Question

我正在玩Mux和net/http。最近，我试图获得一个带有一个端点的简单服务器来接受文件上传。

这是我到目前为止的代码：

server.go

package main

import (
    "fmt"
    "github.com/gorilla/mux"
    "log"
    "net/http"
)

func main() {
    router := mux.NewRouter()
    router.
        Path("/upload").
        Methods("POST").
        HandlerFunc(UploadCsv)
    fmt.Println("Starting")
    log.Fatal(http.ListenAndServe(":8080", router))
}

endpoint.go

package main

import (
    "fmt"
    "net/http"
)

func UploadFile(w http.ResponseWriter, r *http.Request) {
    err := r.ParseMultipartForm(5 * 1024 * 1024)
    if err != nil {
        panic(err)
    }

    fmt.Println(r.FormValue("fileupload"))
}

我认为我已经将问题缩小到实际从UploadFile内的请求中检索正文。当我运行这个cURL命令时：

curl http://localhost:8080/upload -F "fileupload=@test.txt" -vvv

我得到一个空响应（正如预期;我不打印到ResponseWriter），但我只是在我正在运行服务器的提示符处打印一个新的（空）行，而不是请求机构。

我将文件作为multipart发送（AFAIK，在cURL中使用-F隐含而不是-d），并且cURL的详细输出显示已发送502个字节：

$ curl http://localhost:8080/upload -F "fileupload=@test.txt" -vvv
*   Trying ::1...
* TCP_NODELAY set
* Connected to localhost (::1) port 8080 (#0)
> POST /upload HTTP/1.1
> Host: localhost:8080
> User-Agent: curl/7.51.0
> Accept: */*
> Content-Length: 520
> Expect: 100-continue
> Content-Type: multipart/form-data; boundary=------------------------b578878d86779dc5
> 
< HTTP/1.1 100 Continue
< HTTP/1.1 200 OK
< Date: Fri, 18 Nov 2016 19:01:50 GMT
< Content-Length: 0
< Content-Type: text/plain; charset=utf-8
< 
* Curl_http_done: called premature == 0
* Connection #0 to host localhost left intact

使用Go中的net/http服务器接收作为多部分表单数据上传的文件的正确方法是什么？

Answer 1

这是一个简单的例子

func ReceiveFile(w http.ResponseWriter, r *http.Request) {
    var Buf bytes.Buffer
    // in your case file would be fileupload
    file, header, err := r.FormFile("file")
    if err != nil {
        panic(err)
    }
    defer file.Close()
    name := strings.Split(header.Filename, ".")
    fmt.Printf("File name %s\n", name[0])
    // Copy the file data to my buffer
    io.Copy(&Buf, file)
    // do something with the contents...
    // I normally have a struct defined and unmarshal into a struct, but this will
    // work as an example
    contents := Buf.String()
    fmt.Println(contents)
    // I reset the buffer in case I want to use it again
    // reduces memory allocations in more intense projects
    Buf.Reset()
    // do something else
    // etc write header
    return
}

Answer 2

您应该使用FormFile代替FormValue：

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

// copy example
f, err := os.OpenFile("./downloaded", os.O_WRONLY|os.O_CREATE, 0666)
defer f.Close()
io.Copy(f, file)

Answer 3

这是我编写的帮助我上传文件的功能。您可以在此处查看完整版本。 How to upload files in golang

package helpers
import (
    "net/http"
    "os"
    "io"    
 )
 func FileUpload(r *http.Request) (string, error) {
      //this function returns the filename(to save in database) of the saved file or an error if it occurs

     r.ParseMultipartForm(32 << 20)

    //ParseMultipartForm parses a request body as multipart/form-data

     var file_name string
     var errors error
     file, handler, err := r.FormFile("file")//retrieve the file from form data
     defer file.Close() //close the file when we finish

     if err != nil {
         errors = err

     }
    //this is path which  we want to store the file

    f, err := os.OpenFile("./images/"+handler.Filename, os.O_WRONLY|os.O_CREATE, 0666)
    if err != nil {
        errors = err
    }
    file_name = handler.Filename
    defer f.Close()
    io.Copy(f, file)
    //here we save our file to our path

    return file_name, errors
  }

如何使用Golang net / http服务器接收上传的文件？

3 个答案: