GoLang send file via POST request - rest

Question

I am new in GoLang language, and I want to create REST API WebServer for file uploading...
So I am stuck in main function (file uploading) via POST request to my server...
I have this line for calling upload function
router.POST("/upload", UploadFile)
and this is my upload function:
func UploadFile( w http.ResponseWriter, r *http.Request, _ httprouter.Params ) {
io.WriteString(w, "Upload files\n")
postFile( r.Form.Get("file"), "/uploads" )
}
func postFile(filename string, targetUrl string) error {
bodyBuf := &bytes.Buffer{}
bodyWriter := multipart.NewWriter(bodyBuf)
// this step is very important
fileWriter, err := bodyWriter.CreateFormFile("file", filename)
if err != nil {
fmt.Println("error writing to buffer")
return err
}
// open file handle
fh, err := os.Open(filename)
if err != nil {
fmt.Println("error opening file")
return err
}
//iocopy
_, err = io.Copy(fileWriter, fh)
if err != nil {
panic(err)
}
bodyWriter.FormDataContentType()
bodyWriter.Close()
return err
}
but I can't see any uploaded files in my /upload/ directory...
So what am I doing wrong?
P.S I am getting second error => error opening file, so I think something wrong in file uploading or getting file from UploadFile function, am I right? If yes, than how I can teancfer or get file from this function to postFile function?

Answer 1

The multipart.Writer generates multipart messages, this is not something you want to use for receiving a file from a client and saving it to disk.
Assuming you're uploading the file from a client, e.g. a browser, with Content-Type: application/x-www-form-urlencoded you should use FormFile instead of r.Form.Get which returns a *multipart.File value that contains the content of the file the client sent and which you can use to write that content to disk with io.Copy or what not.

Answer 2

os.Open will open a file, since the file doesn't exist you will get an error.
Use os.Create instead it will create a new file and open it. (ref: https://golang.org/pkg/os/#Open)
func Open
func Open(name string) (*File, error)
Open opens the named file for
reading. If successful, methods on the returned file can be used for
reading; the associated file descriptor has mode O_RDONLY. If there is
an error, it will be of type *PathError.
func Create
func Create(name string) (*File, error)
Create creates the named file with mode 0666 (before umask),
truncating it if it already exists. If successful, methods on the
returned File can be used for I/O; the associated file descriptor has
mode O_RDWR. If there is an error, it will be of type *PathError.
EDIT
Made a new handler as an example:
And also using OpenFile as mentioned by: GoLang send file via POST request
func Upload(w http.ResponseWriter, r *http.Request) {
io.WriteString(w, "Upload files\n")
file, handler, err := r.FormFile("file")
if err != nil {
panic(err) //dont do this
}
defer file.Close()
// copy example
f, err := os.OpenFile(handler.Filename, os.O_WRONLY|os.O_CREATE, 0666)
if err != nil {
panic(err) //please dont
}
defer f.Close()
io.Copy(f, file)
}