I know how to upload a single file but now I'm trying to upload multiple files in the same Go function.
Here is my code:
func PhotoCreatePOST(w http.ResponseWriter, r *http.Request) {
var err error
r.ParseMultipartForm(32 << 20) // 32MB is the default used by FormFile
fhs := r.MultipartForm.File["files"]
var fileNames []string
var filename string
var ext string
for _, file := range fhs { //Iterate over multiple uploaded files
if err != nil {
log.Fatal(err)
} else {
dir, err := os.Getwd()
if err != nil {
log.Fatal(err)
}
ext = strings.ToLower(path.Ext(file.Filename))
filename = path.Join(random.RandString(10) + ext)
destFolder := "/media/photos"
if _, err := os.Stat(destFolder); os.IsNotExist(err) {
os.Mkdir(destFolder, 0755)
}
//destination Path. The string which is Saving in DB
savePath := destFolder + "/" + filename
err = ioutil.WriteFile(savePath, file, 0777) //<--Here is the problem
if err != nil {
log.Println(err)
io.WriteString(w, err.Error())
return
}
}
}
//Add file url to the slice
fileNames = append(fileNames, filename)
}
But I get this error:
cannot use file (type *multipart.FileHeader) as type []byte in argument to ioutil.WriteFile
I've tried to read the file into bytes like this:
b, err := ioutil.ReadFile(file)
if err != nil {
fmt.Print(err)
}
And save b instead of file.
But then I get another error that
can not read *multipart.Fileheader as string
How can I fix this?
Alright, thanks to this great blog post, I found my answer and here it the complete solution for the record:
// PhotoCreatePOST saves multiple photo uploads
func PhotoCreatePOST(w http.ResponseWriter, r *http.Request) {
err := r.ParseMultipartForm(100000)
if err != nil {
fmt.Println("error parsing multiplepart form", err)
return
}
files := r.MultipartForm.File["files"]
for i, _ := range files { //Iterate over multiple uploaded files
file, err := files[i].Open()
defer file.Close()
if err != nil {
fmt.Println("error opening file ", err)
return
}
ext := path.Ext(files[i].Filename)
//TODO: Verify extension is valid
filename := GetRandomString(10) + ext
//create destination file making sure the path is writeable.
dst, err := os.Create("media/photos/" + filename)
defer dst.Close()
if err != nil {
fmt.Println("error creating destination ", err)
return
}
//copy the uploaded file to the destination file
if _, err := io.Copy(dst, file); err != nil {
fmt.Println("error copying file", err)
return
}
fmt.Println("Image upload success: ", files[i].Filename)
}
fmt.Println("all are uploaded")
PhotoCreateGET(w, r)
return
}
//Generate random filename (never trust user input!)
var letterRunes = []rune("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ123456789")
// GetString returns a random string
func GetRandomString(n int) string {
rand.Seed(time.Now().UnixNano())
b := make([]rune, n)
for i := range b {
b[i] = letterRunes[rand.Intn(len(letterRunes))]
}
return string(b)
}
And the upload form:
<form method="post" action="/create/photo" enctype="multipart/form-data">
<input type="file" name="files" multiple>
<input type="hidden" name="token" value="{{.token}}">
<button title="submit" type="submit" >Save </button>
</form>
Voila. Hopefully it can save some time for others.
This is roughly how you do it.
fileHeaders := r.MultipartForm.File["files"]
var fileNames []string
for _, fileHeader := range fileHeaders {
file, err := fileHeader.Open()
if err != nil {
return err
}
defer file.Close()
// Generate the destination filename randomly. Using even the passed
// extension is a security vulnerability unless you have a whitelist.
dest, err := ioutil.TempFile("/media/photos", "")
if err != nil {
return err
}
defer dest.Close()
if _, err := io.Copy(dest, file); err != nil {
return err
}
fileNames = append(fileNames, dest.Name())
}