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    2 mai 2011, par

    This page lists some websites based on MediaSPIP.

  • Creating farms of unique websites

    13 avril 2011, par

    MediaSPIP platforms can be installed as a farm, with a single "core" hosted on a dedicated server and used by multiple websites.
    This allows (among other things) : implementation costs to be shared between several different projects / individuals rapid deployment of multiple unique sites creation of groups of like-minded sites, making it possible to browse media in a more controlled and selective environment than the major "open" (...)

  • Les statuts des instances de mutualisation

    13 mars 2010, par

    Pour des raisons de compatibilité générale du plugin de gestion de mutualisations avec les fonctions originales de SPIP, les statuts des instances sont les mêmes que pour tout autre objets (articles...), seuls leurs noms dans l’interface change quelque peu.
    Les différents statuts possibles sont : prepa (demandé) qui correspond à une instance demandée par un utilisateur. Si le site a déjà été créé par le passé, il est passé en mode désactivé. publie (validé) qui correspond à une instance validée par un (...)

Sur d’autres sites (8711)

  • check if file is downloaded in temp in GCF from GCS

    3 septembre 2020, par soshi

    Im making video encoder using GCF

    


    that downloads video file from gcs to /tmp and encode it by ffmpeg.

    


    I want to ask how to check if file is downloaded in tmp ?

    


    I tried console.log(process.cwd()) but tmp folder is not displayed

    


    exports.encodeVideo = async (req, res) => {

 mkdirp('/tmp')

 storage.bucket('video-bucket').file('raw/samplevideo.mp4').download({ 
 destination:'encoded/samplevideo.mp4' })

ffmpeg........
}


    


  • How can I take a user input in python3 and use it for the file name of ffmpeg-python file output ?

    25 novembre 2022, par tylerdurden4285

    I am trying to do something like this :

    


    import ffmpeg

user_input =input("give your output file a name: ")

(
    ffmpeg
    .input('input.mp4')
    .vflip()
    .output('(user_input).mp4')
    .run()
)



    


    I am wondering how I can take the string input from the user and make it the filename. So if they input "tester" the resulting output file would be "tester.mp4".

    


    I am running it on an ubuntu 20.04 WSL. I'm new to here and doing my best to learn python3 and ffmpeg so please be gentle with me if I broke any community rules. I'll improve as I go.

    


    I don't know what to search for to find the answer to this problem.

    


  • ffmpeg issue with file paths and wav file [closed]

    8 septembre 2023, par 101is5

    I'm trying to run :

    


    ffmpeg -i audio_input_files/voz.wav voz.mp3


    


    I'm using Windows 11 and I've tested it in PowerShell and CMD.

    


    The issue here involves path syntax and file format.
    
As shown in an example from ffmpeg docs (ffmpeg -i /tmp/test%d.Y /tmp/out.mpg) and many other sources, I should put a leading slash in relative paths. Here, however, that causes No such file or directory, for both slash and backslash. Unlike everywhere I've looked, removing the leading slash was the solution.

    


    Now, if the input is provided as a path and it is a .wav file, I get Invalid data found when processing input.

    


    Providing just names (and being in the correct folder, obviously), e.g. ffmpeg -i voz.wav voz.mp3 works just fine, though.

    


    Is there a workaround for that, i.e. in case I want to use paths for wav files with ffmpeg ?

    


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