在Golang中访问上传的文件

huangapple
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英文:

Accessing Uploaded Files in Golang

问题

我在使用Golang上传文件时遇到了问题。我对这门语言非常陌生，尝试了很多次，但在网上找不到任何答案。

我做错了什么？在这段代码中，我从未进入列出已上传文件数量的代码块。

  1. func handler(w http.ResponseWriter, r *http.Request) {
  2.   fmt.Println("handling req...")
  4.   if r.Method == "GET" {
  5.     fmt.Println("GET req...")
  6.   } else {
  7.     // 如果有的话，解析多部分内容
  8.     err := r.ParseMultipartForm(15485760)
  10.     if err == nil {
  11.       form := r.MultipartForm
  12.       if form == nil {
  13.         fmt.Println("no files...")
  14.       } else {
  15.         defer form.RemoveAll()
  16.         // 我从未看到这个实际发生
  17.         fmt.Printf("%d files", len(form.File))
  18.       }
  19.     } else {
  20.       http.Error(w, err.Error(), http.StatusInternalServerError)
  21.       fmt.Println(err.Error())
  22.     }
  23.   }
  25.   // fmt.Fprintf(w, "Hi there, I love %s!", r.URL.Path[1:])
  26.   fmt.Println("leaving...")
  27. }

更新

我已经成功让上述代码工作了，这太棒了。下面的答案展示了如何以异步方式完成，这可能是比我的代码示例更好的示例。

英文:

I'm having issues with accessing files i upload w/ golang. I'm really new to the language and have gone through more than a few attempts-- can't find any answers to this online either.

What am i doing wrong? In this code, i never get to the block where it lists the # of files uploaded.

  1. func handler(w http.ResponseWriter, r *http.Request) {
  2.   fmt.Println("handling req...")
  4.   if r.Method =="GET"{
  5. 	fmt.Println("GET req...")
  7.   } else {
  9. 	//parse the multipart stuff if there
  10. 	err := r.ParseMultipartForm(15485760)
  12. 	//
  13. 	if err == nil{
  14. 		form:=r.MultipartForm
  15. 		if form==nil {
  16. 			fmt.Println("no files...")
  18. 		} else {
  19. 			defer form.RemoveAll()
  20. 			// i never see this actually occur
  21. 			fmt.Printf("%d files",len(form.File))
  22. 		}
  23. 	} else {
  24. 		http.Error(w,err.Error(),http.StatusInternalServerError)
  25. 		fmt.Println(err.Error())
  26. 	}
  27.   }
  29.   //fmt.Fprintf(w, "Hi there, I love %s!", r.URL.Path[1:])
  30.   fmt.Println("leaving...")
  31. }

###Update

I was able to get the above code to work. Which is great. The answer below shows how to do it async, which may be a better code sample than mine.

答案1

得分: 12

答案：下载最新的golang版本。

我之前遇到过这个问题，使用旧版本的golang时出现了问题，我不知道发生了什么，但是使用最新的golang版本就可以正常工作了。=)

以下是我的上传处理程序代码...
完整代码请参考：http://noypi-linux.blogspot.com/2014/07/golang-web-server-basic-operatons-using.html

  1. // 解析请求
  2. const _24K = (1 << 10) * 24
  3. if err = req.ParseMultipartForm(_24K); nil != err {
  4.      status = http.StatusInternalServerError
  5.      return
  6. }
  7. for _, fheaders := range req.MultipartForm.File {
  8.      for _, hdr := range fheaders {
  9.           // 打开上传的文件
  10.           var infile multipart.File
  11.           if infile, err = hdr.Open(); nil != err {
  12.                status = http.StatusInternalServerError
  13.                return
  14.           }
  15.           // 打开目标文件
  16.           var outfile *os.File
  17.           if outfile, err = os.Create("./uploaded/" + hdr.Filename); nil != err {
  18.                status = http.StatusInternalServerError
  19.                return
  20.           }
  21.           // 32K 缓冲区拷贝
  22.           var written int64
  23.           if written, err = io.Copy(outfile, infile); nil != err {
  24.                status = http.StatusInternalServerError
  25.                return
  26.           }
  27.           res.Write([]byte("uploaded file:" + hdr.Filename + ";length:" + strconv.Itoa(int(written))))
  28.      }
  29. }
英文:

Answer Download the latest golang release.

