golang POST数据使用Content-Type multipart / form-data [英] golang POST data using the Content-Type multipart/form-data
问题描述
我尝试使用go将图片从我的电脑上传到网站。通常,我使用一个bash脚本向attur发送文件一个键:
I'm trying to upload images from my computer to a website using go. Usually, I use a bash script who send file a key to the serveur:
curl -F "image"=@"IMAGEFILE" -F "key"="KEY" URL
将此请求转换为我的golang程序。
it's work fine, but I'm trying to convert this request into my golang program.
http://matt.aimonetti.net/posts/2013/07/01/golang-multipart-file-upload-example/
我试过这个链接和许多其他。但对于我尝试的每个代码,服务器的响应是没有图像发送。我不知道为什么。
I tried this link and many other. But for each code that I try, the response for the server is "no image sended". And I've no idea why. If someone know what's happend with the exemple above.
感谢
推荐答案
以下是一些示例代码。
简而言之,您需要使用 mime / multipart
包构建表单。
In short, you'll need to use the mime/multipart
package to build the form.
package sample
import (
"bytes"
"fmt"
"io"
"mime/multipart"
"net/http"
"os"
)
func Upload(url, file string) (err error) {
// Prepare a form that you will submit to that URL.
var b bytes.Buffer
w := multipart.NewWriter(&b)
// Add your image file
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
}
// Add the other fields
if 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 request
client := &http.Client{}
res, err := client.Do(req)
if err != nil {
return
}
// Check the response
if res.StatusCode != http.StatusOK {
err = fmt.Errorf("bad status: %s", res.Status)
}
return
}
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