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.
func handler(w http.ResponseWriter, r *http.Request) {
fmt.Println("handling req...")
if r.Method =="GET"{
fmt.Println("GET req...")
} else {
//parse the multipart stuff if there
err := r.ParseMultipartForm(15485760)
//
if err == nil{
form:=r.MultipartForm
if form==nil {
fmt.Println("no files...")
} else {
defer form.RemoveAll()
// i never see this actually occur
fmt.Printf("%d files",len(form.File))
}
} else {
http.Error(w,err.Error(),http.StatusInternalServerError)
fmt.Println(err.Error())
}
}
//fmt.Fprintf(w, "Hi there, I love %s!", r.URL.Path[1:])
fmt.Println("leaving...")
}
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.
Go to localhost:8080/upload, and you will see a form to upload a file. After selecting a file and clicking upload, the file should be created in your local filesystem.
MaxBytesReader() method is used to limit the size of incoming request bodies. For single file uploads, limiting the size of the request body provides a good approximation of limiting the file size. The ParseMultipartForm() method subsequently parses the request body as multipart/form-data up to the max memory argument.
2) Create Golang API as an Uploader Create client uploader instance with pre-defined bucket name, project id (your GCP project ID), and upload path (we will use “test-files/”) Create a POST API (path: “/upload”) that receives “file_input” Form File. Trigger upload file to GCP with concatenated upload path as the prefix.
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
// parse request
const _24K = (1 << 10) * 24
if err = req.ParseMultipartForm(_24K); nil != err {
status = http.StatusInternalServerError
return
}
for _, fheaders := range req.MultipartForm.File {
for _, hdr := range fheaders {
// open uploaded
var infile multipart.File
if infile, err = hdr.Open(); nil != err {
status = http.StatusInternalServerError
return
}
// open destination
var outfile *os.File
if outfile, err = os.Create("./uploaded/" + hdr.Filename); nil != err {
status = http.StatusInternalServerError
return
}
// 32K buffer copy
var written int64
if written, err = io.Copy(outfile, infile); nil != err {
status = http.StatusInternalServerError
return
}
res.Write([]byte("uploaded file:" + hdr.Filename + ";length:" + strconv.Itoa(int(written))))
}
}
If you know they key of the file upload you can make it a bit simpler I think (this is not tested):
infile, header, err := r.FormFile("file")
if err != nil {
http.Error(w, "Error parsing uploaded file: "+err.Error(), http.StatusBadRequest)
return
}
// THIS IS VERY INSECURE! DO NOT DO THIS!
outfile, err := os.Create("./uploaded/" + header.Filename)
if err != nil {
http.Error(w, "Error saving file: "+err.Error(), http.StatusBadRequest)
return
}
_, err = io.Copy(outfile, infile)
if err != nil {
http.Error(w, "Error saving file: "+err.Error(), http.StatusBadRequest)
return
}
fmt.Fprintln(w, "Ok")
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With