如果传输状态为 206 Partial Content，Chrome 无法加载视频

Question

我正在制作一个简单的 Java com.sun.net.httpserver.HttpServer 来提供静态视频文件。如果我返回状态代码 206，部分内容，当我尝试通过浏览器访问它时，视频无法播放（状态代码为 200 的视频可以正常播放，但我希望能够搜索和循环播放视频），这是我的HttpHandler：

final String path = StaticHandler.toPathSafe(httpExchange.getRequestURI().getPath());
System.out.println(path);
final File file = new File(path);

if (file.isFile())
{
    int code = 200;
    long position = 0L;
    long end = file.length();

    if (httpExchange.getRequestHeaders().containsKey("Range"))
    {
        try
        {
            long[] range = StaticHandler.parseRange(httpExchange.getRequestHeaders().get("Range").get(0));

            position = range[0];
            if (range[1] != -1)
                end = range[1];

            // the video loads fine when code = 200;
            code = 206;

            httpExchange.getResponseHeaders().set("Content-Range", "bytes " + position + "-" + end + "/" + file.length());
        }
        catch (Exception e)
        {
            e.printStackTrace();
        }
    }

    httpExchange.getResponseHeaders().set("Accept-Range", "bytes");
    httpExchange.getResponseHeaders().set("Content-Type", "video/mp4");
    httpExchange.getResponseHeaders().set("Content-Length", String.valueOf(end - position));

    System.out.println("Response: " + position + ", " + end);

    httpExchange.sendResponseHeaders(code, 0L);

    final FileChannel fileChannel = new FileInputStream(file).getChannel();
    final WritableByteChannel responseChannel = Channels.newChannel(response.getOutputStream());
    fileChannel.transferTo(position, end - position, responseChannel);

    responseChannel.close();
    fileChannel.close();
}
else
{
    System.out.println("404");
    httpExchange.sendResponseHeaders(404, -1);
}

以上代码无法在 chrome 上加载，但在 Firefox 中运行良好，这是我在 Chrome 中获得的标题：

Response Headers:

Accept-range: bytes
Content-length: 31491166
Content-range: bytes 0-31491166/31491166
Content-type: video/mp4
Date: Sat, 27 Jul 2019 14:32:55 GMT
Transfer-encoding: chunked


Request Headers:

Accept: */*
Accept-Encoding: identity;q=1, *;q=0
Accept-Language: en-US,en;q=0.9
Cache-Control: no-cache
Connection: keep-alive
Host: 192.168.56.1:5000
Pragma: no-cache
Range: bytes=0-
Referer: http://192.168.56.1:5000/30MB.mp4
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/75.0.3770.142 Safari/537.36

我是否遗漏了什么和/或我的代码有什么问题？

Answer 1

范围包括起点-终点。不是起始长度

例如：

Content-range: bytes 0-0/31491166

返回 1 个字节（从零字节到零字节）

您的代码不起作用，因为浏览器正在等待一个永远不会发送的字节。

Answer 2

使用下面的代码 sn-p。它可以在所有浏览器 Safari、Firefox 和 Chrome 上运行和测试。

protected byte[] prepareContent(final HttpHeaders headers, byte[] media,
        String range) throws IOException {
    
    long rangeStart = 0;
    long rangeEnd;

    long fileSize = media.length;
    String[] ranges = range.split("-");
    rangeStart = Long.parseLong(ranges[0].substring(6));

    if (ranges.length > 1) {
        rangeEnd = Long.parseLong(ranges[1]);
    } else {
        rangeEnd = fileSize - 1;
    }
    if (fileSize < rangeEnd) {
        rangeEnd = fileSize - 1;
    }
    
    String contentLength = String.valueOf((rangeEnd - rangeStart) + 1);
    
    headers.add("Content-Length", contentLength);
    headers.add("Content-Range", "bytes " + rangeStart + "-" + rangeEnd + "/" + fileSize);
    headers.add("Content-Type", "video/mp4");
    headers.add("Accept-Ranges","bytes");
    return readByteRange(media, rangeStart, rangeEnd);
}

private byte[] readByteRange(byte[] media, long start, long end) throws IOException {
    
    try (InputStream inputStream = new ByteArrayInputStream(media);
         ByteArrayOutputStream bufferedOutputStream = new ByteArrayOutputStream()) {
        
        int nRead;
        while ((nRead = inputStream.read(media, 0, media.length)) != -1) {
            bufferedOutputStream.write(media, 0, nRead);
        }
        bufferedOutputStream.flush();
        byte[] result = new byte[(int) (end - start) + 1];
        System.arraycopy(bufferedOutputStream.toByteArray(), (int) start, result, 0, result.length);
        return result;
    }
}