Merging other languages
This commit is contained in:
James Miranda
310
en/04.5.md
310
en/04.5.md
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# 4.5 处理文件上传
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你想处理一个由用户上传的文件，比如你正在建设一个类似Instagram的网站，你需要存储用户拍摄的照片。这种需求该如何实现呢？
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要使表单能够上传文件，首先第一步就是要添加form的`enctype`属性，`enctype`属性有如下三种情况:
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application/x-www-form-urlencoded 表示在发送前编码所有字符（默认）
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multipart/form-data 不对字符编码。在使用包含文件上传控件的表单时，必须使用该值。
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text/plain 空格转换为 "+" 加号，但不对特殊字符编码。
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所以，创建新的表单html文件, 命名为upload.gtpl, html代码应该类似于:
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<html>
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<head>
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<title>上传文件</title>
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</head>
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<body>
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<form enctype="multipart/form-data" action="/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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在服务器端，我们增加一个handlerFunc:
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http.HandleFunc("/upload", upload)
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// 处理/upload 逻辑
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func upload(w http.ResponseWriter, r *http.Request) {
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fmt.Println("method:", r.Method) //获取请求的方法
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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) // 此处假设当前目录下已存在test目录
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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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通过上面的代码可以看到，处理文件上传我们需要调用`r.ParseMultipartForm`，里面的参数表示`maxMemory`，调用`ParseMultipartForm`之后，上传的文件存储在`maxMemory`大小的内存里面，如果文件大小超过了`maxMemory`，那么剩下的部分将存储在系统的临时文件中。我们可以通过`r.FormFile`获取上面的文件句柄，然后实例中使用了`io.Copy`来存储文件。
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>获取其他非文件字段信息的时候就不需要调用`r.ParseForm`，因为在需要的时候Go自动会去调用。而且`ParseMultipartForm`调用一次之后，后面再次调用不会再有效果。
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通过上面的实例我们可以看到我们上传文件主要三步处理：
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1. 表单中增加enctype="multipart/form-data"
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2. 服务端调用`r.ParseMultipartForm`,把上传的文件存储在内存和临时文件中
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3. 使用`r.FormFile`获取文件句柄，然后对文件进行存储等处理。
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文件handler是multipart.FileHeader,里面存储了如下结构信息
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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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我们通过上面的实例代码打印出来上传文件的信息如下
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图4.5 打印文件上传后服务器端接受的信息
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## 客户端上传文件
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我们上面的例子演示了如何通过表单上传文件，然后在服务器端处理文件，其实Go支持模拟客户端表单功能支持文件上传，详细用法请看如下示例：
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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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//关键的一步操作
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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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//打开文件句柄操作
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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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defer fh.Close()
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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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上面的例子详细展示了客户端如何向服务器上传一个文件的例子，客户端通过multipart.Write把文件的文本流写入一个缓存中，然后调用http的Post方法把缓存传到服务器。
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>如果你还有其他普通字段例如username之类的需要同时写入，那么可以调用multipart的WriteField方法写很多其他类似的字段。
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## links
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* [目录](<preface.md>)
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* 上一节: [防止多次递交表单](<04.4.md>)
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* 下一节: [小结](<04.6.md>)
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# 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 would you implement that functionality?
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You have to add property `enctype` to the form that you want to use for uploading photos. There are three possible values for this property:
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application/x-www-form-urlencoded Transcode all characters before uploading (default).
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multipart/form-data No transcoding. You must use this value when your form has file upload controls.
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text/plain Convert spaces to "+", but no transcoding for special characters.
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Therefore, the 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 on the server side to handle this form.
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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 function `maxMemory` argument. After you call `ParseMultipartForm`, the file will be saved in the server memory with `maxMemory` size. If the file size is larger than `maxMemory`, the rest of the data will be saved in a system temporary file. You can use `r.FormFile` to get the 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, calling `ParseMultipartForm` once is enough -multiple calls make no difference.
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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` on the server side to save the file either to memory or to a temporary file.
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3. Call `r.FormFile` to get the file handle and save to the file system.
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The file handler is the `multipart.FileHeader`. It uses the 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 on server after receiving file.
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## Clients upload files
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I showed an example of using a form to a upload a file. We can impersonate a client form to upload files 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 a client to upload files. It uses `multipart.Write` to write files into cache and sends them to the server through the POST method.
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If you have other fields that need to write into data, like username, 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)
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