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Asked: 2026-06-15T07:58:56+00:00

I would like to stream my audio/video files to web using servlet. I made

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I would like to stream my audio/video files to web using servlet.

I made an attempt with the following servlet:

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    File file = new File("/Users/myfolder/Documents/workspace/love.mp3");
    response.setContentType(getServletContext().getMimeType(file.getName()));
    response.setContentLength((int) file.length());
    Files.copy(file.toPath(), response.getOutputStream());
}

And the following HTML:

<a href="/media" data-format="mp3 ogg">Click Here!</a>

However, the player is just loading… loading… loading…

How is this caused and how can I solve it?

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1 Answer

  1. Editorial Team

    A lot of media players require that the server supports the so-called HTTP range requests. I.e. it must be able to return specific parts of the media file on request with a Range header. For example, only the bytes at exactly the index 1000 until with 2000 on a file of 10MB long. This is mandatory for many media players in order to be able to skip a certain range of the media stream quickly enough and/or to improve buffering speed by creating multiple connections which each requests different parts of the file.

    This is however a lot of additional code in your servlet which requires a well understanding of the HTTP Range specification. Usually the servletcontainer’s (Tomcat, JBoss AS, Glassfish, etc) own default servlet already supports this out the box. So if there’s a way to publish the media folder into the web by standard means, so that you don’t need to homegrow a servlet for this, then I’d go on this route.

    It’s unclear which servletcontainer you’re using, so I’ll assume Tomcat in this example:

    1. Just drop love.mp3 file in the public web content of the web project, so that it’s just available by <a href="love.mp3"> without the need for a whole servlet.

    2. Or, put the love.mp3 file in a new subfolder of Tomcat/webapps folder, e.g. Tomcat/webapps/media/love.mp3. This way it’s available by <a href="/media/love.mp3">.

    3. Or, put the love.mp3 file elsewhere on disk, e.g. /path/to/media/love.mp3 and add the /media folder as new context by adding the following line to Tomcat’s /conf/server.xml:

       <Context docBase="/path/to/media" path="/media" />

      This way it’s available by <a href="/media/love.mp3"> as well.

    Either way, Tomcat’s own DefaultServlet, which has proper support for Range requests, will be used to stream the content.

    But if there’s absolutely no way to make use of servletcontainer’s own default servlet, then you need to rewrite your servlet code in such way that it properly supports Range requests. You can get inspiration from open source examples such as Tomcat DefaultServlet and OmniFaces FileServlet.

    See also:

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