Golang multipart/form-data File Upload

file-upload-multipart.go

package main

import (
  "net/http"
  "os"
  "bytes"
  "path"
  "path/filepath"
  "mime/multipart"
  "io"
)

func main() {
  fileDir, _ := os.Getwd()
  fileName := "upload-file.txt"
  filePath := path.Join(fileDir, fileName)

  file, _ := os.Open(filePath)
  defer file.Close()

  body := &bytes.Buffer{}
  writer := multipart.NewWriter(body)
  part, _ := writer.CreateFormFile("file", filepath.Base(file.Name()))
  io.Copy(part, file)
  writer.Close()

  r, _ := http.NewRequest("POST", "http://example.com", body)
  r.Header.Add("Content-Type", writer.FormDataContentType())
  client := &http.Client{}
  client.Do(r)
}

Is it possible to compile this in an executable and then whenever you run that '.exe' the server will start to listen on incoming requests?

Hello, if I need to upload a zip file as form-data in POST request body, I am able to find this,

zipWriter := zip.NewWriter(buf)
zipFile, err := zipWriter.Create(fileName)

Unlike multipart.Writer, in case of zip.Writer I can't find any option to create form file in a key-value fashion.

How can I achieve this for zip as well?

Will this code use sendfile to achieve zero copy?

No It doesn't.
If you want to achieve zero copy, you can use os.pipe instead of sendfile.
sendfile is fit to tcp package so, it is hard to use with http package.
See this code snippet and benchmarks of it.