convenient way to parse incoming multipart form data parameters in a servlet

convenient way to parse incoming multipart form data parameters in a servlet  using -'Is,there,any,convenient,way,to,read,and,parse,data,from,incoming,request.






asked Sep 16, 2015 by yashwantpinge
edited Sep 16, 2015 by rajesh
0 votes

3 Answers

0 votes

multipart/form-data encoded requests are indeed not by default supported by the Servlet API prior to version 3.0. The Servlet API parses the parameters by default using application/x-www-form-urlencoded encoding. When using a different encoding, the request.getParameter() calls will all return null. When you're already on Servlet 3.0 (Glassfish 3, Tomcat 7, etc), then you can use HttpServletRequest#getParts() instead. Also see this blog for extended examples.

Prior to Servlet 3.0, a de facto standard to parse multipart/form-data requests would be using Apache Commons FileUpload. Just carefully read its User Guide and Frequently Asked Questions sections to learn how to use it. I've posted an answer with a code example before here (it also contains an example targeting Servlet 3.0).

answered Sep 16, 2015 by virendra.bajaj
0 votes


Solution A:

  1. Download
  2. Invoke getParameters() on com.oreilly.servlet.MultipartRequest

Solution B:

  1. Download
  2. Invoke readHeaders() in org.apache.commons.fileupload.MultipartStream

Solution C:

  1. Download
  2. Invoke getParameter on com.bigfoot.bugar.servlet.http.MultipartFormData

Solution D:

Use Struts. Struts 1.1 handles this automatically.


answered Sep 16, 2015 by android_master
0 votes

Not always there's a servlet before of an upload (I could use a filter for example). Or could be that the same controller ( again a filter or also a servelt ) can serve many actions, so I think that rely on that servlet configuration to use the getPart method (only for Servlet API >= 3.0), I don't know, I don't like.

In general, I prefer indipendent solutions, able to live alone, and in this case is one of that.

List multiparts = new ServletFileUpload(new DiskFileItemFactory()).parseRequest(request);
    for (FileItem item : multiparts) {
        if (!item.isFormField()) {
            //your operations on file
        } else {
            String name = item.getFieldName();
            String value = item.getString();
            //you operations on paramters
answered Sep 16, 2015 by girisha