转到：上传http文件

Question

I am trying to upload a file as an attachment from a client written in Go. I am using http multipart CreateFormFile. When I do this, it sends the data in body to the server which server rejects.

file, err := os.Open(*img.Filepath)
if err != nil {
    return err
}

defer file.Close()

body := &bytes.Buffer{}
writer := multipart.NewWriter(body)
part, err := writer.CreateFormFile("asset", filepath.Base(*img.Filepath))
if err != nil {
    return err
}
_, err = io.Copy(part, file)
if err != nil {
    return err
}

contentType := writer.FormDataContentType()
err = writer.Close()
if err != nil {
    return err
}

request, err := http.NewRequest("POST", API_URL, body)
if err != nil {
    return err
}

request.Header.Set("Content-Type", contentType)

values := request.URL.Query()
values.Add("filename", util.GetFileNameWithoutExtension(*img.FileName))
request.URL.RawQuery = values.Encode()
client := &http.Client{}
resp, err := client.Do(request)
if err != nil {
    return err
} else {
    respBody := &bytes.Buffer{}
    _, err = respBody.ReadFrom(resp.Body)
    if err != nil {
        return err
    }
    resp.Body.Close()
}

When I try the same using curl it works curl -X POST -F "asset=@filepath" "http://localhost/api/v1/xyz?filename=filename"

How can I implement key=@val functionality with Go?

Answer 1

Finally I checked the code for CreateFormFile and the Content-Type is hard coded to application/octet-stream with the mime header. I reimplemented the same method on my own and it works.

func CreateFormFile(w *multipart.Writer, fieldname string, filename string)     (io.Writer, error) {
    h := make(textproto.MIMEHeader)
    h.Set("Content-Disposition", fmt.Sprintf(`form-data; name="%s";  filename="%s"`, escapeQuotes(fieldname), escapeQuotes(filename)))
    h.Set("Content-Type", "image/jpeg")
    return w.CreatePart(h)
}