154 lines
4.8 KiB
Markdown
154 lines
4.8 KiB
Markdown
# 4.5 File upload
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Suppose you have a website like Instagram, and you want users to upload their beautiful photos, how you are going to do?
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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:
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application/x-www-form-urlencoded Trans-coding all characters before upload(default).
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multipart/form-data No trans-coding, you have to use this value when your form have file upload controls.
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text/plain Convert spaces to "+", but no trans-coding for special characters.
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Therefore, HTML content of a file upload form should look like this:
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<html>
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<head>
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<title>Upload file</title>
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</head>
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<body>
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<form enctype="multipart/form-data" action="http://127.0.0.1:9090/upload" method="post">
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<input type="file" name="uploadfile" />
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<input type="hidden" name="token" value="{{.}}"/>
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<input type="submit" value="upload" />
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</form>
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</body>
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</html>
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We need to add a function in server side to handle this affair.
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http.HandleFunc("/upload", upload)
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// upload logic
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func upload(w http.ResponseWriter, r *http.Request) {
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fmt.Println("method:", r.Method)
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if r.Method == "GET" {
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crutime := time.Now().Unix()
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h := md5.New()
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io.WriteString(h, strconv.FormatInt(crutime, 10))
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token := fmt.Sprintf("%x", h.Sum(nil))
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t, _ := template.ParseFiles("upload.gtpl")
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t.Execute(w, token)
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} else {
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r.ParseMultipartForm(32 << 20)
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file, handler, err := r.FormFile("uploadfile")
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if err != nil {
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fmt.Println(err)
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return
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}
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defer file.Close()
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fmt.Fprintf(w, "%v", handler.Header)
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f, err := os.OpenFile("./test/"+handler.Filename, os.O_WRONLY|os.O_CREATE, 0666)
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if err != nil {
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fmt.Println(err)
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return
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}
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defer f.Close()
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io.Copy(f, file)
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}
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}
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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.
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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.
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We use three steps for uploading files as follows:
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1. Add `enctype="multipart/form-data"` to your form.
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2. Call `r.ParseMultipartForm` in server side to save file in memory or temporary file.
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3. Call `r.FormFile` to get file handle and save to file system.
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The file handler is the `multipart.FileHeader`, it uses following struct:
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type FileHeader struct {
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Filename string
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Header textproto.MIMEHeader
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// contains filtered or unexported fields
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}
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Figure 4.5 Print information in server after received file.
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## Clients upload files
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I showed the example of using form to upload file, and we can impersonate a client form to upload file in Go as well.
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package main
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import (
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"bytes"
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"fmt"
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"io"
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"io/ioutil"
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"mime/multipart"
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"net/http"
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"os"
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)
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func postFile(filename string, targetUrl string) error {
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bodyBuf := &bytes.Buffer{}
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bodyWriter := multipart.NewWriter(bodyBuf)
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// this step is very important
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fileWriter, err := bodyWriter.CreateFormFile("uploadfile", filename)
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if err != nil {
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fmt.Println("error writing to buffer")
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return err
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}
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// open file handle
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fh, err := os.Open(filename)
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if err != nil {
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fmt.Println("error opening file")
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return err
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}
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//iocopy
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_, err = io.Copy(fileWriter, fh)
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if err != nil {
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return err
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}
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contentType := bodyWriter.FormDataContentType()
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bodyWriter.Close()
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resp, err := http.Post(targetUrl, contentType, bodyBuf)
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if err != nil {
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return err
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}
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defer resp.Body.Close()
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resp_body, err := ioutil.ReadAll(resp.Body)
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if err != nil {
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return err
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}
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fmt.Println(resp.Status)
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fmt.Println(string(resp_body))
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return nil
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}
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// sample usage
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func main() {
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target_url := "http://localhost:9090/upload"
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filename := "./astaxie.pdf"
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postFile(filename, target_url)
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}
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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.
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If you have other field need to write into data like user name, call `multipart.WriteField` as needed.
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## Links
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- [Directory](preface.md)
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- Previous section: [Duplicate submissions](04.4.md)
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- Next section: [Summary](04.6.md) |