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  • Installation en mode ferme

    4 février 2011, par

    Le mode ferme permet d’héberger plusieurs sites de type MediaSPIP en n’installant qu’une seule fois son noyau fonctionnel.
    C’est la méthode que nous utilisons sur cette même plateforme.
    L’utilisation en mode ferme nécessite de connaïtre un peu le mécanisme de SPIP contrairement à la version standalone qui ne nécessite pas réellement de connaissances spécifique puisque l’espace privé habituel de SPIP n’est plus utilisé.
    Dans un premier temps, vous devez avoir installé les mêmes fichiers que l’installation (...)

  • Websites made ​​with MediaSPIP

    2 mai 2011, par

    This page lists some websites based on MediaSPIP.

  • Possibilité de déploiement en ferme

    12 avril 2011, par

    MediaSPIP peut être installé comme une ferme, avec un seul "noyau" hébergé sur un serveur dédié et utilisé par une multitude de sites différents.
    Cela permet, par exemple : de pouvoir partager les frais de mise en œuvre entre plusieurs projets / individus ; de pouvoir déployer rapidement une multitude de sites uniques ; d’éviter d’avoir à mettre l’ensemble des créations dans un fourre-tout numérique comme c’est le cas pour les grandes plate-formes tout public disséminées sur le (...)

Sur d’autres sites (9900)

  • how to choose ffmpeg video mp3 audio version id ?

    7 juin 2013, par Ulterior

    I am having issues on ffmpeg encoded video files audio tracks. My encoded video contains ID for audio track as extracted from mediainfo :

    I use CODEC_ID_MP3 in guess_format "mov" container for quicktime

    Audio
    ID                                       : 2
    Format                                   : MPEG Audio
    Format version                           : Version 2
    Format profile                           : Layer 3
    Codec ID                                 : .mp3
    Duration                                 : 2s 916ms
    Bit rate mode                            : Constant
    Bit rate                                 : 128 Kbps
    Channel(s)                               : 1 channel
    Sampling rate                            : 16.0 KHz
    Compression mode                         : Lossy
    Stream size                              : 45.3 KiB (2%)
    Language                                 : English

    This is not recognized on a vanilla codecless installation of windows 7, only played by k-lite codec libmad

    I have noticed, that another test file contains similar mp3 track and is played by media player OK :

    Audio
    ID                                       : 2
    Format                                   : MPEG Audio
    Format version                           : Version 1
    Format profile                           : Layer 3
    Mode                                     : Joint stereo
    Mode extension                           : MS Stereo
    Codec ID                                 : 6B
    Duration                                 : 1mn 9s
    Bit rate mode                            : Constant
    Bit rate                                 : 320 Kbps
    Channel(s)                               : 2 channels
    Sampling rate                            : 44.1 KHz
    Compression mode                         : Lossy
    Stream size                              : 2.67 MiB (38%)
    Writing library                          : LAME3.98

    The difference I noticed is in Format version number and Codec ID, which is Version 2 from ffmpeg output - I couldnt locate this version setting in ffmpeg source files, so my question is - is there a way to influence this format version identificator and set the codec id as in above playable video ?

  • Evolution #4080 : Raccourci puce : se débarasser de l’image

    1er octobre 2018

    Je vote pour le caractère
    https://unicode-table.com/fr/2023/

    Voici un test chez moi en local sur SPIP 3.2, la dist et Firefox Windows (bref 100% vanille)

  • Generate waveforms for audio files with large amount of channels

    3 mai 2021, par motio

    I want to generate .png files displaying the waveforms of interleaved audio .wav files using the FFmpeg libraries. http://ffmpeg.org/documentation.html

    


    If the interleaved audio file contains maximum 8 channels, I manage to successfully achieve this using the following command line :

    


    ffmpeg -i 8_channels_input.wav -y -filter_complex "showwavespic=s=1920x1200:split_channels=1" -frames:v 1 8_channels_waveform_output.png


    


    However, if the interleaved audio file contains more than 8 channels, FFmpeg gives the following :

    


    


    Input #0, wav, from '30_channels_input.wav' : Duration : 00:00:02.08,
bitrate : 31752 kb/s Stream #0:0 : Audio : pcm_s24le ([1][0][0][0] /
0x0001), 44100 Hz, 30 channels, s32 (24 bit), 31752 kb/s Stream
mapping : Stream #0:0 (pcm_s24le) -> showwavespic showwavespic ->
Stream #0:0 (png) Press [q] to stop, [?] for help [auto_resampler_0 @
0x7faf5d60a3c0] Cannot select channel layout for the link between
filters auto_resampler_0 and Parsed_showwavespic_0. [auto_resampler_0
@ 0x7faf5d60a3c0] Unknown channel layouts not supported, try
specifying a channel layout using 'aformat=channel_layouts=something'.
Error reinitializing filters ! Failed to inject frame into filter
network : Invalid argument Error while processing the decoded data for
stream #0:0 Conversion failed !

