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  • Publier sur MédiaSpip

    13 juin 2013

    Puis-je poster des contenus à partir d’une tablette Ipad ?
    Oui, si votre Médiaspip installé est à la version 0.2 ou supérieure. Contacter au besoin l’administrateur de votre MédiaSpip pour le savoir

  • Encoding and processing into web-friendly formats

    13 avril 2011, par

    MediaSPIP automatically converts uploaded files to internet-compatible formats.
    Video files are encoded in MP4, Ogv and WebM (supported by HTML5) and MP4 (supported by Flash).
    Audio files are encoded in MP3 and Ogg (supported by HTML5) and MP3 (supported by Flash).
    Where possible, text is analyzed in order to retrieve the data needed for search engine detection, and then exported as a series of image files.
    All uploaded files are stored online in their original format, so you can (...)

  • Keeping control of your media in your hands

    13 avril 2011, par

    The vocabulary used on this site and around MediaSPIP in general, aims to avoid reference to Web 2.0 and the companies that profit from media-sharing.
    While using MediaSPIP, you are invited to avoid using words like "Brand", "Cloud" and "Market".
    MediaSPIP is designed to facilitate the sharing of creative media online, while allowing authors to retain complete control of their work.
    MediaSPIP aims to be accessible to as many people as possible and development is based on expanding the (...)

Sur d’autres sites (7386)

  • Revision ad479a9b3d : multi-res : modify memory allocation code Reverted part of change in memory alll

    23 mai 2012, par Yunqing Wang

    Changed Paths : Modify /vp8/vp8_cx_iface.c multi-res : modify memory allocation code Reverted part of change in memory alllocation code, which ensures that the function returns 0 and encoder works correctly when CONFIG_MULTI_RES_ENCODING isn't turned on. Change-Id : (...)

  • How to run a command in python with file paths containing spaces ?

    24 juillet 2023, par tomatochan

    I'm developing a program in Python that requires the use of FFMPEG. In one step, I need to create a video from a text file containing the list of images (path and duration) that need to be assembled to make a video. To do this, I use the following code :

    


    cmd = [ffmpeg, "-y", "-f", "concat", "-safe", "0", "-i", tmp, "-c:v", "libx264", "-r", "25", "-pix_fmt", "yuv420p", out]
subprocess.run(cmd)
# os.system(" ".join(cmd)) # returns same error


    


    What returns the error : The syntax of the file, directory or volume name is incorrect.

    


    I know that paths can be problematic when they contain spaces, and as my username has one, I've taken care to quote the paths in such a way that :

    


    ffmpeg = '"C:\Users\John Doe\Documents\ffmpeg\bin\ffmpeg.exe"'
tmp = '"C:\Users\John Doe\Desktop\path\to\file.txt"'
out = '"C:\Users\John Doe\Desktop\path\to\video.mp4"'


    


    When I print(" ".join(cmd)), this is what I get in my terminal :

    


    "C:\Users\John Doe\Documentsffmpeg\binffmpeg.exe" -y -f concat -safe 0 -i "C:\Users\John Doe\Desktop\path\to\file.txt" -c:v libx264 -r 25 -pix_fmt yuv420p "C:\Users\John Doe\Desktop\path\to\video.mp4"


    


    However, the problem persists.
Has anyone ever had this problem and managed to solve it ?

    


    I've also tried the method where you have to escape the spaces (taking care to replace the \ with / and then the with \ ) but nothing works... The error persists.

    


    For your information, here are the contents of one of my .txt files

    


    file C:\Users\John Doe\Desktop\path\to\image_1.png
duration 0.04
file C:\Users\John Doe\Desktop\path\to\image_2.png
duration 0.04
file C:\Users\John Doe\Desktop\path\toimage_3.png
duration 0.04


    


    When I quote the paths in the .txt files

    


      

    • subprocess gives me the error :
PermissionError: [WinError 5] Access denied
    • 


    • os.system gives me the error :
    • 


    


    C:\Users\John' is not recognized as an internal or external command, an executable program or a command file.
The syntax of the file, directory or volume name is incorrect.


    


  • Requesting header information of a file

    11 janvier 2012, par 0v3rrid3

    is there anyway I can request only the header information of any media. For example I just want to request header information of any video file so as to find its video length. I tried using ffmpeg -i {video_url} and did the work but I noticed that it actually downloads the given media in local storage and returns back the header information which obviously increases roundtrip time.

    So I would really appreciate if there is any idea for finding the length of media in a fly. BTW I have a ruby on rails application where I need to implement this.