HTTP POST multipart/formdata using HttpClient

Question

当我使用httpclient

发布以下代码时

using var formContent = new MultipartFormDataContent("NKdKd9Yk");
using var stream = new MemoryStream();
file.CopyTo(stream);
var fileBytes = stream.ToArray();
formContent.Headers.ContentType.MediaType = "multipart/form-data";
formContent.Add(new StreamContent(stream), "file", fileName);
var response = await httpClient.PostAsync(GetDocumentUpdateRelativeUrl(), formContent);

这类型的文件是IFormfile

在API端，我按如下方式检索文件

var base64str= "";
using (var ms = new MemoryStream())
{
request.file.CopyTo(ms);
var fileBytes = ms.ToArray();
base64str= Convert.ToBase64String(fileBytes);
// act on the Base64 data
}

得到0字节。我的问题是，这种方法有什么问题?

但是如果我使用下面的代码。然后API工作，我得到我发布的内容。

using var formContent = new MultipartFormDataContent("NKdKd9Yk");
using var stream = new MemoryStream();
file.CopyTo(stream);
var fileBytes = stream.ToArray();
formContent.Headers.ContentType.MediaType = "multipart/form-data";
formContent.Add(new StreamContent(new MemoryStream(fileBytes)), "file", fileName);

不同之处在于我如何添加流内容

formContent.Add(new StreamContent(stream), "file", fileName);
vs
formContent.Add(new StreamContent(new MemoryStream(fileBytes)), "file", fileName);

为什么第一种方法不起作用而第二种方法可以?

Answer 1

您需要添加stream.Seek(0, SeekOrigin.Begin);以便跳转回MemoryStream的开头。你也应该使用CopyToAsync

在第二个版本中，您从byte[]数组中获得了一个新的MemoryStream，它无论如何都位于0上。

using var formContent = new MultipartFormDataContent("NKdKd9Yk");
using var stream = new MemoryStream();
await file.CopyToAsync(stream);
stream.Seek(0, SeekOrigin.Begin);
formContent.Headers.ContentType.MediaType = "multipart/form-data";
formContent.Add(new StreamContent(stream), "file", fileName);
using var response = await httpClient.PostAsync(GetDocumentUpdateRelativeUrl(), formContent);

虽然老实说，MemoryStream在这里似乎完全没有必要。直接从file传递aStream。

using var formContent = new MultipartFormDataContent("NKdKd9Yk");
formContent.Headers.ContentType.MediaType = "multipart/form-data";
using var stream = file.OpenReadStream();
formContent.Add(new StreamContent(stream), "file", fileName);
using var response = await httpClient.PostAsync(GetDocumentUpdateRelativeUrl(), formContent);