Question

我需要从WinRT应用创建POST请求，该应用应包含StorageFile。我需要完全按照这样的方式执行此操作：在文件内部发布请求。可能吗？我知道HttpClient.PostAsync(..)，但我不能将StorageFile放在请求正文中。我想将mp3文件发送到Web Api

在服务器端，我得到这样的文件：

[System.Web.Http.HttpPost]
        public HttpResponseMessage UploadRecord([FromUri]string filename)
        {
            HttpResponseMessage result = null;
            var httpRequest = HttpContext.Current.Request;
            if (httpRequest.Files.Count > 0)
            {
                foreach (string file in httpRequest.Files)
                {
                    var postedFile = httpRequest.Files[file];
                    var filePath = HttpContext.Current.Server.MapPath("~/Audio/" + filename + ".mp3");
                    postedFile.SaveAs(filePath);
                }
                result = Request.CreateResponse(HttpStatusCode.Created);
            }
            else
            {
                result = Request.CreateResponse(HttpStatusCode.BadRequest);
            }

            return result;
        }

Answer 1

您可以使用byte[]类作为第二个参数将其作为ByteArrayContent发送：

StroageFile file = // Get file here..
byte[] fileBytes = null;
using (IRandomAccessStreamWithContentType stream = await file.OpenReadAsync())
{
    fileBytes = new byte[stream.Size];
    using (DataReader reader = new DataReader(stream))
    {
        await reader.LoadAsync((uint)stream.Size);
        reader.ReadBytes(fileBytes);
    }
}

var httpClient = new HttpClient();
var byteArrayContent = new ByteArrayContent(fileBytes);

await httpClient.PostAsync(address, fileBytes);

Answer 2

如果您要上传任意大小的文件，那么最好使用后台传输API，以便在应用暂停时上传不会暂停。具体请参见直接接收StorageFile的BackgroundUploader.CreateUpload。有关此关系的客户端和服务器端，请参阅Background Transfer sample，因为该示例还包括示例服务器。

Answer 3

要使用更少的内存，您可以直接将文件流通过管道传输到 HttpClient 流。

    public async Task UploadBinaryAsync(Uri uri)
    {
        var openPicker = new FileOpenPicker();
        StorageFile file = await openPicker.PickSingleFileAsync();
        if (file == null)
            return;
        using (IRandomAccessStreamWithContentType fileStream = await file.OpenReadAsync())
        using (var client = new HttpClient())
        {
            try
            {
                var content = new HttpStreamContent(fileStream);
                content.Headers.ContentType =
                    new HttpMediaTypeHeaderValue("application/octet-stream");
                HttpResponseMessage response = await client.PostAsync(uri, content);
                _ = response.EnsureSuccessStatusCode();
            }
            catch (Exception ex)
            {
                // Handle exceptions appropriately
            }
        }
    }

通过HttpClient.PostAsync与BodyFile（WinRT）内的POST请求

3 个答案: