使用Java servlet的视频下载/流

Question

当客户端访问URL类似于此时,我试图在我的服务器中下载视频文件:

http://localhost:8088/openmrs/moduleServlet/patientnarratives/videoDownloadServlet?videoObsId=61

我试过这段代码.但它不起作用.当我访问servlet时,它只下载一个空白(0大小)文件.

protected void doGet(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException
{
    try {
        Integer videoObsId = Integer.parseInt(request.getParameter("videoObsId"));

        Obs complexObs = Context.getObsService().getComplexObs(videoObsId, OpenmrsConstants.RAW_VIEW); 
        ComplexData complexData = complexObs.getComplexData();
        Object object2 = complexData.getData(); // <-- an API used in my service. this simply returns an object.

        byte[] videoObjectData = SerializationUtils.serialize(object2);

        // Get content type by filename.
        String contentType = null;
        if (contentType == null) {
            contentType = "application/octet-stream";
        }

        // Init servlet response.
        response.reset();
        response.setBufferSize(DEFAULT_BUFFER_SIZE);
        response.setContentType(contentType);
        response.setHeader("Content-Length", String.valueOf(videoObjectData.length));
        response.setHeader("Content-Disposition", "attachment; filename=\"" + "test.flv" + "\"");

        // Prepare streams.
        BufferedInputStream input = null;
        BufferedOutputStream output = null;

        try {
            // Open streams.
            input = new BufferedInputStream(new ByteArrayInputStream(videoObjectData), DEFAULT_BUFFER_SIZE);              
            output = new BufferedOutputStream(response.getOutputStream(), DEFAULT_BUFFER_SIZE);

            // Write file contents to response.
            byte[] buffer = new byte[DEFAULT_BUFFER_SIZE];
            int length;
            while ((length = input.read(buffer)) > 0) {
                output.write(buffer, 0, length);
            }
        } finally {
            // Gently close streams.
            close(output);
            close(input);
        }
    }

    // Add error handling above and remove this try/catch
    catch (Exception e) {
        log.error("unable to get file", e);
    }
}
private static void close(Closeable resource) {
    if (resource != null) {
        try {
            resource.close();
        } catch (IOException e) {
            // Do your thing with the exception. Print it, log it or mail it.
            e.printStackTrace();
        }
    }
}

我使用过BalusC的fileservlet教程,但在我的情况下,我没有文件对象作为输入流只是字节数组对象.

救命..

Answer 1

您找到的servlet确实不能用于流式传输视频文件.它更像是一个简单的文件下载servlet,用于静态文件,如PDF,XLS等.

许多视频播放器要求服务器支持所谓的HTTP范围请求.即它必须能够通过带有Range标题的请求返回视频文件的特定字节范围.例如,在10000字节长的文件上只有索引1000到2000的字节.这是必须的,以便能够足够快地跳过特定范围的视频流而无需下载整个文件和/或通过创建多个HTTP连接来提高缓冲速度,每个HTTP连接请求视频文件的不同部分.

然而,这是servlet中的许多附加代码,需要很好地理解HTTP Range规范.您找到的文件servlet的同一作者在此扩展文件servlet的风格中提供了一个随时可用的示例.在特定情况下,它可能是值得推荐的,将文件保存到本地磁盘的文件系统基于缓存第一(例如,通过File#createTempFile()在HTTP会话一些关键的),这样你就不必一次又一次地从外部服务获得.