminal and these flags do not throw up an error when building. The header files are also found with no problems as the autocomplete flags them top while typing.
Please see attached pictures for Xcode build settings. Really stuck pease help
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Why is exec('which ffmpeg') returning nothing in my application ?
28 avril 2015, par EchoLogicI’m trying to deal with FFMpeg for my application, but I’m encountering some difficulty in getting it set up. To instantiate or create an instance of FFMpeg, I need to call a
create()
method and pass through the operating locations of FFMpeg and FFProbe. To do this, I was hoping to be able to write :$ffmpeg = FFMpeg\FFMpeg::create([
'ffmpeg.binaries' => exec('which ffmpeg'),
'ffprobe.binaries' => exec('which ffprobe')
]);But, sadly, the command
exec('which ffmpeg')
returns an empty string. If I run the ’which ffmpeg’ command in Terminal, I get a response :/usr/local/bin/ffmpeg
So why does it work in my terminal, but not when my PHP script tries to execute it ?
I’ve tried a few solutions, such as appending
2>&1
to the end of the command, which did not help, running the command inshell_exec()
, and placing the output in the second parameter ofexec()
which also did not solve my issue.I feel like this may be some sort of weird permissions issue...
Addendum
It has been brought to my attention that PHP cannot see my
$PATH
variable. From the console,echo $PATH
produces :/Applications/MAMP/bin/php/php5.6.2/bin:/usr/local/bin:/usr/bin:/bin:/usr/sbin:/sbin:/opt/X11/bin:/usr/local/git/bin:/opt/ImageMagick/bin
While calling the same from my application with
exec()
produces :/usr/bin:/bin:/usr/sbin:/sbin237.671000