Example of doing a multipart upload in Go (golang)

multipart_upload.go

package main

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

// Creates a new file upload http request with optional extra params
func newfileUploadRequest(uri string, params map[string]string, paramName, path string) (*http.Request, error) {
	file, err := os.Open(path)
	if err != nil {
		return nil, err
	}
	defer file.Close()

	body := &bytes.Buffer{}
	writer := multipart.NewWriter(body)
	part, err := writer.CreateFormFile(paramName, filepath.Base(path))
	if err != nil {
		return nil, err
	}
	_, err = io.Copy(part, file)

	for key, val := range params {
		_ = writer.WriteField(key, val)
	}
	err = writer.Close()
	if err != nil {
		return nil, err
	}

	req, err := http.NewRequest("POST", uri, body)
	req.Header.Set("Content-Type", writer.FormDataContentType())
	return req, err
}

func main() {
	path, _ := os.Getwd()
	path += "/test.pdf"
	extraParams := map[string]string{
		"title":       "My Document",
		"author":      "Matt Aimonetti",
		"description": "A document with all the Go programming language secrets",
	}
	request, err := newfileUploadRequest("https://google.com/upload", extraParams, "file", "/tmp/doc.pdf")
	if err != nil {
		log.Fatal(err)
	}
	client := &http.Client{}
	resp, err := client.Do(request)
	if err != nil {
		log.Fatal(err)
	} else {
		body := &bytes.Buffer{}
		_, err := body.ReadFrom(resp.Body)
		if err != nil {
			log.Fatal(err)
		}
                resp.Body.Close()
		fmt.Println(resp.StatusCode)
		fmt.Println(resp.Header)

		fmt.Println(body)
	}
}

I found a problem:
When I simulated uploading a file in golang to the spring restful API, I found that the uploaded file was incorrect when I uploaded it using your method. Although it can get the data.
@PostMapping("/demo") public String postdemo(@RequestParam("userName") String userName, @RequestBody byte[] file) throws IOException { System.out.println(userName); if (file != null) { System.out.println(file.length); String jsonFilePath = "/local/code/12312312.zip"; File jsonFile = new File(jsonFilePath); FileOutputStream outputStream = FileUtils.openOutputStream(jsonFile); outputStream.write(file); outputStream.flush(); outputStream.close(); } return "you send post parame is " + userName; }

The header of the generated file will contain the following information

--45e6a34acb7e1b27165a75bf5052f2f4f00fc96dd04899d1bf66bd782fa1 Content-Disposition: form-data; name="file"; filename="pom.zip" Content-Type: application/octet-stream ����

If I use this code "request.Header.Add("Content-Type", writer.FormDataContentType())", then he will prompt me to get the parameters: username

and then just test it:

	rr := httptest.NewRecorder()
	handler := http.HandlerFunc(API.Create)
	handler.ServeHTTP(rr, req)

	if status := rr.Code; status != http.StatusOK {
		t.Errorf("handler returned wrong status code: got %v want %v",
			status, http.StatusOK)
	}

any examples of posting a file along with a serialized struct?

@mkaz save my day

@mkaz thanks

using io.pipe will be better?

server use multipart to response to client
https://gist.github.com/shuaihanhungry/9c65d3d1e866faf6bef08152b078c519

@themihai I'm wondering it too, do you have solution?

My two cents: using file.Stat is an triggers an unnecessary system call. You could use filepath.Base(path) in order to obtain the filename.

great solution @bacongobbler, big thanks

I'm having similar issue as @vicbaily528 where the content type is an octet stream. Any ideas how to get around this?

var bodyContent []byte
resp.Body.Read(bodyContent)
resp.Body.Close()
fmt.Println(bodyContent)

not work,

body, _ := ioutil.ReadAll(res.Body)
fmt.Println(res)
fmt.Println(string(body))

work well

Didn't work at first. Just used wireshark and found there is no content-type in the request.

Thanks for this example.
It helped a lot.
However, if I may, I'd like to contribute to this discussion in the hope that it'll be of use to others and save them time.

