@MultipartForm How to get the original file name?

Question

I am using jboss\'s rest-easy multipart provider for importing a file. I read here http://docs.jboss.org/resteasy/docs/1.0.0.GA/userguide/html/Content_Marshalling_Providers.html

Answer 1

You could use @PartFilename but unfortunately this is currently only used for writing forms, not reading forms: RESTEASY-1069.

Till this issue is fixed you could use MultipartFormDataInput as parameter for your resource method.

Answer 2

After looking around a bit for Resteasy examples including this one, it seems like there is no way to retrieve the original filename and extension information when using a POJO class with the @MultipartForm annotation.

The examples I have seen so far retrieve the filename from the Content-Disposition header from the "file" part of the submitted multiparts form data via HTTP POST, which essentially, looks something like:

Content-Disposition: form-data; name="file"; filename="your_file.zip"
Content-Type: application/zip

You will have to update your file upload REST service class to extract this header like this:

@POST
@Path("/upload")
@Consumes("multipart/form-data")
public Response uploadFile(MultipartFormDataInput input) {

  String fileName = "";
  Map<String, List<InputPart>> formParts = input.getFormDataMap();

  List<InputPart> inPart = formParts.get("file"); // "file" should match the name attribute of your HTML file input 
  for (InputPart inputPart : inPart) {
    try {
      // Retrieve headers, read the Content-Disposition header to obtain the original name of the file
      MultivaluedMap<String, String> headers = inputPart.getHeaders();
      String[] contentDispositionHeader = headers.getFirst("Content-Disposition").split(";");
      for (String name : contentDispositionHeader) {
        if ((name.trim().startsWith("filename"))) {
          String[] tmp = name.split("=");
          fileName = tmp[1].trim().replaceAll("\"","");          
        }
      }

      // Handle the body of that part with an InputStream
      InputStream istream = inputPart.getBody(InputStream.class,null);

      /* ..etc.. */
      } 
    catch (IOException e) {
      e.printStackTrace();
    }
  }

  String msgOutput = "Successfully uploaded file " + filename;
  return Response.status(200).entity(msgOutput).build();
}

Hope this helps.

Answer 3

It seems that Isim is right, but there is a workaround.

Create a hidden field in your form and update its value with the selected file's name. When the form is submitted, the filename will be submitted as a @FormParam.

Here is some code you could need (jquery required).

<input id="the-file" type="file" name="file">
<input id="the-filename" type="hidden" name="filename">

<script>
$('#the-file').on('change', function(e) {
    var filename = $(this).val();
    var lastIndex = filename.lastIndexOf('\\');
    if (lastIndex < 0) {
        lastIndex = filename.lastIndexOf('/');
    }
    if (lastIndex >= 0) {
        filename = filename.substring(lastIndex + 1);
    }
    $('#the-filename').val(filename);
});
</script>