I\'m trying to upload images from my computer to a website using go. Usually, I use a bash script that sends a file and a key to the server:
curl -F \"image
Here's a function I've used that uses io.Pipe()
to avoid reading in the entire file to memory or needing to manage any buffers. It handles only a single file, but could easily be extended to handle more by adding more parts within the goroutine. The happy path works well. The error paths have not hand much testing.
import (
"fmt"
"io"
"mime/multipart"
"net/http"
"os"
)
func UploadMultipartFile(client *http.Client, uri, key, path string) (*http.Response, error) {
body, writer := io.Pipe()
req, err := http.NewRequest(http.MethodPost, uri, body)
if err != nil {
return nil, err
}
mwriter := multipart.NewWriter(writer)
req.Header.Add("Content-Type", mwriter.FormDataContentType())
errchan := make(chan error)
go func() {
defer close(errchan)
defer writer.Close()
defer mwriter.Close()
w, err := mwriter.CreateFormFile(key, path)
if err != nil {
errchan <- err
return
}
in, err := os.Open(path)
if err != nil {
errchan <- err
return
}
defer in.Close()
if written, err := io.Copy(w, in); err != nil {
errchan <- fmt.Errorf("error copying %s (%d bytes written): %v", path, written, err)
return
}
if err := mwriter.Close(); err != nil {
errchan <- err
return
}
}()
resp, err := client.Do(req)
merr := <-errchan
if err != nil || merr != nil {
return resp, fmt.Errorf("http error: %v, multipart error: %v", err, merr)
}
return resp, nil
}