Rjdlee Rjdlee - 5 months ago 109
HTML Question

Ajax Upload File to GoLang Server with Content Type Multipart

I'm trying to upload an audio file to a Golang server using a multipart form. However, Go returns the error:

multipart: NextPart: bufio: buffer full


I believe this indicates there is something not in the multipart format with my Javascript request. This is my Javascript:

function UploadFile(file) {
var xhr = new XMLHttpRequest();


if (file.type == "audio/mpeg" && file.size <= $id("MAX_FILE_SIZE").value) {
// start upload
var boundary = '---------------------------' + Math.floor(Math.random()*32768) + Math.floor(Math.random()*32768) + Math.floor(Math.random()*32768);

xhr.open("POST", $id("upload").action, true);
xhr.setRequestHeader("Content-Type", "multipart/form-data; boundary=" + boundary);
xhr.setRequestHeader("X_FILENAME", file.name);
xhr.send(file);
}
}


And this is my Golang server handler:

func FileHandler(w http.ResponseWriter, r *http.Request) {
var (
status int
err error
)
defer func() {
if nil != err {
http.Error(w, err.Error(), status)
}
}()
// parse request with maximum memory of _24Kilobits
const _24K = (1 << 20) * 24
if err = r.ParseMultipartForm(_24K); nil != err {
fmt.Println(err)
status = http.StatusInternalServerError
return
}
for _, fheaders := range r.MultipartForm.File {
for _, hdr := range fheaders {
// open uploaded
var infile multipart.File
if infile, err = hdr.Open(); nil != err {
status = http.StatusInternalServerError
return
}
// open destination
var outfile *os.File
if outfile, err = os.Create("./uploaded/" + hdr.Filename); nil != err {
status = http.StatusInternalServerError
return
}
// 32K buffer copy
var written int64
if written, err = io.Copy(outfile, infile); nil != err {
status = http.StatusInternalServerError
return
}
w.Write([]byte("uploaded file:" + hdr.Filename + ";length:" + strconv.Itoa(int(written))))
}
}
}


If anyone has any ideas why I'm getting this error, I'd greatly appreciate it.

Answer

After a long, arduous battle with the Ajax request, I got it to send the right information. Here's the code I used:

var xhr = new XMLHttpRequest(),
    boundary=Math.random().toString().substr(2);

var formdata = new FormData();
formdata.append("file", file);

xhr.open("POST", $id("upload").action, true);
//xhr.setRequestHeader("content-type", "multipart/form-data; charset=utf-8; boundary=" + boundary);
xhr.send(formdata);

Note the header is no longer in use and I found you can attach data to formdata much easier than any other method such as this: How to send multipart/form-data form content by ajax (no jquery)?