Submit multipart and form data

[Justin Howard]

Hi All,

When I submit multipart content with the form data , it does not seem to recognize
the form data. I am trying to upload the file using jquery file upload scripts, is there any other way I can
handle it ?

Thanks

[Bear Bibeault]

When you submit a multi-part form, the request's getParameter() family of methods cannot parse the data (it's in the wrong format). Rather, the file upload library you are using should have an API to make the values available. If you are parsing the request data yourself, instead of using a file upload library, you need to do it yourself.

See the JSP FAQ for more info.

[Justin Howard]

Thanks for the reply.

I tried using org.apache.commons.fileupload.* file upload library.

But still i am not able to retrieve the form values.

How can I parse the values myself?

Thanks

[Bear Bibeault]

Justin Howard wrote:
I tried using org.apache.commons.fileupload.* file upload library.

But still i am not able to retrieve the form values.

Why not?

How can I parse the values myself?

You can't. The library has already read the stream. Answering the above question is more important than finding work-arounds.

[Justin Howard]

When the form gets submit this is the code I have in the servlet.

    
String ccEmails = request.getParameter(Constants.CC_EMAIL);
try {
					boolean isMultipart = ServletFileUpload.isMultipartContent(request);
				
					
					if(isMultipart){
						
						// Create a factory for disk-based file items
						FileItemFactory factory = new DiskFileItemFactory();

						// Create a new file upload handler
						ServletFileUpload upload = new ServletFileUpload(factory);

						// Parse the request
						List /* FileItem */ items = upload.parseRequest(request);
						
						// Process the uploaded items
						Iterator iter = items.iterator();
						while (iter.hasNext()) {
						    FileItem item = (FileItem) iter.next();
						    String comment = ((FileItem)item).getString();
						    logger.debug("comment"+comment);
						    FileItem fi = (FileItem)item;
					        logger.debug("here");// filename on the client
					        String fileName = fi.getName();
					        logger.debug("fileName"+fileName);
					        fi.write(new File("C:/FORMS/Research", fileName));
					        logger.debug("wote");
						    
						}
						
					}

But I everytime I submit the value of ccMails is null. ccMails is the example I gave, all the other form values are null as well.

[Bear Bibeault]

As I have already explained, getParameter() will not work. You need to use the API provided by the upload library.

[Justin Howard]

You were absolutely correct. I changed the code to reflect the new apis.
However now, for the file content it gives me an exception.

    
ServletFileUpload upload = new ServletFileUpload();

				FileItemIterator iter = upload.getItemIterator(request);
				while (iter.hasNext()) {
				    FileItemStream item = iter.next();
				    String name = item.getFieldName();
				    InputStream stream = item.openStream();
				    if (item.isFormField()) {
				        System.out.println("Form field " + name + " with value "
				            + Streams.asString(stream) + " detected.");
				    } else {
                                               String fieldName = item.getFieldName();
					        String fileName = item.getName();
					        System.out.println("fileName"+fileName);
					        String contentType = item.getContentType();
				    	FileItemFactory factory = new DiskFileItemFactory();

						File uploadedFile = new File("C:/FORMS/Research", fileName);
					        ((FileItem) item).write(uploadedFile);
    
				   
					}
				  }

If I cast it (((FileItem) item).write(uploadedFile);) gives me an exception if I dont it gives an error.
Exception: org.apache.commons.fileupload.FileUploadBase$FileItemIteratorImpl$FileItemStreamImpl cannot be cast

Thanks

[Paul Clapham]

I haven't looked at the API documentation myself, but I see you create a FileItemFactory and then don't do anything with it. Shouldn't you use that to create a FileItem object?

I'm not surprised you can't cast a FileItemStream object to a FileItem variable, just based on the names of the types.