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ASP.NET核心2中的流视频文件

memorystream filestream asp.net-core-2.0

unos baghaii · 技术社区 · 6 年前

我想使用ASP.NET核心在浏览器中播放视频

在HTML中我有

<video width="320" height="240" controls>
 <source src="http://localhost:55193/api/VideoPlayer/Download" type="video/mp4">
 Your browser does not support the video tag.
</video>

在ASP.NET核心2中

    [HttpGet]
    [Route("Download")]
    public async Task<IActionResult> Download()
    {
        var path = @"d:\test\somemovie.mp4";
        var memory = new MemoryStream();
        using (var stream = new FileStream(@"d:\test\somemovie.mp4", FileMode.Open, FileAccess.Read, FileShare.ReadWrite, 65536, FileOptions.Asynchronous | FileOptions.SequentialScan))
        {
            await stream.CopyToAsync(memory);
        }
        memory.Position = 0;
        return File(memory, "application/octet-stream", Path.GetFileName(path));
    }

此代码是否通过流媒体播放文件(我的意思是缓冲区文件逐块播放)?

如果我想从用户设置的任何位置播放文件,我该怎么做?

3 回复 | 直到 5 年前

VahidN 6 年前

只需使用Normal Return PhysicalFile here:。


   
    public class homecontroller:controller
{
公共IActionResult下载()
{
返回physicalfile(@“d:\test\somemovie.mp4”,“application/octet stream”);
}

因为它支持流式传输和恢复文件下载所需的范围头文件:


还return file,filestreamresultandvirtualfileresultsupportpartial range requeststoo.即使是静态文件中间件也支持这一点。


因为它支持流式传输和恢复文件下载所需的范围头文件:


阿尔索return File,FileStreamResult和VirtualFileResult支持partial range requests也是。偶数static files middleware也支持这一点。

unos baghaii 6 年前

出了点问题。我的样本不支持简历

[httpget]
[路线(“下载2”)]
公共IActionResult下载2()
{
返回physicalfile(@“d:\test\somemovie.mp4”,“application/octet stream”);
}


响应头中没有接受范围



但当我使用

[httpget]
[路由(“下载”)]
公共异步任务<IActionResult>下载()
{
var path=@“d:\test\somemovie.mp4”;
var memory=new memoryStream();
使用(var stream=new filestream(@“d:\test\somemovie.mp4”,filemode.open,fileaccess.read,fileshare.readwrite,65536,fileoptions.asynchronous fileoptions.sequentialscan))
{
wait stream.copytoAsync(内存);
}
记忆位置=0;
返回文件(内存,“application/octet stream”,path.getfilename(path),true);/enableRangeProcessing=true
}

“EnableRangeProcessing”参数为真



你能解释一下为什么会这样吗?
我应该使用哪种解决方案?我很困惑。




响应头中没有接受范围



但当我使用

[HttpGet]
    [Route("Download")]
    public async Task<IActionResult> Download()
    {
        var path = @"d:\test\somemovie.mp4";
        var memory = new MemoryStream();
        using (var stream = new FileStream(@"d:\test\somemovie.mp4", FileMode.Open, FileAccess.Read, FileShare.ReadWrite, 65536, FileOptions.Asynchronous | FileOptions.SequentialScan))
        {
            await stream.CopyToAsync(memory);
        }
        memory.Position = 0;
        return File(memory, "application/octet-stream", Path.GetFileName(path),true); //enableRangeProcessing = true
    }


“EnableRangeProcessing”参数为真



你能解释一下为什么会这样吗?
我应该使用哪种解决方案?我很困惑。

-1

Rekiem Ulises Xandrya 5 年前

  static MediaStream song = null;
        static byte[] SongArray = null;

song = await client.GetMediaStreamAsync(streamFilter).ConfigureAwait(false);
                        MemoryStream ms = new MemoryStream();
                        song.CopyTo(ms);
                        SongArray = ms.ToArray();

  long fSize = song.Length, startbyte = 0, endbyte = fSize - 1;
                int statusCode = 200;

                var rangeRequest = Request.Headers["Range"].ToString();
                if (!string.IsNullOrEmpty(rangeRequest))
                {
                    string[] range = Request.Headers["Range"].ToString().Split(new char[] { '=', '-' });
                    startbyte = Convert.ToInt64(range[1]);

                    if (range.Length > 2 && range[2] != "")
                    {
                        endbyte = Convert.ToInt64(range[2]);
                    }

                    if (startbyte != 0 || endbyte != fSize - 1 || range.Length > 2 && range[2] == "")
                    {
                        statusCode = 206;
                    }
                }

                long desSize = endbyte - startbyte + 1;
                Response.StatusCode = statusCode;
                //  Response.Headers.Add(ETag = new EntityTagHeaderValue(String.Format("\"{0}\"", _eTag));
                Response.ContentType = "video/" + streamFilter.Container.GetFileExtension();
                Response.Headers.Add("Content-Accept", Response.ContentType);
                Response.Headers.Add("Content-Length", desSize.ToString());
                Response.Headers.Add("Content-Range", string.Format("bytes {0}-{1}/{2}", startbyte, endbyte, fSize));
                return new FileStreamResult(new MemoryStream(SongArray, (int)startbyte, (int)desSize), Response.ContentType);