    


    


    Here is the related documentation (c.f. bottom of the page) :
https://trac.ffmpeg.org/wiki/AudioChannelManipulation

    


    My problem is :
I need to generate the visual waveforms of audio files containing up to 30 channels.
All my attempts were unsuccessful so far (I am trying to define custom channel layouts and I am not sure if I am on the right track here).

    


    To simplify, I need to complete/modify the following command to make it work :

    


    ffmpeg -i 30_channels_input.wav -y -filter_complex "showwavespic=s=1920x1200:split_channels=1" -frames:v 1 30_channels_waveform_output.png 


    


    [EDIT] Remarks :

    


    I manage to generate the waveforms of a 10 channels input by combining existing layouts :

    


    ffmpeg -i 10_channels_input.wav -y -filter_complex "aformat=channel_layouts=7.1+downmix, showwavespic=s=1920x1200:split_channels=1" -frames:v 1 10_channels_waveform_output.png


    


    However, if you attempt to do it for a 30 channels input by combining 5x 6.0 layouts :

    


    ffmpeg -i 30_channels_input.wav -y -filter_complex "aformat=channel_layouts=6.0+6.0+6.0+6.0+6.0, showwavespic=s=1920x1200:split_channels=1" -frames:v 1 30_channels_waveform_output.png


    


    FFmpeg gives the following :

    


    


    [auto_resampler_0 @ 0x7ffd7002a480] [SWR @ 0x7ffd7013a000] Rematrix is
needed between 30 channels and 6.0 but there is not enough information
to do it [auto_resampler_0 @ 0x7ffd7002a480] Failed to configure
output pad on auto_resampler_0 Error reinitializing filters ! Failed to
inject frame into filter network : Invalid argument Error while
processing the decoded data for stream #0:0

    


    


    My assumption is that I need to create a custom layout using 30 unique channel IDs (c.f. https://trac.ffmpeg.org/wiki/AudioChannelManipulation bottom of the page) instead of combining existing layouts.

    


    It seems that only 25 channel IDs are available though. Creating a custom layout with 30 channels is maybe not possible at all...

    


    [EDIT 2] Remarks :

    


    I finally found the documentation I was looking for. But it still seems that generating the waveforms of 30 channels would be difficult.

    


    Here is how to create custom layouts :
https://ffmpeg.org/ffmpeg-utils.html

    


    


    A custom channel layout can be specified as a sequence of terms,
separated by ’+’ or ’|’. Each term can be :

    


    the name of a standard channel layout (e.g. ‘mono’, ‘stereo’, ‘4.0’,
‘quad’, ‘5.0’, etc.) the name of a single channel (e.g. ‘FL’, ‘FR’,
‘FC’, ‘LFE’, etc.) a number of channels, in decimal, followed by ’c’,
yielding the default channel layout for that number of channels (see
the function av_get_default_channel_layout). Note that not all channel
counts have a default layout. a number of channels, in decimal,
followed by ’C’, yielding an unknown channel layout with the specified
number of channels. Note that not all channel layout specification
strings support unknown channel layouts. a channel layout mask, in
hexadecimal starting with "0x" (see the AV_CH_* macros in
libavutil/channel_layout.h. Before libavutil version 53 the trailing
character "c" to specify a number of channels was optional, but now it
is required, while a channel layout mask can also be specified as a
decimal number (if and only if not followed by "c" or "C").

    


    See also the function av_get_channel_layout defined in
libavutil/channel_layout.h.

    


    


    e.g.

    


    Therefore, for 11 channels input :

    


    ffmpeg -i 11_channels_input.wav -y -filter_complex "aformat=channel_layouts=FL+FR+FC+BL+BR+BC+SL+SR+WL+WR+TBL, showwavespic=s=1920x1200:split_channels=1" -frames:v 1 11_waveform_output.png


    


    —> worked for me

    


    while :

    


    ffmpeg -i 11_channels_input.wav -y -filter_complex "aformat=channel_layouts=11c, showwavespic=s=1920x1200:split_channels=1" -frames:v 1 11_waveform_output.png


    


    —> does not work

    


    But :

    


    ffmpeg -i 24_channels_input.wav -y -filter_complex "aformat=channel_layouts=24c, showwavespic=s=1920x1200:split_channels=1" -frames:v 1 24_waveform_output.png


    


    —> does work

    


    and finally, what I am still trying to achieve :

    


    ffmpeg -i 30_channels_input.wav -y -filter_complex "aformat=channel_layouts=30c, showwavespic=s=1920x1200:split_channels=1" -frames:v 1 30_waveform_output.png


    


    —> does not work

    


    p.s.

    


      

    • I execute these commands in sub shells via Ruby scripts (puts %x...)
    • 


    • My system : macOS 10.15.6 | zsh | FFmpeg 4.4 | Ruby 2.6.3
    •