How to extract files from multipart-form

Question

I'm writing a Go client to create backups via a REST-API. The REST-API Response with a multipart form data to a GET-Request. So the content of the response (type *http.Response) body looks like this:

--1ceb25134a5967272c26c9f3f543e7d26834a5967272c26c9f3f595caf08
Content-Disposition: form-data; name="configuration"; filename="test.gz"
Content-Type: application/x-gzip

...

--1ceb25134a5967272c26c9f3f543e7d26834a5967272c26c9f3f595caf08--

How can I extract the zip file from the response body?

I tried to use the builtin (net/http) methods but these requires an Request struct.

blunder · Accepted Answer · 2020-10-19 22:49:12Z

Use the mime/multipart package. Assuming that resp is the *http.Response, use the following code to iterate through the parts.

contentType := resp.Header.Get("Content-Type")
mediaType, params, err := mime.ParseMediaType(contentType)
if err != nil {
    log.Fatal(err)
}
if strings.HasPrefix(mediaType, "multipart/") {
    mr := multipart.NewReader(resp.Body, params["boundary"])
    for {
        p, err := mr.NextPart()
        if err == io.EOF {
            return
        }
        if err != nil {
            log.Fatal(err)
        }
        // p.FormName() is the name of the element.
        // p.FileName() is the name of the file (if it's a file)
        // p is an io.Reader on the part

        // The following code prints the part for demonstration purposes.
        slurp, err := ioutil.ReadAll(p)
        if err != nil {
            log.Fatal(err)
        }
        fmt.Printf("Part %q, %q: %q\n", p.FormName(), p.FileName(), slurp)
    }
}

The code in the answer handles errors by calling log.Fata. Adjust the error handling to meet the needs of your application.

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