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Home/ Questions/Q 746379
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T14:05:03+00:00 2026-05-14T14:05:03+00:00

I’m trying to upload a file using JavaFX using the HttpRequest. For this purpose

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I’m trying to upload a file using JavaFX using the HttpRequest. For this purpose I have written the following function.

function uploadFile(inputFile : File) : Void {
    // check file
    if (inputFile == null or not(inputFile.exists()) or inputFile.isDirectory()) {
        return;
    }    
    def httpRequest : HttpRequest = HttpRequest {
        location: urlConverter.encodeURL("{serverUrl}");
        source: new FileInputStream(inputFile)
        method: HttpRequest.POST
        headers: [
            HttpHeader {
                name: HttpHeader.CONTENT_TYPE
                value: "multipart/form-data"
            }
        ]
    }
    httpRequest.start();
}

On the server side, I am trying to handle the incoming data using the Apache Commons FileUpload API using a Jersey REST service. The code used to do this is a simple copy of the FileUpload tutorial on the Apache homepage.

@Path("Upload")
public class UploadService {

  public static final String RC_OK = "OK";
  public static final String RC_ERROR = "ERROR";

  @POST
  @Produces("text/plain")
  public String handleFileUpload(@Context HttpServletRequest request) {
    if (!ServletFileUpload.isMultipartContent(request)) {
      return RC_ERROR;
    }
    FileItemFactory factory = new DiskFileItemFactory();
    ServletFileUpload upload = new ServletFileUpload(factory);
    List<FileItem> items = null;
    try {
      items = upload.parseRequest(request);
    } 
    catch (FileUploadException e) {
      e.printStackTrace();
      return RC_ERROR;
    }
    ...
  }
}   

However, I get a exception at items = upload.parseRequest(request);:
org.apache.commons.fileupload.FileUploadException: the request was rejected because no multipart boundary was found

I guess I have to add a manual boundary info to the InputStream. Is there any easy solution to do this? Or are there even other solutions?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T14:05:04+00:00Added an answer on May 14, 2026 at 2:05 pm

    Have you tried just using the InputStream from HttpServletRequest like so

    InputStream is = httpRequest.getInputStream();
    BufferedInputStream in = new BufferedInputStream(is);
    //Write out bytes
    
    out.close();
    is.close();
    
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