GoLang send file via POST request

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?

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.

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: https://stackoverflow.com/questions/45541656/golang-send-file-via-post-request/45541764?noredirect=1#comment78043634_45541764

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)
	
}