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我试着用 go 将图像从我的电脑上传到网站。通常,我用 bash 脚本向服务器发送文件和密钥: l- }9 i% v" E& x
curl -F "image"=@"IMAGEFILE" -F "key"="KEY" URL
' K3 J8 Q* I1 @2 _9 K% Z) j5 p- j 它工作正常,但我试图把这个请求转换成我的 golang 程序。
# K% V8 X" q: p! p4 J1 R' C* Chttp://matt.aimonetti.net/posts/2013/07/01/golang-multipart-file-upload-example/+ V' ^2 M: D0 ^# _2 N4 _4 ?7 `% ]9 {5 b
我尝试了这个链接和许多其他链接,但服务器对我尝试的每个代码的响应是没有图像,我不知道为什么。如果有人知道上面的例子发生了什么。7 n8 ~, V& O% o+ H' u
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解决方案:
' o o& o5 V3 s# t4 B) G. d0 R 这是一些示例代码。
0 ?' k- k& `8 X+ S- X+ z# {简而言之,你需要使用它mime/multipart包来构建表单。) c4 }, E% K2 t3 l
package mainimport "bytes" "fmt" "io" "mime/multipart" "net/http" "net/http/httptest" "net/http/httputil" "os" "strings")func main() var client *http.Client var remoteURL string setup a mocked http client. ts := httptest.NewTLSServer(http.HandlerFunc(func(w http.ResponseWriter,r *http.Request) b,err := httputil.DumpRequest(r,true) if err != nil panic(err) fmt.Printf("%s",b) )defer ts.Close() client = ts.Client() remoteURL = ts.URL } //prepare the reader instances to encode values := map[string]io.Reader{ "file": mustOpen("main.go"),// lets assume its this file "other": strings.NewReader("hello world!"), err := Upload(client,remoteURL,values) if err != nil panic(err) func Upload(client *http.Client,url string,values map[string]io.Reader) (err error) / Prepare a form that you will submit to that URL. var b bytes.Buffer w := multipart.NewWriter(&b) for key,r := range values var fw io.Writer if x,ok := r.(io.Closer); ok defer x.Close() Add an image file if x,ok := r.(*os.File); ok if fw,err = w.CreateFormFile(key,x.Name()); err != nil return else // Add other fields if fw,err = w.CreateFormField(key); err != nil return if _,err = io.Copy(fw,r); err != nil return err 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(" OST",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 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}func mustOpen(f string) *os.File r,err := os.Open(f) if err != nil panic(err) } return r}# \, o% W- y. l4 e( j* n
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