dpgu5074
2017-08-07 07:40
浏览 492
已采纳

GoLang通过POST请求发送文件

I am new in GoLang language, and I want to create REST API WebServer for file uploading...

So I am stuck in main function (file uploading) via POST request to my server...

I have this line for calling upload function

router.POST("/upload", UploadFile)

and this is my upload function:

func UploadFile( w http.ResponseWriter, r *http.Request, _ httprouter.Params ) {
    io.WriteString(w, "Upload files
")
    postFile( r.Form.Get("file"), "/uploads" )
}

func postFile(filename string, targetUrl string) error {
    bodyBuf := &bytes.Buffer{}
    bodyWriter := multipart.NewWriter(bodyBuf)

    // this step is very important
    fileWriter, err := bodyWriter.CreateFormFile("file", filename)
    if err != nil {
        fmt.Println("error writing to buffer")
        return err
    }

    // open file handle
    fh, err := os.Open(filename)
    if err != nil {
        fmt.Println("error opening file")
        return err
    }

    //iocopy
    _, err = io.Copy(fileWriter, fh)
    if err != nil {
        panic(err)
    }

    bodyWriter.FormDataContentType()
    bodyWriter.Close()

    return err

}

but I can't see any uploaded files in my /upload/ directory...

So what am I doing wrong?

P.S I am getting second error => error opening file, so I think something wrong in file uploading or getting file from UploadFile function, am I right? If yes, than how I can teancfer or get file from this function to postFile function?

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2条回答 默认 最新

  • dongzhi1949 2017-08-07 07:45
    已采纳

    os.Open will open a file, since the file doesn't exist you will get an error. Use os.Create instead it will create a new file and open it. (ref: https://golang.org/pkg/os/#Open)

    func Open

    func Open(name string) (*File, error)

    Open opens the named file for reading. If successful, methods on the returned file can be used for reading; the associated file descriptor has mode O_RDONLY. If there is an error, it will be of type *PathError.

    func Create

    func Create(name string) (*File, error)

    Create creates the named file with mode 0666 (before umask), truncating it if it already exists. If successful, methods on the returned File can be used for I/O; the associated file descriptor has mode O_RDWR. If there is an error, it will be of type *PathError.

    EDIT

    Made a new handler as an example: And also using OpenFile as mentioned by: GoLang send file via POST request

    func Upload(w http.ResponseWriter, r *http.Request) {
        io.WriteString(w, "Upload files
    ")
    
        file, handler, err := r.FormFile("file")
        if err != nil {
            panic(err) //dont do this
        }
        defer file.Close()
    
        // copy example
        f, err := os.OpenFile(handler.Filename, os.O_WRONLY|os.O_CREATE, 0666)
        if err != nil {
            panic(err) //please dont
        }
        defer f.Close()
        io.Copy(f, file)
    
    }
    
    已采纳该答案
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  • drz5553 2017-08-07 08:22

    The multipart.Writer generates multipart messages, this is not something you want to use for receiving a file from a client and saving it to disk.

    Assuming you're uploading the file from a client, e.g. a browser, with Content-Type: application/x-www-form-urlencoded you should use FormFile instead of r.Form.Get which returns a *multipart.File value that contains the content of the file the client sent and which you can use to write that content to disk with io.Copy or what not.

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