Krishna Chaitanya Krishna Chaitanya - 9 days ago 8
Java Question

@MultipartForm How to get the original file name?

I am using jboss's rest-easy multipart provider for importing a file. I read here regarding @MultipartForm because I can exactly map it with my POJO.

Below is my POJO

public class SoftwarePackageForm {

private File file;

private String contentDisposition;

public File getFile() {
return file;

public void setFile(File file) {
this.file = file;

public String getContentDisposition() {
return contentDisposition;

public void setContentDisposition(String contentDisposition) {
this.contentDisposition = contentDisposition;

Then I got the file object and printed its absolute path and it returned a file name of type file. The extension and uploaded file name are lost. My client is trying to upload a archive file(zip,tar,z)

I need this information at the server side so that I can apply the un-archive program properly.

The original file name is sent to the server in content-disposition header.

How can I get this information? Or atleast how can I say jboss to save the file with the uploaded file name and extension? Is it configurable from my application?


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=""
Content-Type: application/zip

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

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) {

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

Hope this helps.