golang 上传文件到php

mimi886 · · 1247 次点击 · · 开始浏览

这是一个创建于的文章，其中的信息可能已经有所发展或是发生改变。

GO上传文件给PHP，第一篇文章里面少了个request.Header.Set(“Content-Type”, formcontenttype) 希望能够帮助别人

func upimgAction(imgurl string, url string) {

  path := imgurl
  extraParams := map[string]string{
    "param1":    "1",
    "param2": "2",
    "param3":   "3",
  }
  request, err := newfileUploadRequest(url, extraParams, "Filedata", path)//Filedata文件名称
  if err != nil {
    log.Error(err)
  }
  //fmt.Println(request)

  resp, err := Gclient.Do(request)
  if err != nil {
    log.Error(err)
  } else {
    body := &bytes.Buffer{}
    _, err := body.ReadFrom(resp.Body)
    if err != nil {
      log.Error(err)
    }
    resp.Body.Close()
    //fmt.Println(resp.StatusCode)
    //fmt.Println(resp.Header)
    fmt.Println(body)
  }
}
func newfileUploadRequest(uri string, params map[string]string, paramName, path string) (*http.Request, error) {
  file, err := os.Open(path)
  if err != nil {
    return nil, err
  }
  defer file.Close()

  body := &bytes.Buffer{}
  writer := multipart.NewWriter(body)
  for key, val := range params {
    _ = writer.WriteField(key, val)
  }
  formcontenttype := writer.FormDataContentType()
  part, err := writer.CreateFormFile(paramName, filepath.Base(path))
  if err != nil {
    return nil, err
  }
  _, err = io.Copy(part, file)

  err = writer.Close()
  if err != nil {
    return nil, err
  }
  //fmt.Println(body)
  request, err := http.NewRequest("POST", uri, body)
  request.Header.Set("Content-Type", formcontenttype)
  return request, err
}

参考网站 http://matt.aimonetti.net/posts/2013/07/01/golang-multipart-file-upload-example/

http://stackoverflow.com/questions/20205796/golang-post-data-using-the-content-type-multipart-form-data