How do I upload a file?

[pengqian089]

Using HttpClient, it should look like this:

private readonly HttpClient _httpClient;
ctor(HttpClient httpClient)
{
         _httpClient = httpClient;
}

async Task UploadAsync()
{
	var request = new HttpRequestMessage(HttpMethod.Put, "/images/2023/temp.jpg");
    // file content, byte[] or Stream
    byte[] buffer = null; // Todo
    Stream stream = null; // Todo

    request.Content = new ByteArrayContent(buffer);
    // or
    request.Content = new StreamContent(stream);

	var response = await _httpClient.SendAsync(request);
}

So how should Refit be used?

[glennawatson]

Use a Multipart method with one of the part wrapper types. Refit handles building the multipart content for you.

public interface IUploadApi
{
    [Multipart]
    [Put("/images/2023/temp.jpg")]
    Task Upload([AliasAs("file")] StreamPart stream);

    // or for a byte[]
    [Multipart]
    [Put("/images/2023/temp.jpg")]
    Task Upload([AliasAs("file")] ByteArrayPart bytes);
}

Call it:

await api.Upload(new StreamPart(stream, "temp.jpg", "image/jpeg"));
// or
await api.Upload(new ByteArrayPart(buffer, "temp.jpg", "image/jpeg"));

If you do NOT want multipart and instead want the raw body to be the file bytes/stream (matching your ByteArrayContent / StreamContent example), pass a Stream or byte[] as the body without [Multipart]:

[Put("/images/2023/temp.jpg")]
Task Upload([Body] Stream content);

• A Stream body is sent via StreamContent; a byte[] body via ByteArrayContent.
• Wrapper types available for multipart parts: StreamPart, ByteArrayPart, FileInfoPart.

See the README "Multipart uploads" section for the full list and options.