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。
这是到目前为止我得到的代码:
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"))
}
[我认为我已将问题范围缩小到实际上是从UploadFile
中的请求中检索正文。当我运行此cURL命令时:
curl http://localhost:8080/upload -F "[email protected]" -vvv
我得到一个空的响应(按预期;我没有打印到ResponseWriter
),但是我只是在运行服务器的提示而不是请求正文上打印了一个新的(空)行。 。
我将文件作为多部分发送(AFAIK,在cURL中使用-F
而不是-d
隐含,并且cURL的详细输出显示了发送的502个字节:
$ curl http://localhost:8080/upload -F "[email protected]" -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
在Go中使用net/http
服务器接收作为多份表单数据上传的文件的正确方法是什么?
这里是一个简单的例子
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
}
您应该使用FormFile
而不是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)
[我写了一个函数来帮助我上传文件。您可以在此处查看完整版本。 How to upload files in golang
package helpers
import (
"io"
"net/http"
"os"
)
// This function returns the filename(to save in database) of the saved file
// or an error if it occurs
func FileUpload(r *http.Request) (string, error) {
// ParseMultipartForm parses a request body as multipart/form-data
r.ParseMultipartForm(32 << 20)
file, handler, err := r.FormFile("file") // Retrieve the file from form data
defer file.Close() // Close the file when we finish
if err != nil {
return "", 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 {
return "", err
}
// Copy the file to the destination path
io.Copy(f, file)
return handler.Filename, nil
}