转到-接受http发布多部分文件

Question

I am trying to figure out how to accept/receive a HTTP Post in Go. I just want to be able to receive a file, grab its mime-type and save the file locally.

I've been searching all day but all I can find is how to send a file to some remote location but none of the examples I find cover receiving it.

Any help would be appreciated.

using Justinas' example and mixing with my existing experiment I've gotten this far but m.Post never seems to be called.

package main

import (
    "fmt"
    "io"
    "net/http"
    "os"
    "github.com/codegangsta/martini"
    "github.com/codegangsta/martini-contrib/render"
)

func main() {

    m := martini.Classic()

    m.Use(render.Renderer(render.Options{
        Directory: "templates", // Specify what path to load the templates from.
        Layout: "layout", // Specify a layout template. Layouts can call {{ yield }} to render the current template.
        Charset: "UTF-8", // Sets encoding for json and html content-types.
    }))


    m.Get("/", func(r render.Render) {
        fmt.Printf("%v
", "g./")
        r.HTML(200, "hello", "world")
    })

    m.Get("/:who", func(args martini.Params, r render.Render) {
        fmt.Printf("%v
", "g./:who")
        r.HTML(200, "hello", args["who"])
    })

    m.Post("/up", func(w http.ResponseWriter, r *http.Request) {
        fmt.Printf("%v
", "p./up")

        file, header, err := r.FormFile("file")
        defer file.Close()

        if err != nil {
            fmt.Fprintln(w, err)
            return
        }

        out, err := os.Create("/tmp/file")
        if err != nil {
            fmt.Fprintf(w, "Failed to open the file for writing")
            return
        }
        defer out.Close()
        _, err = io.Copy(out, file)
        if err != nil {
            fmt.Fprintln(w, err)
        }

        // the header contains useful info, like the original file name
        fmt.Fprintf(w, "File %s uploaded successfully.", header.Filename)
    })

    m.Run()
}

Answer 1

Go's net/http server handles this pretty, using mime/multipart package behind the scenes. You only need to call r.FormFile() on your *http.Request to get a multipart.File back.

Here's a complete example. And the result of uploading a file with curl:

justinas@ubuntu /tmp curl -i -F file=@/tmp/stuff.txt http://127.0.0.1:8080/
HTTP/1.1 100 Continue

HTTP/1.1 200 OK
Date: Tue, 24 Dec 2013 20:56:07 GMT
Content-Length: 37
Content-Type: text/plain; charset=utf-8

File stuff.txt uploaded successfully.%                                                                                              
justinas@ubuntu /tmp cat file
kittens!