I experienced the problem before, using the old golang versions, I do not know what happened, but with the latest golang its working. =)

My upload handler code below...
Full code at: http://noypi-linux.blogspot.com/2014/07/golang-web-server-basic-operatons-using.html

  1.   // parse request  
  2.   const _24K = (1 &lt;&lt; 10) * 24  
  3.   if err = req.ParseMultipartForm(_24K); nil != err {  
  4.        status = http.StatusInternalServerError  
  5.        return  
  6.   }  
  7.   for _, fheaders := range req.MultipartForm.File {  
  8.        for _, hdr := range fheaders {  
  9.             // open uploaded  
  10.             var infile multipart.File  
  11.             if infile, err = hdr.Open(); nil != err {  
  12.                  status = http.StatusInternalServerError  
  13.                  return  
  14.             }  
  15.             // open destination  
  16.             var outfile *os.File  
  17.             if outfile, err = os.Create(&quot;./uploaded/&quot; + hdr.Filename); nil != err {  
  18.                  status = http.StatusInternalServerError  
  19.                  return  
  20.             }  
  21.             // 32K buffer copy  
  22.             var written int64  
  23.             if written, err = io.Copy(outfile, infile); nil != err {  
  24.                  status = http.StatusInternalServerError  
  25.                  return  
  26.             }  
  27.             res.Write([]byte(&quot;uploaded file:&quot; + hdr.Filename + &quot;;length:&quot; + strconv.Itoa(int(written))))  
  28.        }  
  29.   }

答案2

得分: 5

如果你知道文件上传的密钥，我认为可以简化一下（未经测试）：

  1. infile, header, err := r.FormFile("file")
  2. if err != nil {
  3. 	http.Error(w, "解析上传文件时出错："+err.Error(), http.StatusBadRequest)
  4. 	return
  5. }
  7. // 这样做非常不安全！请不要这样做！
  8. outfile, err := os.Create("./uploaded/" + header.Filename)
  9. if err != nil {
  10. 	http.Error(w, "保存文件时出错："+err.Error(), http.StatusBadRequest)
  11. 	return
  12. }
  14. _, err = io.Copy(outfile, infile)
  15. if err != nil {
  16. 	http.Error(w, "保存文件时出错："+err.Error(), http.StatusBadRequest)
  17. 	return
  18. }
  20. fmt.Fprintln(w, "Ok")
英文:

If you know they key of the file upload you can make it a bit simpler I think (this is not tested):

  1. infile, header, err := r.FormFile(&quot;file&quot;)
  2. if err != nil {
  3. 	http.Error(w, &quot;Error parsing uploaded file: &quot;+err.Error(), http.StatusBadRequest)
  4. 	return
  5. }
  7. // THIS IS VERY INSECURE! DO NOT DO THIS!
  8. outfile, err := os.Create(&quot;./uploaded/&quot; + header.Filename)
  9. if err != nil {
  10. 	http.Error(w, &quot;Error saving file: &quot;+err.Error(), http.StatusBadRequest)
  11. 	return
  12. }
  14. _, err = io.Copy(outfile, infile)
  15. if err != nil {
  16. 	http.Error(w, &quot;Error saving file: &quot;+err.Error(), http.StatusBadRequest)
  17. 	return
  18. }
  20. fmt.Fprintln(w, &quot;Ok&quot;)

huangapple
  • 本文由 发表于 2013年9月5日 23:08:26
  • 转载请务必保留本文链接：https://go.coder-hub.com/18639929.html
  • file-upload
  • go
  • http
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