As was already pointed out by @ghost when it comes to large files, it's inconvenient to use a buffer, as it consumes a lot of resources.

Luckily, there's a solution. After googling for a while I encountered this article where the idea of in-memory piping is leveraged for multipart uploading.
The advantages of this approach are tremendous in my point of view.
Another example can be found here.
I hope it'll be helpful.
Cheers.

https://github.com/030/go-multipart

Thanks a lot!

I had the same problem as @vicbaily528 and @melissafzhang, after take a look at mime/multipart/writer.go I decide to replace the CreateFormFile function for a custom one:

func MyCreateFormFile(fieldname, filename, contentType string) (io.Writer, error) {
        h := make(textproto.MIMEHeader)
	h.Set("Content-Disposition", fmt.Sprintf(`form-data; name="%s"; filename="%s"`, escapeQuotes(fieldname), escapeQuotes(filename)))
	h.Set("Content-Type", escapeQuotes(contentType))
	return w.CreatePart(h)
}

this worked for me, I hope it can help you or others who have the same problem.

@mattetti, thanks for helpful gist and as @mkaz mentioned, adding Content-Type header works for me

Yet another example:

func uploadFileMultipart(url string, path string) (*http.Response, error) {
	f, err := os.OpenFile(path, os.O_RDONLY, 0644)
	if err != nil {
		return nil, err
	}

	// Reduce number of syscalls when reading from disk.
	bufferedFileReader := bufio.NewReader(f)
	defer f.Close()

	// Create a pipe for writing from the file and reading to
	// the request concurrently.
	bodyReader, bodyWriter := io.Pipe()
	formWriter := multipart.NewWriter(bodyWriter)

	// Store the first write error in writeErr.
	var (
		writeErr error
		errOnce  sync.Once
	)
	setErr := func(err error) {
		if err != nil {
			errOnce.Do(func() { writeErr = err })
		}
	}
	go func() {
		partWriter, err := formWriter.CreateFormFile("file", path)
		setErr(err)
		_, err = io.Copy(partWriter, bufferedFileReader)
		setErr(err)
		setErr(formWriter.Close())
		setErr(bodyWriter.Close())
	}()

	req, err := http.NewRequest(http.MethodPut, url, bodyReader)
	if err != nil {
		return nil, err
	}
	req.Header.Add("Content-Type", formWriter.FormDataContentType())

	// This operation will block until both the formWriter
	// and bodyWriter have been closed by the goroutine,
	// or in the event of a HTTP error.
	resp, err := http.DefaultClient.Do(req)

	if writeErr != nil {
		return nil, writeErr
	}

	return resp, err
}

thank you.

Can I upload two or more files one times?

@donnol - jep, just call the "createFormFile" function multiple times

"https://google.com/upload" not found

Thanks a lot.. This is really useful.

Really useful, it works

@sebnyberg Hello, just wanted to understand about the comment "This operation will block until both the formWriter...". Could you please explain, thank you!

@sebnyberg Hello, just wanted to understand about the comment "This operation will block until both the formWriter...". Could you please explain, thank you!

Sure. The request reads from the provided reader until it returns io.EOF. For an io.Pipe, the reader-end will return io.EOF after the write-end is closed. That's what is meant by "blocking". If the bodyWriter is not closed, the the request will last forever (until conn timeout).

https://pkg.go.dev/io#PipeWriter.Close

Closing the formWriter isn't strictly necessary to send the request. However, the formWriters Close() writes a trailer to the multipart message that is required for the request to be valid.

https://pkg.go.dev/mime/multipart#Writer.Close

@sebnyberg Thank you very much! :)

If uploading specific format like gzip or server expect content other than octet-stream, better to use writer.CreatePart instead of formWriter.CreateFormFile

	part, err := writer.CreatePart(textproto.MIMEHeader{
		"Content-Type": []string{"application/x-gzip"},
		"Content-Disposition": []string{fmt.Sprintf(`form-data; name="%s"; filename="%s"`,
			"file", fi.Name())},
	})