英文:
How can you upload files as a []byte in go?
问题
我想使用golang进行POST请求并上传图片，但我不想传递文件路径，只想传递[]byte类型的数据。
以下是我找到的一篇文章，但不符合我的需求，因为它使用了os.Open函数：
https://stackoverflow.com/questions/20205796/golang-post-data-using-the-content-type-multipart-form-data
func Upload(url, file string) (err error) {// 准备一个表单，将要提交到指定的URL。var b bytes.Bufferw := multipart.NewWriter(&b)// 添加图片文件f, err := os.Open(file)if err != nil {return}defer f.Close()fw, err := w.CreateFormFile("image", file)if err != nil {return}if _, err = io.Copy(fw, f); err != nil {return}// 添加其他字段if fw, err = w.CreateFormField("key"); err != nil {return}if _, err = fw.Write([]byte("KEY")); err != nil {return}// 不要忘记关闭multipart writer。// 如果不关闭，请求将缺少终止边界。w.Close()// 现在你有了一个表单，可以将其提交给处理程序。req, err := http.NewRequest("POST", url, &b)if err != nil {return}// 不要忘记设置Content-Type，它将包含边界信息。req.Header.Set("Content-Type", w.FormDataContentType())// 提交请求client := &http.Client{}res, err := client.Do(req)if err != nil {return}// 检查响应if res.StatusCode != http.StatusOK {err = fmt.Errorf("bad status: %s", res.Status)}return}
希望这可以帮到你！
英文:
I would like to use golang post request, upload pictures, but I do not want to pass filepath, just want to pass [] byte
The following article are not what I need because they are used os.Open
https://stackoverflow.com/questions/20205796/golang-post-data-using-the-content-type-multipart-form-data
func Upload(url, file string) (err error) {// Prepare a form that you will submit to that URL.var b bytes.Bufferw := multipart.NewWriter(&b)// Add your image filef, err := os.Open(file)if err != nil {return}defer f.Close()fw, err := w.CreateFormFile("image", file)if err != nil {return}if _, err = io.Copy(fw, f); err != nil {return}// Add the other fieldsif fw, err = w.CreateFormField("key"); err != nil {return}if _, err = fw.Write([]byte("KEY")); err != nil {return}// Don't forget to close the multipart writer.// If you don't close it, your request will be missing the terminating boundary.w.Close()// Now that you have a form, you can submit it to your handler.req, err := http.NewRequest("POST", url, &b)if err != nil {return}// Don't forget to set the content type, this will contain the boundary.req.Header.Set("Content-Type", w.FormDataContentType())// Submit the requestclient := &http.Client{}res, err := client.Do(req)if err != nil {return}// Check the responseif res.StatusCode != http.StatusOK {err = fmt.Errorf("bad status: %s", res.Status)}return}
答案1
得分: 1
由于您使用了以下代码：
if _, err = io.Copy(fw, f); err != nil {return}
您可以对代码进行如下修改：
- 添加新的导入：
"bytes" - 将方法签名修改为
func Upload(url string, file []byte) (err error) - 使用
io.Copy(fw, bytes.NewReader(f))
英文:
Since you use
if _, err = io.Copy(fw, f); err != nil {return}
You may as well edit your code to:
- Add new import:
"bytes" - Change the method signature to
func Upload(url string, file []byte) (err error) - Use
io.Copy(fw, bytes.NewReader(f))
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