Filename stripped of prefix in form data - reactjs

Question

I am sendind files from js to my golang server:
for (var value of formData.values()) {
console.log(value);
}
// File {name: 'img/<hash_key>.png', lastModified: 1635043863231, lastModifiedDate: Sat Oct 23 2021 23:51:03 GMT-0300 (Brasilia Standard Time), webkitRelativePath: '', size: 969, …}
// ...
var request = new Request( serverEndpoint, { body: formData, method: "POST", ... })
return fetch(request).then(response => { ... })
In my golang server, I am using the following code to deal with multipart form data from a request to read files
if err := r.ParseMultipartForm(32 << 20); err != nil {
...
}
for _, fileHeader := range r.MultipartForm.File["files"] {
...
}
I expected to read the files in Go with the same filenames, like img/<hash_key>.png
but my server is reading multipart-form to the following struct:
f = {*mime/multipart.Form | 0xc000426090}
├── Value = {map[string][]string}
└── File = {map[string][]*mime/multipart.FileHeader}
├── 0 = files -> len:1, cap:1
│ ├── key = {string} "files"
│ └── value = {[]*mime/multipart.FileHeader} len:1, cap:1
│ └── 0 = {*mime/multipart.FileHeader | 0xc000440000}
│ ├── Filename = {string} "<hash_key>.png" // notice how FileName is missing 'img/' prefix
│ └── ...
└── ...
I am trying to figure out how this is happening and how to prevent this strip prefix as I need this prefix to correctly resolve upload path for my files
Edit:
Closer inspection revealed that my server IS in fact getting the files with the correct name. After calling r.ParseMultipartForm(32 << 20), I get the following in r.Body.src.R.buf:
------WebKitFormBoundary1uanPdXqZeL8IPUH
Content-Disposition: form-data; name="files"; filename="img/upload.svg"
---- notice the img/ prefix
Content-Type: image/svg+xml
<svg height="512pt" viewBox= ...
However, in r.MultipartForm.File["files"][0].FileName, it shows as upload.svg

Answer 1

The directory is removed in in Part.FileName():
// RFC 7578, Section 4.2 requires that if a filename is provided, the
// directory path information must not be used.
return filepath.Base(filename)
Workaround Part.FileName() by parsing the content disposition header directly.
for _, fileHeader := range r.MultipartForm.File["files"] {
_, params, _ := mime.ParseMediaType(fileHeader.Header.Get("Content-Disposition"))
filename := params["filename"]
if filename == "" {
// TODO: Handle unexpected content disposition
// header (missing header, parse error, missing param).
}