Convenient way to parse incoming multipart/form-data parameters in a Servlet

151,012

Solution 1

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).

Solution 2

Solutions:

Solution A:

  1. Download http://www.servlets.com/cos/index.html
  2. Invoke getParameters() on com.oreilly.servlet.MultipartRequest

Solution B:

  1. Download http://jakarta.Apache.org/commons/fileupload/
  2. Invoke readHeaders() in org.apache.commons.fileupload.MultipartStream

Solution C:

  1. Download http://users.boone.net/wbrameld/multipartformdata/
  2. Invoke getParameter on com.bigfoot.bugar.servlet.http.MultipartFormData

Solution D:

Use Struts. Struts 1.1 handles this automatically.

Reference: http://www.jguru.com/faq/view.jsp?EID=1045507

Solution 3

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 independent solutions, able to live alone, and in this case http://commons.apache.org/proper/commons-fileupload/ is one of that.

List<FileItem> 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
        }
}
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151,012
atuser
Author by

atuser

Updated on July 08, 2022

Comments

  • atuser
    atuser almost 2 years

    Is there any convenient way to read and parse data from incoming request.

    E.g client initiate post request

    URLConnection connection = new URL(url).openConnection();
    connection.setDoOutput(true);
    connection.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);
    PrintWriter writer = null;
    try {
        OutputStream output = connection.getOutputStream();
        writer = new PrintWriter(new OutputStreamWriter(output, charset), true); // true = autoFlush, important!
        // Send normal param.
        writer.println("--" + boundary);
        writer.println("Content-Disposition: form-data; name=\"param\"");
        writer.println("Content-Type: text/plain; charset=" + charset);
        writer.println();
        writer.println(param);
    

    I’m not able to get param using request.getParameter("paramName"). The following code

    BufferedReader reader = new BufferedReader(new InputStreamReader(
        request.getInputStream()));
      StringBuilder sb = new StringBuilder();
      for (String line; (line = reader.readLine()) != null;) {
       System.out.println(line);
    
      }
    

    however displays the content for me

    -----------------------------29772313742745
    Content-Disposition: form-data; name="name"
    J.Doe
    -----------------------------29772313742745
    Content-Disposition: form-data; name="email"
    [email protected]
    -----------------------------29772313742745
    

    What is the best way to parse incoming request? I don’t want to write my own parser, probably there is a ready solution.

  • Cheruvim
    Cheruvim over 10 years
    Be careful. Some of the documentation on the apache site is wrong. For example they says that you can call setRepository() on a FileItemFactory object which is false because any Object that implements FileItemFactory has only one method: createItem(). So be sure you read the javadocs as well.
  • Jafar Ali
    Jafar Ali about 10 years
    Please put up some solutions as this links can get absolete any time in future. This will attract negative rating.
  • basZero
    basZero over 7 years
    getParts() always returns zero items. Why is it so hard to retrieve multipart forms in Java and Servlet 3.0? Can't believe it!
  • BalusC
    BalusC over 7 years
    @basZero: the answer to the duplicate question already explains when it would be empty.
  • basZero
    basZero over 7 years
    @BalusC do you have a link? It's quite a lot of info here on this page
  • Ak S
    Ak S over 5 years
    it works only when you have a request object. But what in case of stream object or any other Object data?
  • Luca Rasconi
    Luca Rasconi over 5 years
    The stream or any other object data will replace the request. So I would parse/read the stream or whatever in order to get the multipart form data. Anyway do you have any real example?