# 4.5 File upload
Suppose you have a website like Instagram, and you want users to upload their beautiful photos, how you are going to do?
You have to add property `enctype` to the form that you want to use for uploading photos, and there are three possibilities for its value:
application/x-www-form-urlencoded Trans-coding all characters before upload(default).
multipart/form-data No trans-coding, you have to use this value when your form have file upload controls.
text/plain Convert spaces to "+", but no trans-coding for special characters.
Therefore, HTML content of a file upload form should look like this:
Upload file
We need to add a function in server side to handle this affair.
http.HandleFunc("/upload", upload)
// upload logic
func upload(w http.ResponseWriter, r *http.Request) {
fmt.Println("method:", r.Method)
if r.Method == "GET" {
crutime := time.Now().Unix()
h := md5.New()
io.WriteString(h, strconv.FormatInt(crutime, 10))
token := fmt.Sprintf("%x", h.Sum(nil))
t, _ := template.ParseFiles("upload.gtpl")
t.Execute(w, token)
} else {
r.ParseMultipartForm(32 << 20)
file, handler, err := r.FormFile("uploadfile")
if err != nil {
fmt.Println(err)
return
}
defer file.Close()
fmt.Fprintf(w, "%v", handler.Header)
f, err := os.OpenFile("./test/"+handler.Filename, os.O_WRONLY|os.O_CREATE, 0666)
if err != nil {
fmt.Println(err)
return
}
defer f.Close()
io.Copy(f, file)
}
}
As you can see, we need to call `r.ParseMultipartForm` for uploading files, the argument means the `maxMemory`. After you called `ParseMultipartForm`, file will be saved in your server memory with `maxMemory` size, if the file size is larger than `maxMemory`, rest of data will be saved in system temporary file. You can use `r.FormFile` to get file handle and use `io.Copy` to save to your file system.
You don't need to call `r.ParseForm` when you access other non-file fields in the form because Go will call it when it's necessary. Also call `ParseMultipartForm` once is enough, and no differences for multiple calls.
We use three steps for uploading files as follows:
1. Add `enctype="multipart/form-data"` to your form.
2. Call `r.ParseMultipartForm` in server side to save file in memory or temporary file.
3. Call `r.FormFile` to get file handle and save to file system.
The file handler is the `multipart.FileHeader`, it uses following struct:
type FileHeader struct {
Filename string
Header textproto.MIMEHeader
// contains filtered or unexported fields
}
Figure 4.5 Print information in server after received file.
## Clients upload files
I showed the example of using form to upload file, and we can impersonate a client form to upload file in Go as well.
package main
import (
"bytes"
"fmt"
"io"
"io/ioutil"
"mime/multipart"
"net/http"
"os"
)
func postFile(filename string, targetUrl string) error {
bodyBuf := &bytes.Buffer{}
bodyWriter := multipart.NewWriter(bodyBuf)
// this step is very important
fileWriter, err := bodyWriter.CreateFormFile("uploadfile", 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 {
return err
}
contentType := bodyWriter.FormDataContentType()
bodyWriter.Close()
resp, err := http.Post(targetUrl, contentType, bodyBuf)
if err != nil {
return err
}
defer resp.Body.Close()
resp_body, err := ioutil.ReadAll(resp.Body)
if err != nil {
return err
}
fmt.Println(resp.Status)
fmt.Println(string(resp_body))
return nil
}
// sample usage
func main() {
target_url := "http://localhost:9090/upload"
filename := "./astaxie.pdf"
postFile(filename, target_url)
}
The above example shows you how to use client to upload file, it uses `multipart.Write` to write file in cache and sends to server through POST method.
If you have other field need to write into data like user name, call `multipart.WriteField` as needed.
## Links
- [Directory](preface.md)
- Previous section: [Duplicate submissions](04.4.md)
- Next section: [Summary](04.6.md)