Stream 视频来自 ZD7EFA19FBE7D3972FD74ADB6024Z223D 中的 Azure Function

Question

我正在尝试编写 Azure HttpTrigger Function ，它将 stream 来自 blob 存储的视频。

目前，我有以下代码，但这不是 stream。

    [FunctionName(nameof(GetVideo))]
    public async Task<IActionResult> Run(
        [HttpTrigger(AuthorizationLevel.Anonymous, "GET", Route = "Video/{videoID}")] HttpRequestMessage request,
        ExecutionContext executionContext,
        ILogger log,
        string videoID)
    {
        IActionResult result;

        string storageConnectionString = _configurationSettings.GetStringValue("StorageConnectionString");

        if (!await _blobStorage.BlobExists(storageConnectionString, "Videos", videoID).ConfigureAwait(false))
        {
            result = new NotFoundResult();
        }
        else
        {
            string videoMimeType = _configurationSettings.GetStringValue("VideoMimeType", DefaultVideoMimeType);

            Stream stream = await _blobStorage.GetBlobAsStream(storageConnectionString, "Videos", videoID).ConfigureAwait(false);
            byte[] videoBytes = GetBytesFromStream(stream);

            result = new FileContentResult(videoBytes, videoMimeType);
        }

        return result;
    }

任何人都可以帮忙吗？

Answer 1

这this does not stream是什么意思，我尝试绑定输入 blob 并返回它，它成功了。 也许你可以参考我下面的代码。

public static async Task<IActionResult> Run(
            [HttpTrigger(AuthorizationLevel.Function, "get", "post", Route = null)] HttpRequest req,
            [Blob("test/test.mp4", FileAccess.Read)] Stream myBlob,
            ILogger log)
        {
            log.LogInformation("C# HTTP trigger function processed a request.");


            byte[] bytes = new byte[myBlob.Length];

            myBlob.Read(bytes, 0, bytes.Length);
            myBlob.Seek(0, SeekOrigin.Begin);

            var result = new FileContentResult(bytes, "video/mp4");
            return result;

        }

更新：假设您想在 HTTP 请求中请求动态 blob 名称，您可以参考以下代码，将容器绑定到获取 blob 并将其下载到字节。 我在 HTTP 请求查询参数中传递了 blob 名称。

public static async Task<IActionResult> Run(
            [HttpTrigger(AuthorizationLevel.Function, "get", "post", Route = null)] HttpRequest req,
            [Blob("test", FileAccess.Read,Connection = "AzureWebJobsStorage")] CloudBlobContainer myBlob,
            ILogger log)
        {
            log.LogInformation("C# HTTP trigger function processed a request.");
            string filename = req.Query["filename"];
            CloudBlockBlob blob= myBlob.GetBlockBlobReference(filename);

            await blob.FetchAttributesAsync();

            long fileByteLength = blob.Properties.Length;
            byte[] fileContent = new byte[fileByteLength];
            for (int i = 0; i < fileByteLength; i++)
            {
                fileContent[i] = 0x20;
            }
            await blob.DownloadToByteArrayAsync(fileContent, 0);

            var result = new FileContentResult(fileContent, "video/mp4");
            return result;

        }

Answer 2

我确实喜欢下面的最小 memory 占用空间（没有将完整的 blob 保存到内存中的字节中）。 Note that instead of binding to stream, I am binding to a ICloudBlob instance (luckily, C# function supports several flavors of blob input binding ) and returning open stream. 使用 memory 分析器对其进行了测试，即使对于大斑点也没有 memory 泄漏。

注意：您不需要寻找 stream position 0 或刷新或处置（处置将在响应结束时自动完成）；

using System.IO;
using System.Threading.Tasks;
using Microsoft.AspNetCore.Mvc;
using Microsoft.Azure.WebJobs;
using Microsoft.Azure.WebJobs.Extensions.Http;
using Microsoft.AspNetCore.Http;
using Microsoft.Extensions.Logging;
using Microsoft.Azure.Storage.Blob;

namespace TestFunction1
{
   public static class MyFunction
   {
        [FunctionName("MyFunction")]
        public static async Task<IActionResult> Run(
            [HttpTrigger(AuthorizationLevel.Anonymous, "get", "post", Route = "video/{fileName}")] HttpRequest req,
            [Blob("test/{fileName}", FileAccess.Read, Connection = "BlobConnection")] ICloudBlob blob,
            ILogger log)
        {
            log.LogInformation("C# HTTP trigger function processed a request.");

            var blobStream = await blob.OpenReadAsync().ConfigureAwait(false);
            return new FileStreamResult(blobStream, "video/mp4");
        }
   }
}